Questions: 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 600 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 475.00 and 690.00?
(A) 0.9999
(B) 0.9579
(C) 0.0001
(D) 0.0421
Transcript text: 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 $\$ 600$ 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 $\$ 475.00$ and $\$ 690.00$ ?
(A) 0.9999
(B) 0.9579
(C) 0.0001
(D) 0.0421
Solution
Solution Steps
Step 1: Define the Normal Distribution Parameters
The personal call expenses (PCE) follow a normal distribution with:
Mean (\( \mu \)) = 600
Standard deviation (\( \sigma \)) = 50
Step 2: Calculate Z-scores
To find the probability that a randomly selected month had a PCE between \( \$475 \) and \( \$690 \), we first calculate the Z-scores for the lower and upper bounds.
Using the Z-scores, we can find the probability that the PCE falls within the specified range:
\[
P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.8) - \Phi(-2.5)
\]
From the calculations, we find:
\[
P = 0.9579
\]
Final Answer
The probability that a randomly selected month had a PCE between \( \$475 \) and \( \$690 \) is:
\[
\boxed{0.9579}
\]