Waiters at a restaurant chain earn an average of 242 per...

Waiters at a restaurant chain earn an average of $242 per shift (regular pay + tips) with a standard deviation of $34. For random samples of 40 shifts at the restaurant, within what range of dollar values will their sample mean earnings fall, with 80% probability? Standard Normal Distribution Table Range: to Round to the nearest cent

Transcript

00:01 There is given the average value.

00:02 So the average is the mean value here.

00:05 So the mean denoted by moot, this is 242 and the standard division, which is denoted by sigma.

00:13 That was given as $34.

00:16 So i can define the random variable x here.

00:19 So for random sample of 40, so we are going to use the standard normal distribution here.

00:24 So that means there is a normal distribution here.

00:26 So i can define the random variable x, which is normally distributed.

00:29 So the mean is 242 and the standard division, which is 34.

00:34 And the random sample size was given here, which is 40.

00:37 So what we need, we need the sample standard deviation, the sample standard deviation with central limit theorem, the population standard division divided by squared by the sample size.

00:49 So let me just use this notation here, which is 34 divided by square root of 40.

00:56 And i can define the random variable x bar for the sample.

01:00 Which is normal of the survey due.

01:01 So the mean is the same with the population, which is 242, and the standard division, which is 34 divided by square root of 40.

01:10 So what we have to do, fall with 80 % probability.

01:14 So that means if i just graph the normal distribution here.

01:19 So there should be some lower boundary and the upper boundary.

01:23 And there is the mean value here.

01:24 So the mean is equal to 242 and x1 and x2 both symmetrical with respect to mean value.

01:30 So this yellow shaded region and the green shaded region, they are equal size to each other...

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