It is also known that customers will spend an average of $178 on additional maintenance. The standard deviation of the expenses is $50. If a simple random sample of 100 customers is taken: What is the probability that the sample mean will be between 166.75 and $170.50

probability that the sample mean will be between 166.75 and $170.50 = 0.42816

Step-by-step explanation:

We are given;

Mean; μ = $178

Standard deviation; σ = $50

Now, we want to find the probability that the sample mean will be between $166.75 and $170.50.

Thus, we'll use the z-score formula;

z = (x - μ)/σ

So;

Lower limit of z is;

z = (166.75 - 178)/50

z = -0.225

Upper limit of z is;

z = (170.50 - 178)/50

z = -0.15

From the z-distribution table attached, the area between -0.225 and -0.15 is;

0.44038 - 0.01222 = 0.42816

b, 20 yards

step-by-step explanation: