Question

(b) The table below indicates the cost of certain air tel kiosk Z 0.15 0.20 0.25 0.75 0.80 0.85 1.85 1.90 1.95 2.00 2.05 QZ 0.0596 0.0793 0.0987 0.2734 0.2881 0.3023 0.4678 0.4713 0.4744 0.4772 0.4798 If it was stated that the number of telephone calls received at a switchboard is approximately normal with a mean of 580 calls and a standard deviation of 10. REQUIRED ind the possibility that on a given day the number of calls received at the switchboard will be; (a) Exactly 578 (6) Less than 57 (c) Between 561 and 600 inclusive​

Answer 1

Using the normal distribution, it is found that there is a

a) 0% probability that the number of calls received at the switchboard will be exactly 578.

b) 0.1586 = 15.86% probability that the number of calls received at the switchboard will be less than 570.

c) 0.9485 = 94.85% probability that the number of calls received at the switchboard will be between 561 and 600.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• The mean is of 580, hence $\mu = 580$.

• The standard deviation is of 10, hence $\sigma = 10$.

Item a:

In the normal distribution, the probability of an exact value is 0%, hence, 0% probability that the number of calls received at the switchboard will be exactly 578.

Item b:

The probability is the p-value of Z when X = 570, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{570 - 580}{10}$

$Z = -1$

$Z = -1$ has a p-value of 0.1586.

0.1586 = 15.86% probability that the number of calls received at the switchboard will be less than 570.

Item c:

This probability is the p-value of Z when X = 600 subtracted by the p-value of Z when X = 561, hence:

X = 600:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{570 - 580}{10}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

X = 561:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{561 - 580}{10}$

$Z = -1.9$

$Z = -1.9$ has a p-value of 0.0287.

0.9772 - 0.0287 = 0.9485.

0.9485 = 94.85% probability that the number of calls received at the switchboard will be between 561 and 600.

You can learn more about the normal distribution at brainly.com/question/24663213