A television cable company receives numerous phone calls...

Name: a television cable company receives numerous phone calls throughout the day from customers reporting service troubles and from would be subscribers to the cable network most of these callers 64387
Uploaded: 2021-10-16T17:50:38-08:00
Duration: 8 min 24 s
Channel: Evelyn Cunningham
Description: a television cable company receives numerous phone calls throughout the day from customers reporting service troubles and from would be subscribers to the cable network most of these callers 64387

A television cable company receives numerous phone calls throughout the day from customers reporting service troubles and from would-be subscribers to the cable network. Most of these callers are put "on hold" until a company operator is free to help them. The company has determined that the length of time a caller is on hold is normally distributed with a mean of 3.3 minutes and a standard deviation 0.7 minutes. Company experts have decided that if as many as 7% of the callers are put on hold more operators should be hired. 7% of the callers are put on hold for less than x minutes. Round to 1 decimal place

Transcript

00:01 So this problem is about a cable company and wait time for customers.

00:05 And it does say that the wait time for customers is normally distributed.

00:10 So we can draw our bell shape curve.

00:12 And the mean was 3 .1 minutes.

00:16 And the standard deviation was 0 .9 minutes.

00:22 And in part a, we are trying to determine the probability or what proportion of the company's callers are put on hold for more than 4 .8 minutes.

00:36 So 4 .8 minutes would be closer to the right tail and we are trying to determine the area into that right tail.

00:47 And in order to do so, we will need to calculate a z score and turn the information into standard scores.

00:55 So to find a z score we would use x minus mu divided by sigma so in this instance we would be using 4 .8 minus the average of 3 .1 divided by the standard deviation of 0 .9 and when you do so you will get an approximate z score of 1 .89 so at that point you would look in the standard normal table in the back of most textbooks, and down the left -hand side of the chart, you would be locating the units digit and the tenths digit of our z scores.

01:38 We would locate the 1 .8.

01:41 And then across the top, you would locate the 100th's digit, so the 9 is in the 100th place.

01:48 And when you line up that row and that column, you are going to get a value out of your standard normal table of 0 .9706.

02:00 Now, what that references is the area of our curve that extends into the left tail.

02:08 So when we are trying to determine the proportion or probability that x is greater than 4 .8.

02:14 It is comparable to saying what's the probability that our z score was greater than 1 .89.

02:22 And since we know that the entire bell curve, the area into the left tail plus the area into the right tail has to add up to one, then if we want the area that is into the right tail, we would take the entire area minus the area of extending into the left tail.

02:41 So we would take 1 minus the probability that z is less than 1 .89 or 1 minus 0 .9706.

02:52 So therefore, the probability that a customer is going to wait more than or be put on hold more than 4 .8 minutes would turn out to be 0 .094, which you could then say 2 .94 % of the time.

03:12 Now the question then is, should this company hire more operators? there would be no need to hire more operators at this time because the percentage of customers on hold for more than 4 .8 minutes is less than 5%.

04:14 Because that was the threshold they decided, the company experts decided that if as many as 5 % were put on hold for 4 .8, then they would hire more.

04:25 So now let's take a look at part b.

04:29 So we are going to start with our same shape because the customer hold times are normally distributed...

Question

Please give Ace some feedback