Question

The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds. Using the empirical rule, what is the probability that a cat will weigh less than 11 pounds?

Answer 1

If the value of the z-score is 1. Then the probability that a cat will weigh less than 11 pounds will be 0.84134.

What is the z-score?

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

The z-score is given as

z = (x - μ) / σ

Where μ is the mean, σ is the standard deviation, and x is the sample.

The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds.

Then the probability that a cat will weigh less than 11 pounds will be

The value of z-score will be

z = (11 – 9) / 2

z = 1

Then the probability will be

P(x < 11) = P(z < 1)

P(x < 11) = 0.84134

Thus, the probability that a cat will weigh less than 11 pounds will be 0.84134.

More about the z-score link is given below.

brainly.com/question/15016913

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