The amount of jen's monthly phone bill is normally distributed with a mean of $65 and a standard deviation of 11. what percentage of her phone bills are between 32 and 98

Respuesta :

It would be 99%. If you draw the normal distribution curve and plot the mean and standard deviation, it might help you visually.

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The percentage of her phone bills are between 32 and 98 will be 99.73%.

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 amount of Jen's monthly phone bill is normally distributed with a mean of $65 and a standard deviation of 11.

Then the percentage of her phone bills are between 32 and 98 will be

The value of z-score for x = 32, we have

z = (32 – 65) / 11

z = –3

The value of z-score for x = 98, we have

z = (98 – 65) / 11

z = 3

Then the percentage will be

P(32 < x < 98) = P(-3 < z < 3) × 100

P(32 < x < 98) = 0.9973 × 100

P(32 < x < 98) = 99.73%

More about the z-score link is given below.

https://brainly.com/question/15016913

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