Question

a survey of a track team concluded that they ran an average of 28miles a week with a standard deviation of 7miles per week. Approximately what percentage of runners in the club ran less than 30 miles per week?

Answer 1

a survey of a track team concluded that they ran an average of 28 miles a week with a standard deviation of 7 miles per week. Approximately what percentage of runners in the club ran less than 30 miles per week?

We have normal distribution with mean 28 and standard deviation 7:

$$m = 28$$ $$\sigma = 7$$

And we need to know what percentage of runners in the club ran less than 30 miles per week ( $P(x < 30)$ ):

$$P(x < 30) = \int_{-\infty}^{30} f(x) \, dx$$

$f(x)$ - probability density function

The normal distribution has probability density:

$$f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x - m)^2}{2\sigma^2}}$$

Therefore:

$$P(x < 30) = \int_{-\infty}^{30} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x - m)^2}{2\sigma^2}} \, dx$$

Calculating this integral:

$$P(x < 30) = 0.6125 = 61.25\%$$

Answer: $61.25\%$