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 randomly selected pregnant woman has an energy need of more than 2625 kcal/day?

Answer 1

Hi, do you know how to use a z-score table?

Answer 2

Yes, but I may be slow with it.

Answer 3

OK. The z-score table tells you the probability that a random sample will be so-many standard deviations above or below the mean. So first we have to convert the problem into a "standard" problem. How many standard deviations above the mean is 2625?

Answer 4

Uhh, that'd be 25, right?

Answer 5

well it's 25 kcal above the mean, but how many standard deviations is that?

Answer 6

Oh, .5?

Answer 7

yes. So look that up in your z-score table; what do you find?

Answer 8

..so, positive Z-table, look under .5 and .00?

Answer 9

yeah, different tables have different layouts but 0.5000 is what you want.

Answer 10

I see 0.6915 there.

Answer 11

right, so that's the probability that a sample from a normal distribution will be *below* 0.5 std.devs above the mean. But the question asked the probability that a sample is *above* that level, so how do we convert it?

Answer 12

1-.6915=.3085?

Answer 13

exactly. That's your answer.

Answer 14

Alright, that's a lot more simple than I thought, thank you very much. Although this is unfortunately the first of many final review questions, so I think I'll be here a while.

Thanks again.

Answer 15

no problem.