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 ?

The proportion of hours will be the number of purchases at the online store exceed 1,400

P(z>1) = 0.1587

Step-by-step explanation:

Step (i):-

Given mean of the Population (μ) = 1200

Standard deviation of the population (σ) = 200

let   X be the random variable in normal distribution

Given x = 1400

Step(ii) :-

The proportion of hours will be the number of purchases at the online store exceed 1,400

P(z>1) = 0.5 - A(1)

           = 0.5 - 0.3413

           = 0.1587

The percentage of hours will be the number of purchases at the online store exceed 1,400  is  15.87%