help with statistics

Question

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

197.5 cm

and a standard deviation of

8.6 cm.

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

210.90

cm.

b. Find the probability that the mean for

20

randomly selected distances is greater than 195.30 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

. 0596.0596.

​(Round to four decimal places as​ needed.)

b. The probability is

nothing.

​(Round to four decimal places as​ needed.)

Answer 1

x = individual overhead reach distance = 210.90

mean = 197.5

standard deviation = 8.6

a)

z = (x - mean)/standard deviation = (210.90 - 197.5)/8.6 = 1.56

P(x > 210.9) = P(z > 1.56) = 1 - P(z < 1.56) = 1 - 0.9406 = 0.0594

b)

xbar = mean overhead reach distance = 195.3

n = sample size = 20

z = (xbar - mean)/(standard deviation/square root(n)) = (195.3 - 197.5)/(8.6/sqrt(20)) = -1.14

P(xbar > 20) = P(z > -1.14) = 1 - P(z < -1.14) = 1 - 0.2171 = 0.8729

c) because the overhead reach distances are normally distributed