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([tex]\mu=10,000,\sigma^{2}=1,500^{2}[/tex])
Now, the z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean
[tex]\sigma[/tex] = 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{13,000-10,000}{1,500}[/tex] ) = P(Z > 2) = 1 - P(Z [tex]\leq[/tex] 2)
= 1 - 0.97725 = 0.0228 or 2.28%
Therefore, 2.28% of the days does he exceed 13,000 steps.
