Answer to Question #219518 in Statistics and Probability for mimi

Question

Assume that the height of South African adult males has normal distribution, with an average of 166 cm and a standard deviation of 9 cm.

(a) What percentage of South African adult males are between 156 cm and 176 cm in height? (4)

(b) Find the first and third quartiles of the height distribution of South African adult males. (8)

(c) A bed manufacturer wants to make a bed long enough to fit at least 99% of South African adult males. What is the height of the tallest person they should accommodate to achieve this? 

Answer 1

Let $X=$ the height of South African adult males: $X\sim N(\mu, \sigma^2).$

Given $\mu=166\ cm ,\sigma=9\ cm.$

a)

$$P(156<X<176)=P(X<176)-P(X\leq 156)$$

$$=P(X<\dfrac{176-166}{9})-P(X\leq \dfrac{156-166}{9})$$

$$\approx0.8667397-0.1332603\approx0.733479$$

$73.35\%$

(b)

$$P(Z<\dfrac{x-166}{9})=0.25$$

$$\dfrac{x-166}{9}\approx-0.67449$$

$$x=Q_1=157.23$$

$$P(Z<\dfrac{x-166}{9})=0.75$$

$$\dfrac{x-166}{9}\approx0.67449$$

$$x=Q_3=174.77$$

(c)

$$P(Z<\dfrac{x-166}{9})=0.99$$

$$\dfrac{x-166}{9}\approx2.32635$$

$$x=197$$