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 ?

Answer 1

Answer: 0.1587

Step-by-step explanation:

Given : The distribution of number of purchases per hour is approximately normal with mean 1,200 purchases and standard deviation 200 purchases.

i.e. $\mu=1200,\ \ \sigma=200$

Let X denotes the number of purchases per hour.

Then ,

$P(X>1400)=P(\dfrac{X-\mu}{\sigma}>\dfrac{1400-1200}{200})\\\\=P(Z>1)\ \ \[Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Zz)=1-P(Z$

So, the proportion of hours will the number of purchases at the online store exceed 1,400 =0.1587