Question

The total energy need during pregnancy is normally distributed, with mean Mu = 2600 kcal/day and standard deviation = 50 kcal/day. What is the probability that a random sample of 20 pregnant women has a mean energy need of more than 2625 kcal/day?

Answer 1

Hi, this is a bit more complicated than the last one we did. If a sample is drawn from a population, then the standard deviation of the sample mean is related to the population mean through the formula stddev/sqrt(n), where stddev is the standard deviation of the population, and n is the number of samples

Answer 2

Yikes, that does sound more complicated.

Answer 3

Okay... so 20 = n, and standard dev = 50, so 50/sqrt20=11.18...

Answer 4

right. so now you just do the same thing as the last problem we did, except using 11.18 as the std dev.

Answer 5

Oh, alright. Gimme a bit.

Answer 6

Okay, so I should do 50/11.18=4.47 and use that as the Z-score?

I apologize for my ineptitude.

Answer 7

..wait. I can't even see that on the table.

Answer 8

You got a little mixed up. The std dev of the sample mean is (orig std dev)/sqrt(n) which is 11.18.

Now you divide the level above the mean, which is 25, by the std dev, which is 11.18, to get the z-score.

Answer 9

dividing 50/11.18 is dividing a standard deviation by another standard deviation, that's not what you want.

Answer 10

Okay, that's 2.23, and that is .9871.

Answer 11

sounds about right

Answer 12

Alrighty. Thanks again.

Answer 13

i just computed it and yeah, I got 0.987326. But remember that's the prob that it'll be *below* that level

Answer 14

take the complement if you want to know the prob it'll be above that level.

Answer 15

Okay.