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

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

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 =
`(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(
`(X-\mu)/(\sigma)` <
`(x-98.2)/(0.62)` ) = 0.12

P(Z <
`(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;

`(x-98.2)/(0.62) = -1.1835`

`{x-98.2}= -1.1835* 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.