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enot [183]
3 years ago
5

A study of a population showed that Male'S body temperature are approximately normally distributed with a mean of 98.4°F and a p

opulation standard deviation ofb0.40°f.What boby temperature does a male have if he is at the 83rd percentile?
SAT
1 answer:
Ilia_Sergeevich [38]3 years ago
8 0

Answer:

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

Explanation:

The <em>n</em>th percentile implies that there are <em>n%</em> value below this percentile value.

That is, if P (<em>X </em><<em> x</em>) = n% then <em>x</em> is the <em>n</em>th percentile.

Let<em> </em><em>X</em> = male body temperature.

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

Let <em>x</em> be the 83rd percentile value.

Then, P (X < x) = 0.83.

The value of <em>x</em> can be computed from the <em>z</em>-score.

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

Compute the <em>z</em>-score related to this probability as follows:

P (Z < z) = 0.83

*Use the <em>z</em>-table for the <em>z</em>-score.

The value of <em>z</em> is 0.95.

Compute the value of <em>x</em> 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.

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