<p>The heights, in metres, of female and male students in a school follow normal distributions with the following data.</p>
<table><thead><tr><th></th><th>Mean Height (metres)</th><th>Variance (metres$^2$)</th></tr></thead><tbody><tr><td>Female</td><td>1.6</td><td>0.04</td></tr><tr><td>Male</td><td>1.75</td><td>0.05</td></tr></tbody></table>
<p>Find the probability that</p>
<ol class="i-list"><li><div class="list-item-paragraph">Two randomly selected male are taller than twice a randomly selected female.</div></li><li><div class="list-item-paragraph">A randomly selected male student is taller than a randomly selected female student by at least $20$ cm.</div></li><li><div class="list-item-paragraph">The difference in height between a randomly selected male and a female is more than $30$ cm.</div></li><li><div class="list-item-paragraph">The heights of two randomly selected female students both exceed $1.5$ m.</div></li><li><div class="list-item-paragraph">The total height of two randomly selected female students is more than $3$ m.</div></li></ol>
<p>Comment on the difference in answers for part <strong>(iv)</strong> and <strong>(v)</strong>.</p>

The heights, in metres, of female and male students in a school follow normal distributions with the following data.

Two randomly selected male are taller than twice a randomly selected female.

A randomly selected male student is taller than a randomly selected female student by at least $20$ cm.

The difference in height between a randomly selected male and a female is more than $30$ cm.

The heights of two randomly selected female students both exceed $1.5$ m.

The total height of two randomly selected female students is more than $3$ m.

Comment on the difference in answers for part **(iv)** and **(v)**.

Normal Distribution

International Baccalaureate Higher Level Mathematics

IB HL Math

Worked Example 7

Let $F$ and $M$ be the heights, in metres, of randomly selected Female and Male respectively.

$F \sim N(1.6, 0.04)$, $M \sim N(1.75, 0.05)$

Required Probability $= P(M_1 + M_2 > 2F_1)$
$= P(M_1 + M_2 - 2F_1 > 0)$

$M_1 + M_2 - 2F_1 \sim N(2(1.75) - 2(1.6), 2(0.05) + 2^2(0.04))$
$M_1 + M_2 - 2F_1 \sim N(0.3, 0.26)$

$= 0.7218$
$= 0.722$ (3 s.f.)

Required Probability $= P(M - F > 0.20)$

$M - F \sim N(1.75 - 1.6, 0.05 + 0.04)$
$M - F \sim N(0.15, 0.09)$

$= 0.43381$
$= 0.434$ (3 s.f.)

Required Probability $= P(|M - F| > 0.3)$
$= 1 - P(|M - F| < 0.3)$
$= 1 - P(-0.3 < M - F < 0.3)$

Using GC: normalcdf(lower: -0.3, upper: 0.3, μ: 0.15, σ: √0.09)

$= 1 - 0.624655$
$= 0.375$ (3 s.f.)

Required Probability $= [P(F > 1.5)]^2$

Using GC: normalcdf(lower: 1.5, upper: 10^99, μ: 1.6, σ: √0.04)

$= 0.4781$
$= 0.478$ (3 s.f.)

$F_1 + F_2 \sim N(3.2, 0.08)$

Required Probability $= P(F_1 + F_2 > 3)$
$= 0.760$ (3 s.f.)

Cases in (iv) is a subset of cases in (v). Thus, the probability of case (v) is higher.

Worked Example 7

Worked Example 7

Video Solution: