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Human body temperatures are normally distributed with a mean of 98.20degreesF and a standard deviation of 0.62 degrees
f. Find the temperature that separates the top​ 7% from the bottom​ 93%. Round to the nearest hundredth of a degree.

Respuesta :

MrEquation
Answer: The temperature that separates the top 7% with the bottom 93% is 99.12 degrees.

The z-score that corresponds to 93% is about 1.48. If we write the equation for the z-score, we can solve for the missing temperature.

The equation would be:
(x - 98.2) / 0.62 = 1.48
x - 98.2 = 0.9176
x = 99.12 degrees

Answer with explanation:

The formula of Z score is

  [tex]Z_{Score}=\frac{\Bar X-\mu}{\sigma}\\\\ Z_{7 \text{Percent}}<\frac{\Bar X- 98.20}{0.62}<Z_{93 \text{Percent}}\\\\Z_{0.07}<\frac{\Bar X- 98.20}{0.62}<Z_{0.93}\\\\0.52790<\frac{\Bar X- 98.20}{0.62}<0.82831\\\\0.327298< \Bar X -98.20 < 0.5135522\\\\0.33 < \Bar X - 98.20 < 0.52\\\\ 0.33 +98.20 < \Bar X < 98.20+0.52\\\\98.53 < \Bar X < 98.72[/tex]

The value of temperatures are that separates the top​ 7% from the bottom​ 93% are:⇒

 98.53 < Temperature in degrees < 98.72

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