The average amount customers at a certain grocery store spend yearly is $635.55. Assume the
Variable is normally distributed. If the standard deviation is $94.41, find the probability that a
Randomly selected customer spends between $569.46 and $758.28.
A)0.784 = 78.4%
B)0.133 = 13.3%
C)0.145 = 14.5%
D)0.661 = 66.1%

Question

Quizplus · Accepted Answer

To find the probability, we calculate the z-scores for $569.46 and $758.28 using the formula z = (X - μ) / σ. For $569.46, z = ($569.46 - $635.55) / $94.41 ≈ -0.70. For $758.28, z = ($758.28 - $635.55) / $94.41 ≈ 1.30. Using the standard normal distribution table, the probability for z = -0.70 is approximately 0.2420 and for z = 1.30 is approximately 0.9032. The probability that a customer spends between these amounts is 0.9032 - 0.2420 = 0.6612, or 66.1%.

The Average Amount Customers at a Certain Grocery Store Spend