Question

The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 3915 grams and a variance of 429,025. If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4570 grams. Round your answer to four decimal places.

Answer 1

0.8413 = 84.13% probability that the weight will be less than 4570 grams.

Explanation:

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

In a set with mean
`\mu` and standard deviation(which is the square root of the variance)
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 3915, \sigma = √(429025) = 655`

If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4570 grams.

This is the pvalue of Z when X = 4570. So

`Z = (X - \mu)/(\sigma)`

`Z = (4570 - 3915)/(655)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

0.8413 = 84.13% probability that the weight will be less than 4570 grams.