Question

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.9 cm .

a. Find the probability that an individual distance is greater than 217.50 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 202.80 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 = = 205

standard deviation = = 8.9

a.

P(x > 217.50) = 1 - P(x < 217.50)

= 1 - P[(x - ) / < (217.50 - 205) / 8.9)

= 1 - P(z < 1.40)

= 1 - 0.9192

= 0.0808

Probability = 0.0808

b.

= / n = 8.9 / 15 = 2.2980

P( > 202.80) = 1 - P( < 202.80)

= 1 - P[( - ) / < (202.80 - 205) / 2.2980]

= 1 - P(z < -0.96)

= 1 - 0.1685

= 0.8315

Probability = 0.8315

c.

since the original population has a normal distribution , the distribution of sample means is a normal distribution for any sample size.