Question

The overhead reach distances of adult females are normally distributed with a mean of 197.5 and a standard deviation of 8.9 cm.

a. Find the probability that an individual distance is greater than 206.80 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 195.30

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

Answer 1

Solution :

Given that ,

a) P(x > 206.80) = 1 - p( x< 206.80)

=1- p P[(x - ) / < (206.80 - 197.5) / 8.9]

=1- P(z < 1.04)

= 1 - 0.8508

= 0.1492

b) = 197.5

= / n = 8.9 / 15

P( > 195.30) = 1 - P( < 195.30)

= 1 - P[( - ) / < (195.30 - 197.5) / 8.9 / 15 ]

= 1 - P(z < -0.96)   

= 1 - 0.1685

= 0.8315

c) Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size