Question

A study of a population showed that Male'S body temperature are approximately normally distributed with a mean of 98.4°F and a population standard deviation ofb0.40°f.What boby temperature does a male have if he is at the 83rd percentile?

Answer 1

Answer:

The body temperature of a male at the 83rd percentile is 98.8°F.

Explanation:

The nth percentile implies that there are n% value below this percentile value.

That is, if P (X < x) = n% then x is the nth percentile.

Let X = male body temperature.

The random variable X follows a Normal distribution with mean, μ = 98.4°F and standard deviation, σ = 0.40°F.

Let x be the 83rd percentile value.

Then, P (X < x) = 0.83.

The value of x can be computed from the z-score.

$z=\frac{x-\mu}{\sigma}$

Compute the z-score related to this probability as follows:

P (Z < z) = 0.83

*Use the z-table for the z-score.

The value of z is 0.95.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\0.96=\frac{x-98.4}{0.40} \\x=98.4+(0.96\times0.40)\\=98.784\\\approx98.8$

Thus, the body temperature of a male at the 83rd percentile is 98.8°F.