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GREYUIT [131]
3 years ago
13

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%.
Mathematics
1 answer:
Nady [450]3 years ago
3 0

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 = <u><em>human body temperatures</em></u>

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.

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