Question

The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 8.3 cm. a. Find the probability that an individual distance is greater than 204.30 cm. b. Find the probability that the mean for 20 randomly selected distances is greater than 192.80 cm. c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30

Answer 1

Answer:

Mean : $\mu = 195$

$\sigma = 8.3 cm$

a) Find the probability that an individual distance is greater than 204.30 cm.

We are supposed to find P(Z>204.30)

x = 204.30

Formula : $z=\frac{x-\mu}{\sigma}$

Substitute the value sin the formula :

$z=\frac{204.40-195}{8.3}$

$z=1.13253$

P(Z>204.30)=1-P(Z<204.30)

Refer the z table

P(Z>1.13253)=1-P(Z<1.13253)

P(Z>1.13253)=1-0.8708

P(Z>1.13253)=0.1292

The probability that an individual distance is greater than 204.30 cm is 0.1292.

b) Find the probability that the mean for 20 randomly selected distances is greater than 192.80 cm

We are supposed to find P(Z>192.80)

x = 192.80

Formula : $z=\frac{x-\mu}{\sigma}$

Substitute the value sin the formula :

$z=\frac{ 192.80-195}{8.3}$

$z=-0.265$

P(Z>192.80)=1-P(Z<192.80)

Refer the z table

P(Z>-0.265)=1-P(Z<-0.265)

P(Z>-0.265)=1-0.3974

P(Z>-0.265)=0.6026

The probability that the mean for 20 randomly selected distances is greater than 192.80 cm is 0.6026

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

The normal distribution is used because the original population has normal distribution.