The mean and standard deviation of the amount customers...

The mean and standard deviation of the amount customers spend in a high end clothing store is $600 and $150 respectively. Assume the amount of money customers spend is normally distributed. (a) What is the probability that 5 randomly selected customers spend over $3500 combined? (b) If we randomly sample 49 people, what is the probability they spend less than $500 on average? (c) What is the 90th percentile of total amount spent for 10 randomly selected people? (d) Did we need to assume the population was normally distributed in parts (a) and (b)? Explain

Transcript

00:01 Once again, welcome to a new problem.

00:04 This time we're dealing with sampling distribution.

00:09 So we're looking at sampling distribution of sample proportions.

00:21 So we're looking at sampling distributions of sample proportions of sample means, not sample proportions, sample means.

00:32 So for a standard.

00:34 For standard normal distribution we have a standard normal distribution what's going to happen is that the mean of sampling distribution of sample means is the population mean and this is based on the central limit theorem and the standard deviation of the sampling distribution of sample means is the standard error this is the standard error you're looking at.

01:11 And within one standard deviation of the mean, we have 68 % of the data, and then within two standard deviations of the mean, we have 95 % of the data.

01:32 And then, of course, within three standard deviations of the mean, we have 99 .7%.

01:40 Of the data.

01:42 We're looking at a new problem where we're given the mean and the standard deviation.

01:48 The sample mean or the population mean is 600 and the population standard deviation is 150.

01:56 This would be equivalent to the main of the sampling distribution of sample means.

02:06 And then the distribution for customer expenditure happens to be normal.

02:16 600 is the main expenditure and 150 is the standard deviation.

02:22 In part a, we want to determine the probability that a sample of five randomly selected customers spend a combined value of more than 3 ,500.

02:40 So five customers are spending more than 3 ,500 together means that on average they spend.

03:13 So we want to get the average for these customers.

03:16 So x bar equals to 3 ,500 over 5, which is 700.

03:23 So on average, they're spending more than, so it says more than 700.

03:36 So we want to determine that probability.

03:39 We know what the mean is.

03:42 It's the same as if we go back, look up 600.

03:48 We also have to get the standard error, which is the way to.

03:52 Standard deviation.

03:55 So this is 150 divided by radical 5 because that's a sample and then at 700 x bar is 700 so we want to find this probability when it's more than 700.

04:13 So in that sense we're saying probability that x bar is greater than 700.

04:18 This is the same as probability that x bar minus mu of x over signal.

04:24 Sub x is greater than 700 minus 600 over 150 square root 150 square root of 5 and this simply means that the probability of the z score is greater than so we want to do the map for that 700 minus 600 divide by square root or rather divide by 150 radical 5...

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