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The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 3950 grams and a standard deviation of 374 grams. If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4473 grams. Round your answer to four decimal places.

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Answer:

0.9192 = 91.92% probability that the weight will be less than 4473 grams.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean weight of 3950 grams and a standard deviation of 374 grams.

This means that
\mu = 3950, \sigma = 374

Find the probability that the weight will be less than 4473 grams.

This is the pvalue of Z when X = 4473. So


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


Z = (4473 - 3950)/(374)


Z = 1.4


Z = 1.4 has a pvalue of 0.9192

0.9192 = 91.92% probability that the weight will be less than 4473 grams.

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