The average amount customers at a certain grocery store spend yearly is $703.92. Assume the
Variable is normally distributed. If the standard deviation is $99.00, find the probability that a
Randomly selected customer spends between $515.82 and $892.02.
A)1.065 = 106.5%
B)0.000 = 0.0%
C)-0.012 = -1.2%
D)0.942 = 94.2%

Question

Quizplus · Accepted Answer

We need to find the probability that a customer spends between $515.82 and $892.02. 
First, we need to standardize these values by subtracting the mean and dividing by the standard deviation:

Z1 = (515.82 - 703.92) / 99.00 = -1.897 
Z2 = (892.02 - 703.92) / 99.00 = 1.891

Using a standard normal distribution table or calculator, we can find the probability that a randomly selected customer spends between these values:

P(-1.897 < Z < 1.891) = 0.942

Therefore, the answer is D, 0.942 = 94.2%. First, we need to standardize the values using the formula: z = (x - μ) / σ
For $515.82, z = (515.82 - 703.92) / 99.00 = -1.897
For $892.02, z = (892.02 - 703.92) / 99.00 = 1.888
Next, we use the standard normal table or calculator to find the probability between these two z-scores.
P(-1.897 < z < 1.888) = 0.942
Therefore, the probability that a randomly selected customer spends between $515.82 and $892.02 is 0.942 or 94.2%.

The Average Amount Customers at a Certain Grocery Store Spend