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The number of books borrowed from a library each week follows a normal distribution. When a sample is taken for several weeks, the mean is found to be 190 and the standard deviation is 30. What is the percentage of the chance that more than 250 books were borrowed in a week?

User Xaruth
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Final answer:

A z-score calculation is used to find that there is a 2.28% chance that more than 250 books were borrowed from the library in a given week.

Step-by-step explanation:

The student’s question asks for the percentage chance that more than 250 books were borrowed in a week given that the number of books borrowed follows a normal distribution, where the sample mean is 190, and the sample standard deviation is 30.

To answer this, we’ll need to calculate the z-score for 250 books using the formula:

Z = (X - μ) / σ

Where X is 250 books, μ (mu) is the mean (190 books), and σ (sigma) is the standard deviation (30 books). Substituting these values, we get:

Z = (250 - 190) / 30 = 60 / 30 = 2

After calculating the z-score, we look up this value in the z-table to find the probability to the left of z=2. The table gives us the area under the curve to the left of the z-score. Since we’re interested in the probability of borrowing more than 250 books, we need to find the area to the right of z=2. This is equal to 1 minus the area to the left. Typically, the area to the left of z=2 is approximately 0.9772, which when subtracted from 1 gives us about 0.0228 or 2.28%.

Therefore, there is a 2.28% chance that in any given week, more than 250 books were borrowed.

User Marcie
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