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? 68% B 32% 16% 5% E 2.5%
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? 68% B 32% 16% 5% E 2.5%
Solution
To answer this question, we need to use the properties of a normal distribution.
Step 1: Calculate the Z score. The Z score is a measure of how many standard deviations an element is from the mean.
Z = (X - μ) / σ
Where: X = value = 1400 purchases μ = mean = 1200 purchases σ = standard deviation = 200 purchases
So, Z = (1400 - 1200) / 200 = 1
Step 2: Look up this Z score in the Z table or use a Z score calculator. The value associated with Z=1 is 0.8413. This means that 84.13% of the data falls below 1400 purchases.
Step 3: Since we want to find the proportion of hours where the purchases exceed 1400, we need to subtract the percentage we found from 100%.
100% - 84.13% = 15.87%
So, the closest answer to this percentage is 16%.
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