To find the proportion of hours in which the number of purchases exceeds 1,400 at the online store, given the number of purchases per hour is approximately normally distributed with:
- Mean μ=1200\mu =1200μ=1200 purchases
- Standard deviation σ=200\sigma =200σ=200 purchases
Step 1: Calculate the z-score for 1,400 purchases
z=X−μσ=1400−1200200=200200=1z=\frac{X-\mu}{\sigma}=\frac{1400-1200}{200}=\frac{200}{200}=1z=σX−μ=2001400−1200=200200=1
This means 1,400 purchases is 1 standard deviation above the mean. Step 2: Find the proportion of the distribution above this z-score From standard normal distribution tables or calculators, the cumulative probability for z=1z=1z=1 is approximately 0.84 (or 84%). This represents the proportion of hours with purchases less than or equal to 1,400. Step 3: Calculate the proportion exceeding 1,400
1−0.84=0.161-0.84=0.161−0.84=0.16
So, about 16% of the hours will have purchases exceeding 1,400. Final answer: The proportion of hours with purchases exceeding 1,400 is approximately 16%
