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Statistics suggest that software developer with 2 years of experience in a town earn an average of 70,000 per year, with a standard deviation of 5,000. To verify this salary level, a random sample of 100 software developers with 2 years of experience was selected from a personnel database for all software developers in the town.

a) Describe the sampling distribution of the sample mean of the average salary of these 100 software developers.

b) Calculate the probability that the sample mean is greater than or equal to 71,200.

c) If the random sample actually produced a sample mean of 71,200, would you consider this rather unusual? What conclusion might you draw then?

My work:

a) the sampling distribution of the sample mean follows a normal distribution since n>30.

b) z = (71,200 - μ)/[σ / sqrt(n)] = 2.4, I then used this value for z in order to find the probability: P(x ≥ 71,200) = P(z ≥ 2.4) = 0.0082

c) I would consider this unusual since the probability that the sample mean is greater than or equal to 71,200 is 0.82%. I'm not too sure about the "What conclusion should you draw then" part..

I'm not sure if I am solving this correctly. Any help is greatly appreciated!!

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In part b) you should divide $\sigma$ by $\sqrt n$, dont subtract it from $\sigma$.so, you need to recompute z-statistics and your conclusion on part c will depend on the new z-statistics.

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  • $\begingroup$ That typo went completely unnoticed to me! Thank you for pointing that out! $\endgroup$
    – Astag
    Nov 16, 2015 at 19:34

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