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Chapter 6: Problem 86

A study of U.S. births published on the website Medscape from WebMD reported that the average birth length of babies was \(20.5\) inches and the standard deviation was about \(0.90\) inch. Assume the distribution is approximately Normal. Find the percentage of babies who have lengths of 19 inches or less at birth.

Short Answer

Expert verified
4.75% of the babies are expected to have lengths of 19 or fewer inches at birth.

Step by step solution

01

Identify known parameters

The first step would be to identify and write down the given parameters in the problem. The mean (average) length of babies \(\mu = 20.5\) inches, the standard deviation \(\sigma = 0.90\) inches and the given length is \(X = 19\) inches.
02

Calculate Z-Score

The next step is to calculate the z-score. The z-score measures the number of standard deviations that a given data point is from the mean. It is calculated by using the following formula: \(Z = \frac{X - \mu}{\sigma}\). Substituting the given values, we get \(Z = \frac{19 - 20.5}{0.90} = -1.67\). This means that \(X = 19\) inches is 1.67 standard deviations to the left of the mean.
03

Find the percentage

The final step is to figure out the percentage using the Z-table, also known as the standard normal distribution table. The value for \(Z = -1.67\) is read directly from the Z-table or calculated using statistical software, yielding 0.0475 or 4.75%. This means that 4.75% of the babies are expected to have lengths of 19 or fewer inches at birth.

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