Fe mechanical exam 01 question 8 of 100 hide images mark...

FE Mechanical Exam 01 QUESTION 8 OF 100 Hide Images Mark Translate En Suppose the lengths of telephone calls form a normal distribution with a mean length of 8.0 min and standard deviation of 2.5 min. The probability that a telephone call selected at random will last more than 15.5 min is most nearly: A) 0.0013 B) 0.0026 C) 0.26 D) 0.9987 Whiteboard Calculator Pause End Remaining Time: 05:06:13 << All Questions

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00:01 For this problem, in question 10, we are looking for the probability that a customer's waiting time is less than 3 minutes.

00:09 Probability that x is less than 3 we can find by taking the proportion of the standard normal distribution to the left of the z -score corresponding to 3.

00:16 We find that by taking 3 minus mu, the population mean, divided by sigma, the population standard deviation.

00:22 So that will be 3 minus 4 .6 divided by 2 .1.

00:26 So we can see that the z -score there is negative 0 .76.

00:31 Proportion to the left of negative 0 .76 is 0 .2236.

00:42 So we can see that the value that that is closest to is 0 .224, option c.

00:49 Then for question 11, we're looking for the minimum amount of time that would be considered as a long wait.

00:57 So that means that we want to first find the z -score that has a right tail proportion of 10%, 0 .1, or a left tail proportion, a cumulative proportion of 0 .9.

01:08 I'll find that using my table of values over here.

01:11 It might be a little bit small on the recording, but we have cumulative probability or one tail proportion, 0 .1, or pardon me, cumulative probability 0 .9, one tail proportion 0 .1, and i have the z -score down here.

01:25 So we can see that the z -score is 1 .282.

01:29 So we'll have that the minimum value to be considered a long wait is 1 .282 standard deviations above the mean.

01:38 So we find that by taking, or by writing x equals mu plus z times sigma.

01:45 So that is going to be 4 .6, the mean, plus 1 .282 times 2 .1, the standard deviation, giving us a result of 7 .29.

01:54 You can see that that is closest to option a, 7 .3 minutes.

02:00 For question 12, let's see here.

02:05 We're looking for probability that three or four out of four have a wait of more than 4 .6 minutes.

02:10 So i'll note that since the mean value is equal to 4 .6, the probability that an individual has a wait more than 4 .6 minutes is just equal to 0 .5, because the mean of the normal distribution is also the median.

02:31 So to find the probability that there are three or four out of four with x greater than 4 .6, we'd have four choose three times the probability of having three out of four with a wait time more than 4 .6 minutes, which would be, well, actually in this case, since the probability of wait time more than 4 .6 is the same as the probability of wait time less than 4 .6, we'd have that, we're just multiplying by 0 .5 to the power of four...

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