According to payscale survey of mba graduates those with one...

According to a PayScale survey of MBA graduates, those with one to four years of experience had an average salary of $60,550 in 2017. Treat the result of this survey of 4,000 MBA graduates as population data. Assume the standard deviation is σ = $18,250. What is the probability that a random sample of n = 80 graduates will provide a sample mean within $3,000 of the population mean? What is the middle interval that captures 95% of the mean salaries from samples of size 80 graduates? Keep the error probability at α = 0.05. We want to build an interval that captures 95% of the sample means within ±$2,000 from the population mean. What is the minimum sample size that would yield such an interval?

Transcript

00:01 Hi, i'm david and i'm hitchiem you answering your question.

00:03 Now let me bring up your question here.

00:06 In the question here, we are going to discuss about the sample mean and let me remind you that by the center limit theorem, if when the sample size n -quiet equal to the 30, and then the same will be approximately to the normal.

00:24 In such a way that whether the x -pile will minus the mean of a standard division the rainbow square of n, we have 10.

00:30 Understand that no more.

00:33 And for the confidence interval there will be formula, will happen to find the samples as needed, the end will be at least, the z unfar over 2, then with the sigma divided by merging of the error totally square.

00:49 Now in this question here, we are given the mean it will equal to 60 ,550, and sigma equal to 18 ,000, and sigma equal to 250 and when the n equal to the 80 so it's greater than 30 already now the question asked me to find the probability that the sample mean will be within the 3000 the population mean so that will be the number one so when to find the absolute the x bar minus the mean will be within the 3 ,000 now to find this probability i need to divide bauxite by this quantity here will be the sigma over square root to the end, will be the 18 ,250 over square root of e .t.

01:42 And if we compute it, you see the quantity on the right here.

01:46 It will be the z.

01:48 There will have the absolute of the z, smaller equal to.

01:52 Now to compute this one we will have, it will be 3 ,000, dividing by 18 ,000 ,000 ,000 times with the square root of the 80 then we get equal to the 1 .47 and this probability will equal to the probability of the z will be between the minus 1 .47 and the 1 .47 and to find this probability i need to bring up the z table let me copy the z table and i will put the table on the right here let me make a pick notice that the table i bring here, it is the negative g score with the probability in the left tail.

03:09 So let me explain this probability in terms of the area under the standard normal curve.

03:15 So the probability equals to the total probability will be 1, we will minus the two tails.

03:23 So it will try to draw it on the graph here.

03:26 It will be 0 in the center...

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