Question

for a certain online store, the distribution of number of purchases per hour is approximately normal with mean 1,200 purchases and standard deviation 200 purchases. for what proportion of hours will the number of purchases at the online store exceed 1,400? a 68% b 32% c 16% d 5% e 2.5%

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 1400$, $\mu=1200$, and $\sigma = 200$. So $z=\frac{1400 - 1200}{200}=\frac{200}{200}=1$.

Step2: Use the properties of the normal distribution

In a normal distribution, about 68% of the data lies within $z=- 1$ and $z = 1$. So the proportion of data outside of this range (either $z>1$ or $z < - 1$) is $100\%-68\% = 32\%$. Since the normal distribution is symmetric, the proportion of data with $z>1$ is $\frac{32\%}{2}=16\%$.

Answer:

C. 16%