Question

Birthweights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz.

Answer 1

125 is 1 dev above, so it's  13.5 + 2.35,
so it's B. 0.159

And i think the proportion of infants with birthweights between 125 oz and 140 oz is :
D. 0.136

Hope this helps

Answer 2

Answer:    B. 0.159.

D. 0.136.

Explanation:

Given : Birthweights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz.

i.e. $\mu=110\ ;\ \sigma=15$

Formula to find z-score :-

$z=\dfrac{x-\mu}{\sigma}$

(a) To find the proportion of infants with birthweights above 125 oz, we find the z-score corresponds to 125 will be :-

$z=\dfrac{125-110}{15}=1$

By using the standard normal distribution table , The probability of infants with birthweights above 125 oz :-

$P(x>125)=P(z>1)=1-P(z$

Hence, the proportion of infants with birthweights above 125 oz is 0.159 .

(b) The z-score corresponds to 140 , $z=\dfrac{140-110}{15}=2$

The probability of infants with birthweights between 125 oz and 140 oz  :-

$P(125$

Hence, the proportion of infants with birthweights between 125 oz and 140 oz is 0.136.