Question

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 7.8 cm.

a. Find the probability that an individual distance is greater than 210.00 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 195.70 cm.

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

a. The probability is (Round to four decimal places as​ needed.): 0.0548

b. The probability is (Round to four decimal places as​ needed.):

Answer 1

Solution :

Given that ,

mean = = 197.5

standard deviation = = 7.8

(a)

P(x > 210) = 1 - P(x < 210)

= 1 - P((x - ) / < (210 - 197.5) / 7.8)

= 1 - P(z < 1.60)

= 1 - 0.9452

= 0.0548

Probability = 0.0548

b)

n = 15

= 197.50 and

= / n = 7.8 / 15 = 2.014

P( > 195.70) = 1 - P( < 195.70)

= 1 - P(( - ) / < (195.70 - 197.50) / 2.014)

= 1 - P(z < -0.89)

= 1 - 0.1867

= 0.8133

Probability = 0.8133

C)

Original distribution is normally distributed .