1. ## 97th percentile

"suppose the moisture content per pound of a dehydrated protein concentrate is normally distributed with a mean of 3.5 and standard deviation of 0.5. A random sample of 16 specimens each consisting I a one pound concentrate is to be tested. Letting x bar denote the sample mean of these measurements of moisture content:
....find the 97th percentile of x bar.

I am a little confused...do I look up 0.9700 in the body of the table an then use x = u + (z•standard deviation)?
So it would be x = 3.5 + (1.88•0.5)?

I got the others correct I believe..
One stated find the probability that x bar will exceed 3.7
I did 3.7-3.5/0.125=1.6 checked the table and got 0.9452 then 1-0.9452=0.0548

But this percentile one is confusing me...

Any help?

2. ## Re: 97th percentile

That isn't quite right since you're taking a sample of 16. Is the standard deviation of the mean for a sample of size 16 the same as the standard deviation of the population? (Hint: The answer is no).

3. ## Re: 97th percentile

But it gives me the standard deviation of 0.5 right in the question. I don't need to determine it....

4. ## Re: 97th percentile

Oh wait...

Should I be doing 3.5 + (1.88•0.125)?

5. ## Re: 97th percentile

That looks much better to me.

6. ## The Following User Says Thank You to Dason For This Useful Post:

kelly g (03-26-2012)

7. ## Re: 97th percentile

WooHoo!! Thanks for the guidance!

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