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PSLE

PSLE Foundation Mathematics

<p>At a school canteen, the average queuing time (in minutes) for a student to be served at a stall is known to be $4.5$ minutes. The school operation manager claims that with a recent change in students’ timetable, there is an improvement in the average queuing time.</p>
<p>A random sample of $10$ students’ queuing time is observed after the change in students’ timetable and the data is shown below.</p>
<p style="text-align: center">\[4.2, \quad 5.2, \quad 3.1, \quad 3.5, \quad 4.6, \quad 4.8, \quad 3.8, \quad 3.7, \quad 4.5, \quad 5.0\]</p>
<p>Given that the population variance of queuing time is known to be $0.48$ minutes$^2$, a one-tailed test is carried out to determine whether the manager’s claim is justified.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">State an assumption needed to carry out the hypothesis test.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Test whether the manager’s claim is justified at $5$% level of significance.</span><span class="inline-marks">[5]</span></div></li></ol>
<p>A student leader Jason suspected that with the change in students’ timetable, the average queuing time for a student to be served at a stall is in fact more than $\lambda$ minutes and also that the population variance of queuing time is no longer $0.48$ minutes$^2$. To investigate this, Jason recorded the queuing time of the first $50$ students who visited the stall from $12$ pm to $1$ pm.</p>
<ol class="abc-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Give a reason why the sample collected is not a random sample.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>Jason decided to collect another set of data to do his investigation. He recorded the queuing time, $x$, in minutes, of a random sample of $50$ students. The summarized data is as shown below.</p>
<p style="text-align: center">\[\bar{x} = 4.6 \quad \text{and} \quad \sum x^2 = 1068\]</p>
<ol class="abc-list" start="4"><li><div class="list-item-content"><span class="list-item-text">Find an unbiased estimate of the population variance.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>Jason carried out a hypothesis test at $2$% level of significance with an alternative hypothesis of $\mu &gt; \lambda$ where&nbsp;$\mu$ is the population mean queuing time.</p>
<ol class="abc-list" start="5"><li><div class="list-item-content"><span class="list-item-text">Explain why the assumption in (a) is no longer necessary.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">By finding the critical region in terms of $z$-values, find the range of values of $\lambda$ such that the Jason’s suspicion is justified.</span><span class="inline-marks">[3]</span></div></li></ol>

At a school canteen, the average queuing time (in minutes) for a student to be served at a stall is known to be $4.5$ minutes. The school operation manager claims that with a recent change in students’ timetable, there is an improvement in the average queuing time.

A random sample of $10$ students’ queuing time is observed after the change in students’ timetable and the data is shown below.

\[4.2, \quad 5.2, \quad 3.1, \quad 3.5, \quad 4.6, \quad 4.8, \quad 3.8, \quad 3.7, \quad 4.5, \quad 5.0\]

Given that the population variance of queuing time is known to be $0.48$ minutes$^2$, a one-tailed test is carried out to determine whether the manager’s claim is justified.

State an assumption needed to carry out the hypothesis test.

Test whether the manager’s claim is justified at $5$% level of significance.

A student leader Jason suspected that with the change in students’ timetable, the average queuing time for a student to be served at a stall is in fact more than $\lambda$ minutes and also that the population variance of queuing time is no longer $0.48$ minutes$^2$. To investigate this, Jason recorded the queuing time of the first $50$ students who visited the stall from $12$ pm to $1$ pm.

Give a reason why the sample collected is not a random sample.

Jason decided to collect another set of data to do his investigation. He recorded the queuing time, $x$, in minutes, of a random sample of $50$ students. The summarized data is as shown below.

\[\bar{x} = 4.6 \quad \text{and} \quad \sum x^2 = 1068\]

Find an unbiased estimate of the population variance.

Jason carried out a hypothesis test at $2$% level of significance with an alternative hypothesis of $\mu > \lambda$ where $\mu$ is the population mean queuing time.

Explain why the assumption in (a) is no longer necessary.

By finding the critical region in terms of $z$-values, find the range of values of $\lambda$ such that the Jason’s suspicion is justified.

Assumptions and Validity of Mean Tests

Estimating Parameters from Summarised and Coded Data

Central Limit Theorem

One-Tailed Hypothesis Tests

Parameter Bounds from Hypothesis Tests

Sampling Techniques and Bias

Mean Tests for Normal Populations with Known Variance

Assumptions for Sampling

Large-Sample Mean Tests with Estimated Variance

Critical Values and Critical Regions

2025 TJC P2 Q9

Hypothesis Testing

Sampling

<p>Based on observations over a long period, the mean time taken by male students in Griffles Junior College (GJC) to complete a $2.4$ km run is known to be $11.3$ minutes with a standard deviation of $2.2$ minutes. As part of a review of the current physical training programme, the Physical Education (PE) department in GJC decided to test, at the $2.5\%$ level of significance, whether the mean time taken by male students in GJC to complete a $2.4$ km run has changed. The time taken, $x$ minutes, by a random sample of $8$ Year $5$ male students from GJC to complete a $2.4$ km run, are as follows:</p>
<p style="text-align: center">${11 \quad 11.5 \quad 10.8 \quad 11.2 \quad 11.4 \quad 11 \quad 11.8 \quad 12.5}$</p>
<ol class="abc-list"><li data-list-item-id="item-a-4" data-list-item-path="item-a-4"><div class="list-item-paragraph">Write down the null and alternative hypotheses for this test, defining any symbols you use.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-4" data-list-item-path="item-b-4"><div class="list-item-paragraph">Stating a necessary assumption, find the critical region for this test. Hence state the conclusion of the test in the context of the question.<span class="inline-marks">[6]</span></div></li></ol>
<p>The PE department in GJC also conducted a second test, at the $3\%$ level of significance, to determine whether the mean time taken by female students in GJC to complete a $2.4$ km run is less than $14.5$ minutes. A random sample of $n$ female students from GJC is taken, where&nbsp;$n$ is large. The mean and standard deviation of the time taken by this sample to complete a $2.4$ km run are found to be $14.2$ minutes and $1.5$ minutes respectively.</p>
<ol class="abc-list" start="3"><li data-list-item-id="item-c-4" data-list-item-path="item-c-4"><div class="list-item-paragraph">Given that the PE department concludes that the mean time taken by female students to complete a $2.4$ km run is less than $14.5$ minutes, find the range of values that&nbsp;$n$ can take.<span class="inline-marks">[5]</span></div></li></ol>

Based on observations over a long period, the mean time taken by male students in Griffles Junior College (GJC) to complete a $2.4$ km run is known to be $11.3$ minutes with a standard deviation of $2.2$ minutes. As part of a review of the current physical training programme, the Physical Education (PE) department in GJC decided to test, at the $2.5\%$ level of significance, whether the mean time taken by male students in GJC to complete a $2.4$ km run has changed. The time taken, $x$ minutes, by a random sample of $8$ Year $5$ male students from GJC to complete a $2.4$ km run, are as follows:

${11 \quad 11.5 \quad 10.8 \quad 11.2 \quad 11.4 \quad 11 \quad 11.8 \quad 12.5}$

Write down the null and alternative hypotheses for this test, defining any symbols you use.

Stating a necessary assumption, find the critical region for this test. Hence state the conclusion of the test in the context of the question.

The PE department in GJC also conducted a second test, at the $3\%$ level of significance, to determine whether the mean time taken by female students in GJC to complete a $2.4$ km run is less than $14.5$ minutes. A random sample of $n$ female students from GJC is taken, where $n$ is large. The mean and standard deviation of the time taken by this sample to complete a $2.4$ km run are found to be $14.2$ minutes and $1.5$ minutes respectively.

Given that the PE department concludes that the mean time taken by female students to complete a $2.4$ km run is less than $14.5$ minutes, find the range of values that $n$ can take.

Estimating Population Variance from Sample Variance

Null and Alternative Hypotheses

Two-Tailed Hypothesis Tests

Interpreting Hypothesis Test Conclusions

2025 RI P2 Q9

<p>The life, in hours, of a randomly chosen light bulb produced by a manufacturer is normally distributed with mean $1100$ and standard deviation $70$. How large a sample is required such that the probability that the mean life in the sample shall exceed $1120$ is not more than $5\%$? <br></p>
<p>Do you need to use Central Limit Theorem in your working? <br></p>

The life, in hours, of a randomly chosen light bulb produced by a manufacturer is normally distributed with mean $1100$ and standard deviation $70$. How large a sample is required such that the probability that the mean life in the sample shall exceed $1120$ is not more than $5\%$? 


Do you need to use Central Limit Theorem in your working? 


Determining Sample Size from Probability Conditions

Normal Models for Sample Means and Averages

Continuous Random Variables

<p>The amount of rainfall $X$ in a certain city per day has mean $1.3$ metres and variance $2.8$ square metres. The amount of rainfall for a sample of $60$ days is collected.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Calculate the mean and variance of $\overline{X}$.</div></li><li><div class="list-item-paragraph">Find the probability that the mean amount of rainfall exceeds $0.8$ metres.</div></li><li><div class="list-item-paragraph">State two assumptions about the sample used in obtaining this estimate in <strong>(ii)</strong>.</div></li></ol>

The amount of rainfall $X$ in a certain city per day has mean $1.3$ metres and variance $2.8$ square metres. The amount of rainfall for a sample of $60$ days is collected.

Sample Mean and Its Sampling Distribution

<p><strong>In this question you should state the parameters of any distributions you use.</strong></p>
<p>At a burger shop, the wait time, $W$ (in minutes), is defined to be the time from when a customer places an order at the counter until the food is collected. It was proposed that $W$ is modelled by $N(2,{{1.5}^{2}})$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Give a reason why this model is not suitable.</div><p style="text-align: right;">[1]</p></li></ol>
<p>A new model for $W$ is given by $N(5,{{1}^{2}})$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">Find the range of values of $k$ such that at least $90\%$ of customers experience a wait time longer than $k$ minutes.</div><p style="text-align: right;">[2]</p></li></ol>
<p>A customer bought burgers from the shop on three independent occasions, with wait times denoted by ${{W}_{1}}$, ${{W}_{2}}$ and ${{W}_{3}}$ respectively. Let $\bar{W}=\frac{{{W}_{1}}+{{W}_{2}}+{{W}_{3}}}{3}$.</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">Find the values of $\mathrm{Var}(\bar{W}-{{W}_{1}})$ and $\mathrm{Var}(\bar{W})+\mathrm{Var}({{W}_{1}})$ and hence, determine whether $\mathrm{Var}(\bar{W}-{{W}_{1}})=\mathrm{Var}(\bar{W})+\mathrm{Var}({{W}_{1}})$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the probability that the mean wait time is within one minute of the wait time on the first occasion.</div><p style="text-align: right;">[3]</p></li></ol>
<p>On the 4th occasion, the customer went to the restroom immediately after the order was placed. The time spent in the restroom, in $T$ minutes, is modelled by $T\sim N(7,{{1.5}^{2}})$.</p>
<p>Assume that the time taken to walk to the restroom is negligible and $W$ and $T$ follow independent normal distributions.</p>
<ol class="abc-list" start="5"><li><div class="list-item-paragraph">The customer is in a rush and will not wait for more than 3 minutes after leaving the restroom. Given that the burger is not ready for collection after the customer leaves the restroom, find the probability that the customer will leave the shop without collecting the burger.</div><p style="text-align: right;">[4]</p></li></ol>
<p>To reduce the wait time for customers, the shop later installs self-order kiosks. The wait time from when a customer places an order at the kiosk until the food is collected is modelled as being reduced by $20\%$ compared to that at the counter.</p>
<ol class="abc-list" start="6"><li><div class="list-item-paragraph">8 customers who ordered at the counter and 8 customers who ordered at the kiosk are randomly selected and their wait times are recorded. The wait time of each customer is independent of the others. Find the probability that exactly 2 of these 16 customers have a wait time of at least 5 minutes.</div><p style="text-align: right;">[4]</p></li></ol>

**In this question you should state the parameters of any distributions you use.**

At a burger shop, the wait time, $W$ (in minutes), is defined to be the time from when a customer places an order at the counter until the food is collected. It was proposed that $W$ is modelled by $N(2,{{1.5}^{2}})$.

A new model for $W$ is given by $N(5,{{1}^{2}})$.

A customer bought burgers from the shop on three independent occasions, with wait times denoted by ${{W}_{1}}$, ${{W}_{2}}$ and ${{W}_{3}}$ respectively. Let $\bar{W}=\frac{{{W}_{1}}+{{W}_{2}}+{{W}_{3}}}{3}$.

On the 4th occasion, the customer went to the restroom immediately after the order was placed. The time spent in the restroom, in $T$ minutes, is modelled by $T\sim N(7,{{1.5}^{2}})$.

Assume that the time taken to walk to the restroom is negligible and $W$ and $T$ follow independent normal distributions.

To reduce the wait time for customers, the shop later installs self-order kiosks. The wait time from when a customer places an order at the kiosk until the food is collected is modelled as being reduced by $20\%$ compared to that at the counter.

Conditional Probability

Normal Probabilities and Quantiles

Probabilities Involving Multiple Binomial Variables

Linear Transformations of a Normal Variable

Linear Combinations of Independent Normal Variables

Assessing the Suitability of a Normal Model

2025 HCI P2 Q10

Probability

Discrete Random Variables (DRV)

<p>The number of employees of a company, classified by department and gender, is shown below.</p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 500px;"><colgroup><col style="width: 100px;"><col style="width: 100px;"><col style="width: 100px;"><col style="width: 100px;"><col style="width: 100px;"></colgroup><thead><tr style="height: 25px;"><th style="height: 25px; min-height: 25px; text-align: left; --cell-justify: flex-start"></th><th style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">Production</th><th style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">Development</th><th style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">Administration</th><th style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">Finance</th></tr></thead><tbody><tr style="height: 25px;"><td style="height: 25px; min-height: 25px; text-align: left; --cell-justify: flex-start">Male</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$2345$</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$1013$</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$237$</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$344$</td></tr><tr style="height: 25px;"><td style="height: 25px; min-height: 25px; text-align: left; --cell-justify: flex-start">Female</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$867$</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$679$</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$591$</td><td style="height: 25px; min-height: 25px; text-align: center; --cell-justify: center">$523$</td></tr></tbody></table></div></div>
<ol class="i-list"><li><div class="list-item-paragraph">The directors wish to survey a sample of $100$ of the employees. This sample is to be a stratified sample,based on department and gender.</div><ol class="abc-list"><li><div class="list-item-paragraph">How many males should be in the sample?<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">How many females from the Development department should be in the sample?<span class="inline-marks">[1]</span></div></li></ol></li></ol>
<p>The Managing Director knows that, some years ago, the mean age of employees was $37$ years. He believes that the mean age of employees now is less than $37$ years.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">State why the stratified sample from part <strong>(i)</strong> should not be used for a hypothesis test of the Managing Director's belief.<span class="inline-marks">[1]</span></div></li></ol>
<p>The Company Secretary obtains a suitable sample of $80$ employees in order to carry out a hypothesis test of the Managing Director’s belief that the mean age of the employees now is less than $37$ years. You are given that the population variance of the ages is $140$ years$^{2}$.</p>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">Write down appropriate hypotheses to test the Managing Director’s belief. You are given that the result of the test, using a $5%$ significance level, is that the Managing Director’s belief should be accepted. Determine the set of possible values of the mean age of the sample of employees.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">You are given instead that the mean age of the sample of employees is $35.2$ years, and that the result of a test at the $\alpha %$ significance level is that the Managing Director’s belief should not be accepted. Find the set of possible values of $\alpha $.<span class="inline-marks">[3]</span></div></li></ol>

The number of employees of a company, classified by department and gender, is shown below.

The Managing Director knows that, some years ago, the mean age of employees was $37$ years. He believes that the mean age of employees now is less than $37$ years.

The Company Secretary obtains a suitable sample of $80$ employees in order to carry out a hypothesis test of the Managing Director’s belief that the mean age of the employees now is less than $37$ years. You are given that the population variance of the ages is $140$ years$^{2}$.

Significance Levels and $p$-Values

Stratified Sampling

Large-Sample Mean Tests with Known Variance

N2016 P2 Q6

<p>Birth weight can be used to predict short and long-term health complications for babies. Studies show that the birth weight of babies born to mothers who do not smoke in a certain hospital can be assumed to follow a normal distribution with mean ${3.05}$ kg and variance ${{{\sigma }^{2}}}$kg${^{2}}$.</p>
<p>The hospital classifies babies based on their birth weight as shown in the table below.</p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 400px;"><colgroup><col style="width: 200px;"><col style="width: 200px;"></colgroup><thead><tr><th style="text-align: center; --cell-justify: center">Birth weight</th><th style="text-align: center; --cell-justify: center">Birth weight</th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">Less than $1.5$ kg</td><td style="text-align: center; --cell-justify: center">Very low birth weight</td></tr><tr><td style="text-align: center; --cell-justify: center">$1.5$ kg to $2.5$ kg</td><td style="text-align: center; --cell-justify: center">Low birth weight</td></tr><tr><td style="text-align: center; --cell-justify: center">$2.5$ kg to $4.0$ kg</td><td style="text-align: center; --cell-justify: center">Normal birth weight</td></tr><tr><td style="text-align: center; --cell-justify: center">More than $4.0$ kg</td><td style="text-align: center; --cell-justify: center">High birth weight</td></tr></tbody></table></div></div>
<ol class="i-list"><li><div class="list-item-paragraph">A sample showed that&nbsp;$20.2\%$ of the babies born to mothers who do not smoke have low birth weight. If this is true for the entire population, find two possible values of $\sigma $, corrected to ${2}$ decimal places. Explain clearly why one of the values of $\sigma $ found should be rejected.<span class="inline-marks">[4]</span></div></li></ol>
<p>Studies also show that babies born to mothers who smoke have a lower mean birth weight of ${2.80}$ kg.</p>
<p>For the remaining of the question, you may assume the birth weight of babies born to mothers who smoke is also normally distributed and that the standard deviations for the birth weight of babies born to mothers who do not smoke and mothers who smoke are both&nbsp;${0.86}$ kg.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Three pregnant mothers-to-be who smoke are randomly chosen. Find the probability that all their babies will not be of normal birth weight. State an assumption that is needed for your working.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find the probability that the average birth weight of two babies from mothers who do not smoke differs from twice the birth weight of a baby from a mother who smokes by less than ${2}$ kg.<span class="inline-marks">[3]</span></div></li></ol>
<p>Babies whose birth weight not classified as normal will have to remain in hospital for further observation until their condition stabilises. Depending on the treatment received and length of stay, the mean hospitalisation cost per baby is ${\$2800}$ and standard deviation is ${\$500}$.</p>
<ol class="i-list" start="4"><li><div class="list-item-paragraph">Find the probability that the average hospitalisation cost from a random sample of ${50}$ babies, whose birth weight not classified as normal, exceeds ${\$3000}$.<span class="inline-marks">[2]</span></div></li></ol>

Birth weight can be used to predict short and long-term health complications for babies. Studies show that the birth weight of babies born to mothers who do not smoke in a certain hospital can be assumed to follow a normal distribution with mean ${3.05}$ kg and variance ${{{\sigma }^{2}}}$kg${^{2}}$.

The hospital classifies babies based on their birth weight as shown in the table below.

Studies also show that babies born to mothers who smoke have a lower mean birth weight of ${2.80}$ kg.

For the remaining of the question, you may assume the birth weight of babies born to mothers who smoke is also normally distributed and that the standard deviations for the birth weight of babies born to mothers who do not smoke and mothers who smoke are both ${0.86}$ kg.

Babies whose birth weight not classified as normal will have to remain in hospital for further observation until their condition stabilises. Depending on the treatment received and length of stay, the mean hospitalisation cost per baby is ${\$2800}$ and standard deviation is ${\$500}$.

Finding Unknown Normal Mean and Standard Deviation

2019 ASRJC P2 Q6

<p><strong>In this question you should state clearly the values of the parameters of any normal distribution(s) that you use.</strong></p>
<p>Papayas and watermelons are sold by weight. The masses, in kg, of papayas and watermelons are modelled as having normal distributions with means and standard deviations as shown in the table.</p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 450px;"><colgroup><col style="width: 150px;"><col style="width: 150px;"><col style="width: 150px;"></colgroup><thead><tr><th style="text-align: left; --cell-justify: flex-start"></th><th style="text-align: center; --cell-justify: center">Mean mass</th><th style="text-align: center; --cell-justify: center">Standard deviation</th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">Papayas</td><td style="text-align: center; --cell-justify: center">${1.85}$</td><td style="text-align: center; --cell-justify: center">$0.12$</td></tr><tr><td style="text-align: center; --cell-justify: center">Watermelons</td><td style="text-align: center; --cell-justify: center">${6.50}$</td><td style="text-align: center; --cell-justify: center">$0.72$</td></tr></tbody></table></div></div>
<ol class="i-list"><li data-list-item-id="item-1760848683345-vidftpwsn" data-list-item-path="item-1760848683345-vidftpwsn"><div class="list-item-paragraph">Sketch the distribution for the masses of papayas between $1.4$kg and $2.3$kg.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760848684110-wna5g9j40" data-list-item-path="item-1760848684110-wna5g9j40"><div class="list-item-paragraph">Find the probability that the mass of a randomly chosen papaya is less than $1.7$kg.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760848684785-cq8norbtu" data-list-item-path="item-1760848684785-cq8norbtu"><div class="list-item-paragraph">Find the probability that the mass of a randomly chosen watermelon exceeds the mass of a randomly chosen papaya by more than $4.5$kg.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760848686773-vmg1yo46d" data-list-item-path="item-1760848686773-vmg1yo46d"><div class="list-item-paragraph">The mean mass of $n$ randomly chosen watermelons is denoted by $\bar{Y}$ kg. Given that&nbsp;$\mathrm{P}\left( \bar{Y}&gt;k \right)=0.2$, express $k$ in terms of $n$.<span class="inline-marks">[3]</span></div></li></ol>
<p>Papayas are sold at $\$2.20$ per kg and watermelons at $\$1.45$ per kg.</p>
<ol class="i-list" start="5"><li data-list-item-id="item-1760848742550-w7zz2tiol" data-list-item-path="item-1760848742550-w7zz2tiol"><div class="list-item-paragraph">Calculate the probability that the total selling price of $3$ randomly chosen papayas and $4$randomly chosen watermelons exceed $\$50$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760848752848-g6ex0vnjp" data-list-item-path="item-1760848752848-g6ex0vnjp"><div class="list-item-paragraph">State an assumption for your calculations in parts <strong>(iii)</strong>, <strong>(iv)</strong> and <strong>(v)</strong>.<span class="inline-marks">[1]</span></div></li></ol>

**In this question you should state clearly the values of the parameters of any normal distribution(s) that you use.**

Papayas and watermelons are sold by weight. The masses, in kg, of papayas and watermelons are modelled as having normal distributions with means and standard deviations as shown in the table.

Papayas are sold at $\$2.20$ per kg and watermelons at $\$1.45$ per kg.

Normal Distribution: Model and Parameters

2024 TMJC P2 Q10

<p>The average waiting time (in minutes) for a customer to be served at a fast-food restaurant is $k$ minutes. The manager claims that with subsequent increased staff strength, the restaurant can now better serve the customers and the average waiting time has improved. The population variance of waiting time is known to be $0.22$ minutes$^{2}$.</p>
<p>The observed waiting times (in minutes) of a random sample of $9$ customers after the increased staff strength are:</p>
<p style="text-align: center">${3.6, \qquad 2.5, \qquad 2.4, \qquad 3.5, \qquad 2.8, \qquad 2.9, \qquad 3.1, \qquad 3.7, \qquad 3.0}$</p>
<ol class="abc-list"><li data-list-item-id="item-1760684824142-jq0g84ts5" data-list-item-path="item-1760684824142-jq0g84ts5"><div class="list-item-paragraph">Find the unbiased estimates of the population mean and variance of the waiting time.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760684824907-8w247vqp9" data-list-item-path="item-1760684824907-8w247vqp9"><div class="list-item-paragraph">Using the value $k=3.3$.</div><ol class="i-list"><li data-list-item-id="subitem-1760684841991-37dkm5kbj" data-list-item-path="item-1760684824907-8w247vqp9/subitem-1760684841991-37dkm5kbj"><div class="list-item-paragraph">Carry out a one-tailed test at $5\%$ level of significance to determine whether the manager’s claim is valid, stating clearly the $p$-value and any assumption(s) you have made.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="subitem-1760684843476-u3qbyt7l3" data-list-item-path="item-1760684824907-8w247vqp9/subitem-1760684843476-u3qbyt7l3"><div class="list-item-paragraph">Explain the meaning of $p$-value in the context of the question.<span class="inline-marks">[1]</span></div></li></ol></li><li data-list-item-id="item-1760684827232-z625w5ft8" data-list-item-path="item-1760684827232-z625w5ft8"><div class="list-item-paragraph">Using the same assumption in part <strong>(b)(i)</strong>, a one-tailed test was carried out at $10\%$ and it was found that the manager’s claim was justified. Find the range of possible values of $k$.<span class="inline-marks">[4]</span></div></li></ol>

The average waiting time (in minutes) for a customer to be served at a fast-food restaurant is $k$ minutes. The manager claims that with subsequent increased staff strength, the restaurant can now better serve the customers and the average waiting time has improved. The population variance of waiting time is known to be $0.22$ minutes$^{2}$.

The observed waiting times (in minutes) of a random sample of $9$ customers after the increased staff strength are:

${3.6, \qquad 2.5, \qquad 2.4, \qquad 3.5, \qquad 2.8, \qquad 2.9, \qquad 3.1, \qquad 3.7, \qquad 3.0}$

Calculating and Interpreting $p$-Values

Unbiased Estimates of Population Parameters

Hypothesis Tests Using Critical Regions and $p$-Values

2023 DHS P2 Q10

<p>The manufacturer of Mola snack biscuits claimed that the mean biscuit content in a packet is $15$g. A sample of $80$ packets is taken and the weight, $x$g, of the content in each packet is measured. The results are summarised by</p>
<p style="text-align: center">$\sum{\left( x-15 \right)=53.6}$ and $\sum{{{\left( x-15 \right)}^{2}}=795.7}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Calculate unbiased estimates of the population mean and variance.</div></li><li><div class="list-item-paragraph">Stating a necessary assumption, test whether the mean weight differs from $15$g at $4%$ level of significance.</div></li><li><div class="list-item-paragraph">If the test is conducted at $\alpha %$ level of significance and the null hypothesis is rejected in favour of the alternative hypothesis based on the values from this sample, determine the set of values of $\alpha $.</div></li><li><div class="list-item-paragraph">Another large random sample of $n$ packets is taken. Assume now that the population standard deviation is $3$g. Find the range of values of the sample mean, in terms of $n$, for which the claim that the mean weight differs from $15$g is supported at $4%$ level of significance.</div></li></ol>

The manufacturer of Mola snack biscuits claimed that the mean biscuit content in a packet is $15$g. A sample of $80$ packets is taken and the weight, $x$g, of the content in each packet is measured. The results are summarised by

$\sum{\left( x-15 \right)=53.6}$ and $\sum{{{\left( x-15 \right)}^{2}}=795.7}$.

2015 TJC P2 Q7 [Modified]

<ol class="i-list"><li><div class="list-item-paragraph">A multiple choice test consists of $20$ questions, each with four possible answers, of which only one is correct. If a student randomly chooses the answer to each question, find the probability of getting at least four but less than nine correct answers.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Suppose that each correct answer is awarded five marks and each incorrect answer carries a penalty of one mark, what is the expected score obtained by a student?</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Using a suitable approximation, find the probability that the mean score of $50$ randomly selected students is more than $12$ marks. State an assumption necessary for your answer to be valid.</div><p style="text-align: right;">[4]</p></li></ol>

Binomial Probabilities Using Technology

Mean and Variance of a Binomial Distribution

Linear Combinations of Random Variables

NJC Sampling Tutorial Q4

<p>A virologist claims that due to the introduction of vaccine, the recovery period from a certain type of flu-like virus is on average $7$ days. However, an analyst who works with such patients on a day-to-day basis believes that the average recovery period quoted by the virologist is understated. He carries out a survey to investigate this by recording the recovery period of $120$ randomly chosen patients. The recovery period of each of these $120$ patients are denoted by $x$ days. The summarised results are given by</p>
<p style="text-align: center">$\sum{\left( x-7 \right)=24}$ and $\sum{{{\left( x-7 \right)}^{2}}=\mathrm{14908}}$</p>
<ol class="i-list"><li><div class="list-item-paragraph">Explain in context, the meaning of a random sample of $120$ patients.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Calculate the unbiased estimates of the population mean and variance of the recovery period of the patients.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Carry out a test at $5\%$ significance level, on the claim that the average recovery period of $7$ days is understated.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Explain if the test in part <strong>(iii)</strong> would still be valid if the analyst had asked for recovery periods from a random sample of $15$ patients.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Suppose instead, the analyst wants to test if the average recovery period differs from $7$ days. The analyst collects data from another random sample of $120$ patients. Given that the population standard deviation is now known to be $1$ day, find the range of values of the sample mean, $\overline{x}$, such that the null hypothesis is rejected at $1\%$ level of significance.</div><p style="text-align: right;">[4]</p></li></ol>

A virologist claims that due to the introduction of vaccine, the recovery period from a certain type of flu-like virus is on average $7$ days. However, an analyst who works with such patients on a day-to-day basis believes that the average recovery period quoted by the virologist is understated. He carries out a survey to investigate this by recording the recovery period of $120$ randomly chosen patients. The recovery period of each of these $120$ patients are denoted by $x$ days. The summarised results are given by

$\sum{\left( x-7 \right)=24}$ and $\sum{{{\left( x-7 \right)}^{2}}=\mathrm{14908}}$

Population and Random Samples

2022 CJC P2 Q12

<p>On a remote island a zoologist measures the tail lengths of a random sample of $20$ squirrels. In a species of squirrel known to her, the tail lengths have mean $14.0$ cm. She carries out a test, at the $5\%$ significance level, of whether squirrels on the island have the same mean tail length as the species known to her. She assumes that the tail lengths of squirrels on the island are normally distributed with standard deviation $3.8$ cm.</p>
<ol class="i-list"><li><div class="list-item-paragraph">State appropriate hypotheses for the test.</div><p style="text-align: right;">[1]</p></li></ol>
<p>The sample mean tail length is denoted by $\overline{x}$ cm.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Use an algebraic method to calculate the set of values of $\overline{x}$ for which the null hypothesis would not be rejected. </div><div class="list-item-paragraph">(Answers obtained by trial and improvement from a calculator will obtain no marks.)</div><p style="text-align: right;">[3]</p></li></ol>
<p>Five years later, the zoologist took a sample of $20$ squirrels and found that $\overline{x}=15.8$. He wants to make a claim that the mean is at least ${{\mu }_{0}}$ and he does not want his claim to be rejected at $5\%$ significance level.</p>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">Find the range of values of ${{\mu }_{0}}$ that he can claim.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">In order to carry out this test, the zoologist made two assumptions. State these two assumptions.</div><p style="text-align: right;">[2]</p></li></ol>

On a remote island a zoologist measures the tail lengths of a random sample of $20$ squirrels. In a species of squirrel known to her, the tail lengths have mean $14.0$ cm. She carries out a test, at the $5\%$ significance level, of whether squirrels on the island have the same mean tail length as the species known to her. She assumes that the tail lengths of squirrels on the island are normally distributed with standard deviation $3.8$ cm.

The sample mean tail length is denoted by $\overline{x}$ cm.

Five years later, the zoologist took a sample of $20$ squirrels and found that $\overline{x}=15.8$. He wants to make a claim that the mean is at least ${{\mu }_{0}}$ and he does not want his claim to be rejected at $5\%$ significance level.

N2012 P2 Q6 [Modified]

<p>In a polyclinic, the waiting time, in minutes, for dentist $A$ and dentist $B$, are modelled as having normal distributions with means and standard deviations as shown in the table below.</p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 400px;"><colgroup><col style="width: 200px;"><col style="width: 100px;"><col style="width: 100px;"></colgroup><thead><tr><th style="text-align: left; --cell-justify: flex-start"></th><th style="text-align: center; --cell-justify: center">Mean Duration</th><th style="text-align: center; --cell-justify: center">Standard Deviation</th></tr></thead><tbody><tr><td style="text-align: left; --cell-justify: flex-start">Waiting time for Dentist $A$</td><td style="text-align: center; --cell-justify: center">$15.8$</td><td style="text-align: center; --cell-justify: center">$5.00$</td></tr><tr><td style="text-align: left; --cell-justify: flex-start">Waiting time for Dentist $B$</td><td style="text-align: center; --cell-justify: center">$35.4$</td><td style="text-align: center; --cell-justify: center">$7.21$</td></tr></tbody></table></div></div>
<ol class="i-list"><li data-list-item-id="item-1760259599257-ptz524qw7" data-list-item-path="item-1760259599257-ptz524qw7"><div class="list-item-paragraph">Find the probability that on a randomly chosen day, the waiting time for a randomly chosen patient waiting to see dentist $B$ differs from twice the waiting time for a randomly chosen patient waiting to see dentist $A$ by more than $5$ minutes.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760259599797-wv4dz2a90" data-list-item-path="item-1760259599797-wv4dz2a90"><div class="list-item-paragraph">Two randomly chosen patients who were each queueing to see dentist $A$ and one randomly chosen patient who was queueing to see dentist $B$ each asked for their waiting time. Find the probability that the average waiting time for these three patients to see their dentists is at most half an hour.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760259600450-r07m166js" data-list-item-path="item-1760259600450-r07m166js"><div class="list-item-paragraph">State an assumption needed for your calculations in <strong>(i)</strong> and <strong>(ii)</strong>.<span class="inline-marks">[1]</span></div></li></ol>
<p>The random variable $V$ denotes the waiting time, in minutes, for a patient to make payment right after the consultation with any dentist in the polyclinic. The random variable $V$ follows a normal distribution with mean $\mu $ and standard deviation $\sigma $.</p>
<ol class="i-list" start="4"><li data-list-item-id="item-1760259658725-vxn20qhfy" data-list-item-path="item-1760259658725-vxn20qhfy"><div class="list-item-paragraph">If $\mathrm{P}\left( \mu -4 &lt; V &lt; \mu -\frac{1}{2} \right)=2\mathrm{P}\left( \mu -\frac{1}{2} &lt; V &lt; \mu +\frac{1}{2} \right)$ and $\mathrm{P}\left( \mu -4 &lt; V &lt; \mu +\frac{1}{2} \right)=\mathrm{P}\left( V&gt;\mu \right)$, find the value of $\sigma $.<span class="inline-marks">[3]</span></div></li></ol>
<p>Peter started waiting to make payment at 11.30 am.</p>
<ol class="i-list" start="5"><li data-list-item-id="item-1760259683490-fmrlvdxle" data-list-item-path="item-1760259683490-fmrlvdxle"><div class="list-item-paragraph">Hence find the range of values of $\mu $ if the chance for him to leave the polyclinic after 11.35 am is at least $90\%$.<span class="inline-marks">[3]</span></div></li></ol>

In a polyclinic, the waiting time, in minutes, for dentist $A$ and dentist $B$, are modelled as having normal distributions with means and standard deviations as shown in the table below.

The random variable $V$ denotes the waiting time, in minutes, for a patient to make payment right after the consultation with any dentist in the polyclinic. The random variable $V$ follows a normal distribution with mean $\mu $ and standard deviation $\sigma $.

Peter started waiting to make payment at 11.30 am.

Solving Normal Probability Inequalities

2018 SRJC MYE P2 Q12

<p>The waiting times, in minutes, for orders of Chili Crab and Pepper Crab at a popular hawker stall are modelled as having independent normal distributions with means and standard deviations as shown in the following table:</p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 450px;"><colgroup><col style="width: 150px;"><col style="width: 150px;"><col style="width: 150px;"></colgroup><thead><tr style="height: 23px;"><th style="height: 23px; min-height: 23px; text-align: left; --cell-justify: flex-start"></th><th style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">Mean Waiting Time</th><th style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">Standard Deviation</th></tr></thead><tbody><tr style="height: 23px;"><td style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">Chili Crab</td><td style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">${\mu}$</td><td style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">$3$</td></tr><tr style="height: 23px;"><td style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">Pepper Crab</td><td style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">$10$</td><td style="height: 23px; min-height: 23px; text-align: center; --cell-justify: center">$2$</td></tr></tbody></table></div></div>
<ol class="i-list"><li><div class="list-item-paragraph">If the probability that one randomly chosen order of Pepper Crab takes a longer waiting time than one randomly chosen order of Chili Crab is $0.290$, show that&nbsp;$\mu =12.0$, correct to $3$ significant figures.<span class="inline-marks">[3]</span></div></li></ol>
<p>Customers who drive park their cars at a nearby car park at $\$0.05$ per minute. The walking time between the car park and the hawker stall is negligible. The time taken by a randomly chosen customer to finish his meal is normally distributed with mean $20$ minutes and standard deviation $4$ minutes.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Find the probability that a randomly chosen customer who parked at the car park, bought one order of Pepper Crab, finished his meal and immediately drove off, spent more than $\$2$ on car park charges.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">A random sample of $n$ customers who each make one order of Pepper Crab is taken. Given that the probability that their mean waiting time exceeds $10.5$ minutes is at most $0.0385$, find the smallest possible value of $n$. Explain briefly if any assumption is needed for the above calculation.<span class="inline-marks">[4]</span></div></li></ol>

The waiting times, in minutes, for orders of Chili Crab and Pepper Crab at a popular hawker stall are modelled as having independent normal distributions with means and standard deviations as shown in the following table:

Customers who drive park their cars at a nearby car park at $\$0.05$ per minute. The walking time between the car park and the hawker stall is negligible. The time taken by a randomly chosen customer to finish his meal is normally distributed with mean $20$ minutes and standard deviation $4$ minutes.

2013 CJC P2 Q12

<p>The weight, $Y$, of a Yummy cereal bar is normally distributed with mean $\left( 120-k \right)$g and standard deviation 10g. The weight, F, of a Fullness cereal bar is normally distributed with mean 180 g and standard deviation 20 g.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Given that $\mathrm{P}\left( Y<80 \right)=\mathrm{P}\left( Y>150 \right)$, show that the value of $k$ is 5.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">5 Yummy cereal bars are randomly chosen. Find the probability that exactly one bar weighs lesser than the lower quartile weight and exactly one bar weighs more than the median weight.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the probability that the weight of 2 randomly chosen Yummy cereal bars differs from one and a half times the weight of a randomly chosen Fullness cereal bar by at most 5g. State the assumption needed for your working.</div><p style="text-align: right;">[4]</p></li></ol>
<p>The Fullness cereal bars are sold at $2 per 100g.</p>
<ol class="i-list" start="4"><li><div class="list-item-paragraph">Find the probability that a randomly chosen Fullness cereal bar costs more than $3.50.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">A random sample of 100 Fullness cereal bars is taken. Find the probability that there are at least 55 bars but at most 65 bars with each costing more than $3.50.</div><p style="text-align: right;">[3]</p></li></ol>

The weight, $Y$, of a Yummy cereal bar is normally distributed with mean $\left( 120-k \right)$g and standard deviation 10g. The weight, F, of a Fullness cereal bar is normally distributed with mean 180 g and standard deviation 20 g.

The Fullness cereal bars are sold at $2 per 100g.

Symmetry and Properties of the Normal Distribution

Probability Using Permutations and Combinations

2010 PJC P2 Q12

<p>Vesla, an electric vehicle (EV) car manufacturer claims that their EV cars have an average travelling distance of $650$ km on a single charge. A car reviewer on the MeTube online video sharing platform wants to test if the EV car manufacturer overstated its claim.</p>
<p>The MeTube car reviewer buys a Vesla EV car to conduct his own test drives. He records the travelling distance on a single charge, $x$ km, on $50$ different days over a year. The travelling distance on a single charge, $x$ km, is summarised below:</p>
<p style="text-align: center">$\sum{\left( x-650 \right)}=-34.39$ $\sum{{{\left( x-650 \right)}^{2}}}=22769.98$</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Calculate unbiased estimates of the population mean and variance of the travelling distance of an EV car on a single charge.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">State hypotheses that the MeTube reviewer can use to test if Vesla has overstated its claim about travelling distance on a single charge, defining any symbols you use. Work out the test statistic in this case, and use it to carry out the test at the $5\%$ level of significance, giving your conclusion in the context of the question.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Explain if the MeTube car reviewer needs to apply Central Limit Theorem for the test to be valid.</div><p style="text-align: right;">[2]</p></li></ol>
<p>Another car reviewer on the TokTik online video sharing platform wants to test if the EV car manufacturer’s claim is true. He rents a Vesla EV car to conduct his test drives where he records the travelling distance on a single charge. He records a random sample of $6$ test drives and calculates the mean travelling distance on a single charge.</p>
<ol class="abc-list" start="4"><li><div class="list-item-paragraph">Explain whether the TokTik car reviewer should use a $1$-tail or a $2$-tail test.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">State two assumptions that allow the TokTik car reviewer to carry out his test.</div><p style="text-align: right;">[2]</p></li></ol>

Vesla, an electric vehicle (EV) car manufacturer claims that their EV cars have an average travelling distance of $650$ km on a single charge. A car reviewer on the MeTube online video sharing platform wants to test if the EV car manufacturer overstated its claim.

The MeTube car reviewer buys a Vesla EV car to conduct his own test drives. He records the travelling distance on a single charge, $x$ km, on $50$ different days over a year. The travelling distance on a single charge, $x$ km, is summarised below:

$\sum{\left( x-650 \right)}=-34.39$ $\sum{{{\left( x-650 \right)}^{2}}}=22769.98$

Another car reviewer on the TokTik online video sharing platform wants to test if the EV car manufacturer’s claim is true. He rents a Vesla EV car to conduct his test drives where he records the travelling distance on a single charge. He records a random sample of $6$ test drives and calculates the mean travelling distance on a single charge.

2023 CJC P2 Q10

<p>A sample of $n$ observations was taken from a population of mean $\mu$ and variance 10. Find the least sample size required such that the probability of the sample mean lying between $\mu -0.5$ and $\mu +0.5$ is more than 0.89. State an assumption that you have to make in order to proceed with this calculation.</p><p style="text-align: right;">[6]</p>

A sample of $n$ observations was taken from a population of mean $\mu$ and variance 10. Find the least sample size required such that the probability of the sample mean lying between $\mu -0.5$ and $\mu +0.5$ is more than 0.89. State an assumption that you have to make in order to proceed with this calculation.

2012 CJC P2 Q11

<p>Each morning, Tony drives from his home to his office and has to pass $4$ traffic lights on his way. He has to reach his office by 8.50 am. The driving time to his office and the time held up at a traffic light junction, in minutes, may be assumed to follow normal distributions with means and standard deviations as summarized below:</p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 440px;"><colgroup><col style="width: 220px;"><col style="width: 70px;"><col style="width: 150px;"></colgroup><thead><tr><th style="text-align: left; --cell-justify: flex-start"></th><th style="text-align: center; --cell-justify: center">Mean</th><th style="text-align: center; --cell-justify: center">Standard deviation</th></tr></thead><tbody><tr><td style="text-align: left; --cell-justify: flex-start">Driving time</td><td style="text-align: center; --cell-justify: center">$14$</td><td style="text-align: center; --cell-justify: center">$2.1$</td></tr><tr><td style="text-align: left; --cell-justify: flex-start">Time held up at a traffic light junction</td><td style="text-align: center; --cell-justify: center">${\mu}$</td><td style="text-align: center; --cell-justify: center">$0.2$</td></tr></tbody></table></div></div>
<p>Tony leaves home at 8.30 am. If the probability that Tony is not late is $0.713$, show that&nbsp;$\mu =1.2$, correct to $1$ decimal place. State an assumption needed in your calculations.<span class="inline-marks">[4]</span></p>
<ol class="i-list"><li data-list-item-id="item-1760153491132-wagc0pcw1" data-list-item-path="item-1760153491132-wagc0pcw1"><div class="list-item-paragraph">For&nbsp;$10$ mornings, Tony leaves home at 8.30 am. Find the probability that he arrives late at his office for the third time on the ${{10}^{\mathrm{th}}}$ day.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760153492317-vrzqyser8" data-list-item-path="item-1760153492317-vrzqyser8"><div class="list-item-paragraph">Find the probability that Tony’s driving time to his office is less than $10$ times the time he is held up at a traffic light junction.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760153493007-f1zf4bvtt" data-list-item-path="item-1760153493007-f1zf4bvtt"><div class="list-item-paragraph">Find the probability that Tony’s driving time to his office and the total time he is held up at the $4$ traffic light junctions differs by more than $8$ minutes.<span class="inline-marks">[3]</span></div></li></ol>
<p>Assume $\mu $ is unknown.</p>
<ol class="i-list" start="4"><li data-list-item-id="item-1760153525504-hkq9o5u5b" data-list-item-path="item-1760153525504-hkq9o5u5b"><div class="list-item-paragraph">The time held up at a traffic light junction is recorded on n randomly chosen occasions. Find the smallest $n$ so that it is at least $98\%$ certain that the sample mean time Tony is held up at a traffic light junction is within $5$ seconds of $\mu $.<span class="inline-marks">[3]</span></div></li></ol>

Each morning, Tony drives from his home to his office and has to pass $4$ traffic lights on his way. He has to reach his office by 8.50 am. The driving time to his office and the time held up at a traffic light junction, in minutes, may be assumed to follow normal distributions with means and standard deviations as summarized below:

Tony leaves home at 8.30 am. If the probability that Tony is not late is $0.713$, show that $\mu =1.2$, correct to $1$ decimal place. State an assumption needed in your calculations.

Probabilities of Successive Events

Binomial Probabilities Using the Probability Mass Function

Waiting-Time and Stopping-Rule Probabilities

2018 HCI P2 Q10

<p>A popular investment manager claims that the mean dividend yield of his clients is $12\%$. A research firm investigates further by collecting a random sample of $40$ of the manager’s clients. The sample was found to have a mean dividend yield of $10.9\%$ and a standard deviation of $2.5\%$</p>
<ol class="i-list"><li data-list-item-id="item-1760146831141-merkf1sc2" data-list-item-path="item-1760146831141-merkf1sc2"><div class="list-item-paragraph">State, giving a reason, whether there is a need to make any assumption about the distribution of the dividend yield earned by the manager’s clients in order for a hypothesis test to be valid.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760146831808-06hgjwt33" data-list-item-path="item-1760146831808-06hgjwt33"><div class="list-item-paragraph">Test, at the $5%$ level of significance, whether the manager overstated his claim.<span class="inline-marks">[5]</span></div></li></ol>
<p>It was later known that the standard deviation of the dividend yield of the popular investment manager’s clients is $2.3\%$. A potential investor collected the dividend yield data of a small sample size, $n$, of the manager’s clients. The sample was found to have a mean dividend yield of $10.2\%$.</p>
<ol class="i-list" start="3"><li data-list-item-id="item-1760146857450-6s5b0h49a" data-list-item-path="item-1760146857450-6s5b0h49a"><div class="list-item-paragraph">State, in context, two assumptions that are needed for you to carry out a test to examine the manager’s claim that the mean dividend yield of his clients is $12\%$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760146858095-htzb9i4vk" data-list-item-path="item-1760146858095-htzb9i4vk"><div class="list-item-paragraph">Hence find the least value of $n$ such that the manager’s claim at the $5\%$ level of significance, will be rejected. You should state your hypothesis clearly.<span class="inline-marks">[4]</span></div></li></ol>

A popular investment manager claims that the mean dividend yield of his clients is $12\%$. A research firm investigates further by collecting a random sample of $40$ of the manager’s clients. The sample was found to have a mean dividend yield of $10.9\%$ and a standard deviation of $2.5\%$

It was later known that the standard deviation of the dividend yield of the popular investment manager’s clients is $2.3\%$. A potential investor collected the dividend yield data of a small sample size, $n$, of the manager’s clients. The sample was found to have a mean dividend yield of $10.2\%$.

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