Answer to Question #349910 in Statistics and Probability for jemima

Question

The overhead reach distances of adult females are normally distributed with a mean of 195 and a standard deviation of 8.9. a. Find the probability that an individual distance is greater than 208.40 cm. b. Find the probability that the mean for 20 randomly selected distances is greater than 192.80 c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

a.

$$P(X>208.4)=1-P(Z\le\dfrac{208.4-\mu}{\sigma})$$

$$=1-P(Z\le\dfrac{208.4-195}{8.9})$$

$$\approx1-P(Z\le1.505618)$$

$$\approx0.066083$$

b.

$$P(X>192.8)=1-P(Z\le\dfrac{192.8-\mu}{\sigma\sqrt{n}})$$

$$=1-P(Z\le\dfrac{192.8-195}{8.9/\sqrt{20}})$$

$$\approx1-P(Z\le-1.105472)$$

$$\approx0.865522$$

c)

If the population is normal, then the Central Limit Theorem holds true even for samples smaller than 30.