Question

A health conscious student faithfully wears a device that tracks his steps. Suppose that the distribution of the number of steps he takes in a day is normally distributed with a mean of 10,000 and a standard deviation of 1,500 steps. What percent of the days does he exceed 13,000 steps

Answer 1

Answer:

2.28% of the days does he exceed 13,000 steps.

Step-by-step explanation:

We are given that  the distribution of the number of steps health conscious student takes in a day is normally distributed with a mean of 10,000 and a standard deviation of 1,500 steps.

Let X = number of steps a health conscious student takes in a day

So, X ~ N($\mu=10,000,\sigma^{2}=1,500^{2}$)

Now, the z score probability distribution is given by;

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

where, $\mu$ = population mean

            $\sigma$ = standard deviation

So, percent of the days does he exceed 13,000 steps is given by = P(X > 13,000 steps)

    P(X > 13,000) = P( $\frac{X-\mu}{\sigma}$ > $\frac{13,000-10,000}{1,500}$ ) = P(Z > 2) = 1 - P(Z $\leq$ 2)

                                                                  = 1 - 0.97725 = 0.0228 or 2.28%

Therefore, 2.28% of the days does he exceed 13,000 steps.