Problem 6 35 pts draw the bell curve for each question...

Name: problem 6 35 pts draw the bell curve for each question normal approximation to the binomial distribution a small company uses a parcel service to ship packages of special chocolates ordered 02245
Uploaded: 2022-07-28T11:45:03-08:00
Duration: 7 min 28 s
Channel: Frank Deng
Description: problem 6 35 pts draw the bell curve for each question normal approximation to the binomial distribution a small company uses a parcel service to ship packages of special chocolates ordered 02245

Problem # 6. (35 pts) (Draw the bell curve for each question). (Normal Approximation to the Binomial distribution). A small company uses a parcel service to ship packages of special chocolates ordered as gifts. The company has found that 65% of all orders are delivered on time. A batch of 200 packages is sent out. Let X be the number of packages delivered on time. a) Do the assumptions of the normal approximation to the binomial distribution seem reasonable in this situation? b) Assuming that the assumptions hold, find the following probabilities: (i) That at most 135 packages are delivered on time. (ii) That more than 139 packages are delivered on time. (iii) That exactly 134 packages are delivered on time. (iv) That at least 139 packages are delivered on time.

Transcript

00:01 Now, here in this question, we look at four later questions about normal distribution, basically.

00:06 The first question, you have an airline which estimates that 9 % of their, 9 % of people, 9 % of the people booked their flights were actually show up, right? so let me call this p equals 9%.

00:21 So the books are 74 people of a flight.

00:24 So asked what is the probability that actually more than 70 people is scale.

00:30 Actually to show up so that the capacity plan actually exceeded.

00:34 Well, so let's look at the probability, right? so i would denote by x the number of people that will actually show up out of the 74, right? so actually the x is going to be, it's going to be approximately normal distributed because the number of people booked here is 74, which is much less than 30, right? so you can approximate the binominal distribution with the normal distribution.

00:59 And the mean of this normal disease is given by 74 times p.

01:04 While the standard deviation is given by square root of 74 times p and times 1 minus p.

01:10 So that is actually given by n.

01:14 So 74 times 9%, that's actually given by 74 times 0 .9, gives you 66 .6 .6 .6.

01:26 266 .6 .6.

01:28 And then the standard deviation, which is given by the square root of 74 times 0 .9 times 0 .1, and that you find this to be actually 2 .5 .8.

01:42 2 .5 .8.

01:44 Now you ask the probability px actually larger than actually larger than 71, right? so let's look at the distribution of x rate.

01:54 So the distribution of x, as we know, is a proximate norma disputed, and it has a mean which is 66 .6, right? now you look at the probability, which is 71 and the larger than 71.

02:06 So basically look at the area, right, of this pair, right, to the right, to the right to the right.

02:12 And the probability they were simply given by alpha, and alpha is the area of the pair, right? and that area can be found from the score.

02:20 So the d score at arpa, it's given by actually 71 minus 66 .6 over the standard division, which is 2 .58.

02:26 And you do it, you find this to pay 1 .7, okay? and you look up the z table and you find this alpha to pay 4 .5%...

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