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 ?

a 68%

b 32%

c 16%

d 5%

e 2.5%

Explanation:

Step1: Identify the mean and standard deviation

The mean ($\mu$) is 1200 and the standard deviation ($\sigma$) is 200. We want to find the proportion of hours where purchases exceed 1400.

Step2: Calculate the z - score

The z - score formula is $z=\frac{x - \mu}{\sigma}$. Substituting $x = 1400$, $\mu=1200$ and $\sigma = 200$ into the formula:
$z=\frac{1400 - 1200}{200}=\frac{200}{200}=1$

Step3: Use the empirical rule for normal distribution

The empirical rule (68 - 95 - 99.7 rule) states that for a normal distribution:

• Approximately 68% of the data lies within $\mu\pm\sigma$ (i.e., between $1200 - 200=1000$ and $1200 + 200 = 1400$)
• The total area under the normal curve is 1 (or 100%). The area to the left of $z = 1$ (i.e., less than 1400) is the area within $\mu-\sigma$ to $\mu+\sigma$ plus the area less than $\mu-\sigma$. But since the data is symmetric, the area within $\mu-\sigma$ to $\mu+\sigma$ is 68%, so the area less than $\mu-\sigma$ and greater than $\mu+\sigma$ is $100\% - 68\%=32\%$. And since the distribution is symmetric, the area greater than $\mu+\sigma$ (i.e., greater than 1400) is half of 32%, which is $16\%$.

Answer:

C. 16%