Question

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

197.5 cm

and a standard deviation of

8.3 cm.

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

206.80

cm.

b. Find the probability that the mean for

20

randomly selected distances is greater than 196.20 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 = $\mu$ = 197.5

standard deviation = $\sigma$ = 8.3

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

= 1 - P[(x - $\mu$ ) / $\sigma$ < (206.80 - 197.5) /8.3 )

= 1 - P(z < 1.12)

= 1 - 0.8686

0.1314

Probability = 0.1314

b) n = 20

$\sigma$
_T
= $\sigma$ / $\sqrt$ n = 8.3/ $\sqrt$ 20 = 1.8559

P(
T
> 196.20) = 1 - P(
T
< 196.20)

= 1 - P[(
T
- $\mu$
_T
) / $\sigma$
_T
< (196.20 - 197.5) / 1.8559]

= 1 - P(z < -0.70)

= 1 - 0.242

0.7580

Probability = 0.7580

c) The normal distribution of part (b) is used because,the since original population has normal distribution,the distribution of sample means is normal distribution for any sample size.