Question

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.8 minutes and the standard deviation was 0.50 minutes.
1. What is the probability that calls last between 4.8 and 5.6 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.)
2. What is the probability that calls last more than 5.6 minutes? (RoundYour z value to 2 decimal places and final answer to 4 decimal places.)
3. What is the probability that calls last between 5.6 and 6.5 minutes? (Round Your z value to 2 decimal places and final answer to 4 decimal places.)
4. What is the probability that calls last between 4.5 and 6.5 minutes? (Round Your z value to 2 decimal places and final answer to 4 decimal places.)
5. As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4 percent of the calls. What is this time? (Round Your z value to 2 decimal places and final answer to 2 decimal places.)

Answer 1

Answer:

1. 0.4452; 2. 0.0548; 3. 0.0545; 4. 0.7254; 5. Time is about 5.68 minutes or more.

Step-by-step explanation:

We have a normal distribution with known parameters, that is, for the population mean and for the population standard deviation:

$\\ \mu = 4.8\;minutes$

$\\ \sigma = 0.50\;minutes$

For each of the questions, we have to transform either raw score to a z-score as a way to use the cumulative standard normal distribution table for consulting the probabilities values for each z-score.

$\\ z-score = \frac{x - \mu}{\sigma}$

We also need to have into account that the normal distribution is symmetrical, an indispensable property to calculate negative z-scores, that is, values below the population mean.

1. The probability that calls last between 4.8 and 5.6 minutes

The corresponding z-scores for 4.8 minutes and 5.6 minutes are respectively:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z_{4.8} = \frac{4.8 - 4.8}{0.50} = 0$

$\\ z_{5.6} = \frac{5.6 - 4.8}{0.50} = 1.60$

As we can see, the z-score for a raw value of 4.8 coincides with the population mean, thus the z-score = 0. The probability between 4.8 and 5.6 minutes is the resulting difference between these two values. As a result, consulting the associated probability for each z-score in the cumulative standard normal table, we have:

For  z = 0, P(z<0) = 0.50000.

For z = 1.60, P(z<1.60) = 0.94520

So, the probability that calls last between 4.8 and 5.6 minutes is:

P(4.8<x<5.6) = (z<1.60) - P(z<0) = 0.94520 - 0.50000 = 0.44520 = 0.4452 (four decimal places).

2. The probability that calls last more than 5.6 minutes

We already know from previous answer that the cumulative probability for a raw score = 5.6 minutes (that corresponds to a z-score = 1.60) is P(z<1.60) = 0.94520.

Thus, the probability for call that last more than 5.6 minutes is:

P(x>5.6) = P(z>1.60) = 1 - P(z<1.60) = 1 - 0.94520 = 0.0548 (four decimal places).

3. The probability that calls last between 5.6 and 6.5 minutes

We know the probability for 5.6 minutes from former answers (P(z<1.60) = 0.94520. We still need to consult the probability for 6.5 minutes, following the same procedure:

$\\ z_{6.5} = \frac{6.5 - 4.8}{0.50}$

$\\ z_{6.5} = 3.40$

This corresponds with a cumulative probability of P(z<3.40) = 0.99966.

So, the probability that calls last between 5.6 and 6.5 minutes is:

P(5.6<x<6.5) = P(z<3.40) - P(z<1.60) = 0.99966 - 0.94520 = 0.05446 or 0.0545 (rounding to four decimal places).

4. The probability that calls last between 4.5 and 6.5 minutes

We already know that probability for P(x<6.5) = P(z<3.40) = 0.99966.

For the probability of P(x<4.5), we can see that the value is slightly below the population mean (x = 4.8 minutes). The z-score is:

$\\ z_{4.5} = \frac{4.5 - 4.8}{0.50} = -0.60$

Since we have a negative value and the cumulative standard normal table only permits us consult positive values, that is, z = 0.60, we can find the probability for this z-score, and then subtract it from 1 to find the corresponding value for P(z<-0.60). Then

P(z<0.60) = 0.72575; P(z<-0.60) = 1 - P(z<0.60) = 1 - 0.72575 = 0.27425.

The probability that calls last between 4.5 and 6.5 minutes is:

P(4.5<x<6.5) = P(-0.60<z<3.40) = 0.99966 - 0.27425 = 0.72541 or 0.7254.

5. The length of the longest (in duration) 4 percent of the calls. What is this time?

In the cumulative standard normal table, we can consult the 1 - 0.04 = 0.96 probability that corresponds to a z-score. The value for z is approximately z = 1.75. Then, using the formula for z-scores:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ 1.75 = \frac{x - 4.8}{0.50}$

$\\ 1.75*0.50 = x - 4.8$

$\\ 1.75*0.50 + 4.8 = x$

$\\ x = 5.675 \approx 5.68$

Then, the length of the longest (in duration) 4 percent of the calls is about 5.68 minutes or more.

The corresponding graphs can be seen below.