Question

Birth weights in the United States are normally distributed with a mean of 3420 grams and a standard deviation of 495 grams. If we randomly select 36 babies in the U.S., what is the probability that their mean (average) birth weight will be greater than 3500 grams?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 3420

standard deviation = $\sigma$ = 495

$\sigma$
_T
= $\sigma$ / $\sqrt$ n = 495 / $\sqrt$ 36 = 82.5

P(
T
> 3500) = 1 - P(
T
< 3500)

= 1 - P[(
T
- $\mu$
_T
) / $\sigma$
_T
< (3500 - 3420) / 82.5]

= 1 - P(z < 0.97)

= 1 - 0.834

0.1660

Probability = 0.1660