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Find the indicated probability.

Empirical evidence shows that the "normal" resting adult body temperature is normally distributed with a mean of 98.2 degrees Fahrenheit and a standard deviation of 0.7 degrees Fahrenheit.

Answer :

The probability that a randomly selected adult has a "normal" resting body temperature between 98 and 99 degrees Fahrenheit is 0.487.

How the probability is computed:

We can determine the probability using the normal distribution calculator that computes the z-score between the given ranges as follows:

Mean = 98.2

Standard deviation = 0.7

Probability between 98 and 99 = P (98 < X < 99)

Z = (x - u)/std

= (98 - 98.2)/0.7 = 0.29

and

= (99 - 98.2)/0.7 = 1.14

Therefore, P (98 <98 < 99)

Using the standard normal table:

P (-0.29 < Z < 1.14)

P = 0.487

= 48.7%

Learn more about probability at https://brainly.com/question/23417919.

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Complete Question:

Find the indicated probability. Empirical evidence shows that the "normal" resting adult body temperature is normally distributed with a mean of 98.2 degrees Fahrenheit and a standard deviation of 0.7

What is the probability that a randomly selected adult has a "normal" resting body temperature between 98 and 99 degrees Fahrenheit?

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