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Help with a math question please math 101 college math

A population of all college alumni had a mean of starting salary on their first job of $30,000 with a standard deviation of $1,800. A sample of 36 alumni is taken. What is the probability the mean of their salary will be more than $31,000?

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nerdynerd · 61-69, M

If the sample is a simple random sample, then the sample mean, xbar, has a distribution that is approximately normal with mean 30000 and standard deviation 1800/sqrt(36) = 300.
So the standard normal variable is z = (xbar-30000)/300.

Thus, P(xbar > 31000) = P(z > (31000-30000)/300 ) = P( z > 3.33).

From a normal table, or from software, we find that this probability is about .0004.

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Candypie2010 · 26-30, F

I got 10.35

Candypie2010 · 26-30, F

nerdynerd · 61-69, M

@Candypie2010 You found the probability that a single alum has a salary that exceeds 31000. The problem asked for the probability that the mean salary of a sample of 36 alums exceeds 31000. The means of samples of size 36 have less variability from sample to sample than salaries have from individual to individual, which is why it is is less probably for a sample mean to exceed 31000 than for an individual salary to exceed 31000.

What you did at the end is not valid. The event "mean exceeds 31000" is not found by multiplying by 36. Your answer was 10.35, which cannot be a probability, since it is greater than one. (Or, if you like, 36 times 28.77% is more than 100%)

You have to use the central limit theorem for this.

MarmeeMarch · M

Are they going on welfare ? cause that will disrupt the numbers.

xixgun · M

Do your own homework.

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