At a customer service call center for a large company, the number of calls received per hour is normally distributed with a mean of 120 calls and a standard deviation of 15 calls. What is the probability that during a given hour of the day there will be between 145 calls and 157 calls, to the nearest thousandth?
To find the probability, we need to calculate the z-scores for 145 and 157 calls and then use a standard normal distribution table.
First, let's find the z-scores for each:
Z-score for 145 calls:
Z = (X - μ) / σ
Z = (145 - 120) / 15
Z = 25 / 15
Z = 1.667
Z-score for 157 calls:
Z = (X - μ) / σ
Z = (157 - 120) / 15
Z = 37 / 15
Z = 2.467
Now, we can look up the probabilities associated with these z-scores in a standard normal distribution table. The probability of receiving between 145 and 157 calls during an hour would be the difference between the probability of Z = 2.467 and the probability of Z = 1.667.
Using the standard normal distribution table, we find:
P(Z < 1.667) = 0.9525
P(Z < 2.467) = 0.9939
Therefore, the probability of receiving between 145 and 157 calls during an hour is:
P(145 < X < 157) = P(1.667 < Z < 2.467) = P(Z < 2.467) - P(Z < 1.667) = 0.9939 - 0.9525 = 0.0414
So, the probability is approximately 0.0414 or 4.14% to the nearest thousandth.