3 a survey indicates that for each trip to a supermarket a...

3-)A survey indicates that for each trip to a supermarket, a shopper spends an average of 43 minutes with a standard deviation of 12 minutes in the store. The lengths of time spent in the store are normally distributed and are represented by the variable X. A shopper enters the store. (a) Find the probability that the shopper will be in the store for each interval of time listed below. (b) When 200 shoppers enter the store, how many shoppers would you expect to be in the store for each interval of time listed below? 4-)What is the probability that the shopper in Example 3 will be in the supermarket between 31 and 58 minutes? When 200 shoppers enter the store, how many shoppers would you expect to be in the store between 31 and 58 minutes?

Transcript

00:01 Hi, i'm david and i'm here to have you answer your question.

00:04 Now let me pick up your question here.

00:07 In this question we discuss about the normal distribution.

00:10 Let me remind you that if we have the x followed by the normal with the mean and the standard deviation, if we think the x minus the mean of a standard deviation, we should get the normal.

00:23 Now in this question here we're given the x followed by the normal when the mean equal to 12.

00:30 At 45 standard division 22 12 now the question i asking us to find the probability that the customer will stay in the star for the time between 24 to 45 minutes so means now we're trying to find the probability x between the 24 and the 54 to find this probability we need to turn the x here into the the z.

01:03 To do that we need to take the 24 minus the mean divided by standard division.

01:09 We need to do the same thing on the right and if we compute it, we get the z must be between 24 minus 45 divided by 12 minus 1 .75, 54 minus 45 divided by 12, 54 minus 45 divided 12, 0 .75.

01:35 This one, it will try to draw the graph here, it will just the area in the middle.

01:42 Where we have this one will be the 0 .75, this will be the minus 1 .75.

01:48 There will be the 0 .75.

01:52 So we want to find the middle area.

01:55 So it will be easier to turn the 1, minus the 1 on the left, will be the proper pituit disease smaller than minus 1 .75, minus the one on the right but because of the symmetry we can put the z smaller than minus 1 0 .75 now to find this probability i will bring up the z table and will show you how to use the z table here let me put the table on the right here for you now we see the first one want to find the z smaller than minus 1 .7 and 5 so the corresponding probability will be the 0401 for the z small and then minus 0 .75 it will be here so zupon 2266 if we do the calculation get equal to the 0 .7333 this will be the answer for the a 1 now for the a 2 we will have one to find the probability of the x will be greater than 39.

03:18 Again we need to turn the x here into the z.

03:22 Then we need to turn the 39 minus the mean of a standard deviation...

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