Question

The overhead reach distances of adult females are normally distributed with a mean of

197.5 cm197.5 cm

and a standard deviation of

8.9 cm8.9 cm.

a. Find the probability that an individual distance is greater than

207.50207.50

cm.

b. Find the probability that the mean for

2020

randomly selected distances is greater than 196.20 cm.196.20 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

Solution :

Let X be a random variable which represents the overhead reach distances of adult females.

Given that, X ~ N(197.5, 8.9²)

μ = 197.5 cm and  σ = 8.9 cm

a) We have to find P(X > 207.50 cm).

We know that if X ~ N(μ, σ²) then,

Using "pnorm" function of R we get, P(Z > 1.1236) = 0.1306

Hence, the probability that an individual distance is greater than 207.50 cm is 0.1306.

b) We have to find P(x̅ > 196.20).

(Where, x̅ is mean for 20 randomly selected distances.)

We know that if X ~ N(μ, σ²) then, x̅ ~ N(μ, σ²/n).

And if x̅ ~ N(μ, σ²/n) then,

Using "pnorm" function of R we get, P(Z > -0.6532) = 0.7432

Hence, the probability that the mean of the 20 randomly selected distances is greater than 196.20 cm is 0.7432.

c) The normal distribution be used in part​ (b), even though the sample size does not exceed​ 30 because the population is normally distributed.