I was given this question in my probability homework.

Given a normal population whose

mean is 355355

standard deviation is 4242

Find the probability that a random sample of 55 has a mean between 362362 and 372372

According to the formula I am using:

P(Z>362−35542√5)P(Z >\frac{362-355}{\frac{42}{\sqrt{5}}})

P(Z>372−35542√5)P(Z >\frac{372-355}{\frac{42}{\sqrt{5}}})

From these calculations I get:

0.37−0.910.37-0.91

now I looked up these values in my z-table, which yields:

0.3557−0.18140.3557-0.1814

The result is 0.17430.1743

I have this as my answer but the system won’t accept it as a solution. So I don’t know where the mistake is. I had someone else do the same problem nd we both got the same answer.

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Is 0.5371 from the z-table?

– user274065

2 days ago

and yes. Extra help would be great if you are willing to provide

– user274065

2 days ago

Sorry, I made a mistake. Check the answers. Those are much better explained. It seems the system has a mistake. If you would like further explanation, I’ll be glad to help.

– O. Von Seckendorff

2 days ago

I think these answers will do it. Thank you though.

– user274065

2 days ago

Sure. Sorry I might have confused you.

– O. Von Seckendorff

2 days ago

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2 Answers

2

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If observations are drawn from a population with mean μ=355\mu = 355 and standard deviation σ=42\sigma = 42, then a sample of size n=5n = 5 will have a sample mean ˉX\bar X that is normally distributed with mean μ=355\mu = 355 and standard deviation σ/√n=42/√5\sigma/\sqrt{n} = 42/\sqrt{5}. Then the probability that the sample mean is between 362362 and 372372 is Pr[362<ˉX<372]=Pr[362−35542/√5<ˉX−μσ/√n<372−35542/√5]=Pr[√56