A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees.
Before the new system was installed, the amount being spent on personal calls followed a normal distribution with an average of $700 per month and a standard deviation of $50 per month. Refer to such expenses as PCE's (personal call expenses).
Using the distribution above, what is the probability that a randomly selected month had a PCE of between $375.00 and $550.00?
Show work
a) 0.9987
b) 0.214, 3
c) 0.0013
d) 0.785, 7
Guide On Rating System
Vote
To solve this problem, we need to convert the dollar amounts to z-scores and then use the standard normal distribution to find the probabilities.
The z-score is calculated as follows:
z = (X - μ) / σ
where X is the value, μ is the mean, and σ is the standard deviation.
First, calculate the z-scores for $375 and $550:
z1 = ($375 - $700) / $50 = -6.5
z2 = ($550 - $700) / $50 = -3
Next, we use the standard normal distribution table (or a calculator with a normal distribution function) to find the probabilities associated with these z-scores.
The probability of a z-score being less than -6.5 is essentially 0 (it's so small it's off the charts), and the probability of a z-score being less than -3 is approximately 0.0013.
Therefore, the probability that a randomly selected month had a PCE of between $375.00 and $550.00 is approximately 0.0013 - 0 = 0.0013.
So, the answer is (c) 0.0013.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99