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

8.6 cm8.6 cm.

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

206.80206.80

cm.

b. Find the probability that the mean for

2525

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

Answer a)

P(Z > 1.08) = 1 - P(Z<1.08)

P(Z > 1.08) = 1 - P(Z<1.08) = 1 - 0.8599

P(Z > 1.08) = 0.1401

The probability that an individual distance is greater than 206.80 is 0.1401

Answer b)

Mean of sampling distribution = 197.5

Standard deviation of sampling distribution = 8.6/sqrt(25) = 8.6/5 = 1.72

P(Z > -0.76) = P(Z < 0.76) = 0.7764

Answer c)

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