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.
<h3>Normal Probability Distribution</h3>
In a <em>normal distribution</em> with mean and standard deviation , the z-score of a measure X is given by:
- 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 .
- The standard deviation is of 10, hence .
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 <u>p-value of Z when X = 570</u>, hence:
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 <u>p-value of Z when X = 600 subtracted by the p-value of Z when X = 561</u>, hence:
X = 600:
has a p-value of 0.9772.
X = 561:
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