Question

The weights of books in a bookcase are normally distributed with a mean of 13. 5 pounds and a standard deviation of 2. 3 pounds. There are 145 books on the bookshelf how many of the books will weigh 11 pounds or more

Answer 1

89.09% of books will weigh 11 pounds or more, then the number of books will be 129.

What is a z- score?

The z-score is a numerical measurement used in statistics to refer to how much a given value differs from the standard deviation.

The weights of books in a bookcase are normally distributed with a mean of 11.6 pounds and a standard deviation of 1.3 pounds. There are 136 books on the bookshelf.

Then the z-score when the books will weigh 11 pounds or more will be

$z= \frac{11-13.5}{2.3} \\\\z= -1.0869$

Then the probability will be

$P(x > 11) = 1-P(x < 11 ) = 0.8909$

89.09% of books will weigh 11 pounds or more, then the number of the books will be

=145 × 0.8909

=129.1805 ≈ 129

More about the z-score link is given below.

brainly.com/question/15016913

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