Birth weights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz. The proportion of infants with birth weights between 125 oz and 140 oz is.

a. 0.477.
b. 0.819.
c. 0.136.
d. 0.636.

c. 0.136.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The proportion of infants with birth weights between 125 oz and 140 oz is

This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So

X = 140

has a pvalue of 0.977

X = 125

has a pvalue of 0.841

0.9772 - 0.841 = 0.136

So the correct answer is:

c. 0.136.

c. 0.136.

Step-by-step explanation:

Since the Birth weights at a local hospital have a Normal distribution,

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = Birth weights.

µ = mean weight

σ = standard deviation

From the information given,

µ = 110 oz

σ = 15 oz

We want to find the proportion of infants with birth weights between 125 oz and 140 oz. It is expressed as

P(125 ≤ x ≤ 140)

For x = 125,

z = (125 - 110)/15 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.841

For x = 140,

z = (140 - 110)/15 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.977

Therefore,

P(125 ≤ x ≤ 140) = 0.977 - 0.841 = 0.136