Question

Human body temperatures are normally distributed with a mean of 98.2oF and a standard deviation of 0.62oF. Find the temperature that separates the bottom 12% from the top 88%.

Answer 1

Answer:

The temperature that separates the bottom 12% from the top 88% is 97.5°F.

Step-by-step explanation:

We are given that human body temperatures are normally distributed with a mean of 98.2°F and a standard deviation of 0.62°F.

Let X = human body temperatures

So, X ~ Normal($\mu= 98.2,\sigma^{2} = 0.62^{2}$)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean human body temperature = 98.2°F

           $\sigma$ = stnadard deviation = 0.62°F

Now, we have to find the temperature that separates the bottom 12% from the top 88%, that means;

        P(X < x) = 0.12       {where x is the required temperature}

        P( $\frac{X-\mu}{\sigma}$ < $\frac{x-98.2}{0.62}$ ) = 0.12

        P(Z < $\frac{x-98.2}{0.62}$ ) = 0.12

Now, the critical value of x that represents the bottom 12% of the area in the z table is given as -1.1835, that is;

                    $\frac{x-98.2}{0.62} = -1.1835$

                    ${x-98.2}= -1.1835\times 0.62$

                     $x = 98.2 -0.734$ = 97.5°F

Hence, the temperature that separates the bottom 12% from the top 88% is 97.5°F.