Question

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

202.5 cm202.5 cm

and a standard deviation of

8.6 cm8.6 cm.

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

215.90215.90

cm.

b. Find the probability that the mean for

2020

randomly selected distances is greater than 200.70 cm.200.70 cm.

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 ,

mean = = 202.5

standard deviation = = 8.6

(a)

P(x > 215.90) = 1 - P(x < 215.90)

= 1 - P((x - ) / < (215.90 - 202.5) / 8.6)

= 1 - P(z < 1.56)

= 1 - 0.9406   

= 0.0594

Probability = 0.0594

(b)

n = 20

= 202.5 and

= / n = 8.6 / 20 = 1.9230

P( > 200.70) = 1 - P( < 200.70)

= 1 - P(( - ) / < (200.70 - 202.5) / 1.9230)

= 1 - P(z < -0.94)

= 1 - 0.1736

= 0.8264

Probability = 0.8264

(c)

Original distribution is normal .