Answer to Question #227841 in Statistics and Probability for Carol

Question

Assume that the height of South African adult males has the normal distribution, with an average

of 166 cm and a standard deviation of 9 cm. Let X be the sample mean of the heights of a sample

of 20 men.

(a) Describe the distribution of X; giving the name of the distribution, its expected value and

variance.

(b) What is the probability that the sample mean is between 156 cm and 176 cm?

(c) Find the first and third quartiles of the distribution of the sample mean

Answer 1

(a) Suppose that a random sample of $n$ observations is taken from a normal population with mean $\mu$ and variance $\sigma^2.$ Then by the Central Limit Theorem we conclude that  the distribution of the sample means will be normally distributed with mean $\mu$ and variance $\sigma^2/n.$

$$X\sim N(\mu,\sigma^2/20)$$

Given $\mu=166, \sigma=9, n=20$

$$X\sim N(166, 9^2/20)$$

$$E(X)=\mu=166$$

$$Var(X)=\sigma^2/n=9^2/20=4.05$$

(b)

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

$$=P(Z<\dfrac{176-166}{4.05})-P(X\leq \dfrac{156-166}{4.05})$$

$$\approx P(Z<1.1111)-P(X\leq -1.1111)$$

$$\approx0.86674-0.13326\approx0.7335$$

(c)

$Q_1:$

$$P(Z<\dfrac{X-166}{4.05})=0.25$$

$$\dfrac{X-166}{4.05}\approx-0.6745$$

$$X\approx163.27$$

$$Q_1=163.27\ cm$$

$Q_3:$

$$P(Z<\dfrac{X-166}{4.05})=0.75$$

$$\dfrac{X-166}{4.05}\approx0.6745$$

$$X\approx168.73$$

$$Q_1=168.73\ cm$$