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. [tex]\mu=1200,\ \ \sigma=200[/tex]
Let X denotes the number of purchases per hour.
Then ,
[tex]P(X>1400)=P(\dfrac{X-\mu}{\sigma}>\dfrac{1400-1200}{200})\\\\=P(Z>1)\ \ \[Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z<1)\ \ \ [P(Z>z)=1-P(Z<z)]\\\\\ =1-0.8413\ \ [\text{By p-value table}]\\\\= 0.1587[/tex]
So, the proportion of hours will the number of purchases at the online store exceed 1,400 =0.1587
