2017 ACJC P2 Q10

2017 ACJC P2 Q10

8 marks
Prelims
  1. An examination taken by a large number of students is marked out of a total score of \(100\). It is found that the mean is \(73\) marks and that the standard deviation is \(15\) marks.
    1. Give a reason why the normal distribution is not a good model for the distribution of marks for the examination.[1]
    2. The marks for a random sample of \(50\) students is recorded. Find the probability that the mean mark of this sample lies between \(70\) and \(75\).[2]
  2. The interquartile range of a distribution is the difference between the upper and lower quartile values for the distribution. The lower quartile value, \(l\), of a distribution \(X\), is such that \(\mathrm{P}\left( X < l \right)=0.25\). The upper quartile value, \(u\), of the same distribution is such that \(\mathrm{P}\left( X < u \right)=0.75\).
    The marks of another examination is known to follow a normal distribution. If a student who scores \(51\) marks is at the \(80\)th percentile, and the interquartile range is found to be \(10.8\) marks, find the mean mark and the standard deviation of the marks scored by students who took the examination.[5]
  3. In a third examination, the marks scored by students are normally distributed with a mean of \(52\) marks and a standard deviation of \(13\) marks.
    1. If \(50\) is the passing mark and \(289\) students are expected to pass, how many candidates are there?[2]
    2. CFind the smallest integer value of \(m\) such that more than \(90\%\) of the candidates will score within \(m\) marks of the mean.[3]

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Answer:(a)(i) A normal model assigns probability \(0.0359\) to scores above \(100\), which are impossible. (ii) \(0.748\) (b) \(\mu=44.3,\ \sigma=8.01\) (c)(i) \(515\) candidates (ii) \(m=22\)

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