BrandiMR

02-25-2010, 01:38 PM

I'm doing a sample test for a Stats quiz I have on Tuesday and I've run into a question that I just cannot get. :confused: I'm sure the method to use is right under my nose, but its driving me crazzzzy. :eek:

HDL cholesterol levels of males aged 20-29 are Normally distributed with a mean of 5.5 and a standard deviation of 1.2 For the nine male employees of a certian company, what is the probability that exactly five of them have HDL cholesterol levels exceeding 5.5?

Since it uses "exactly 5", I went to the binomial distribution and tried to use formulas I know to fill in the various parts of the binomial.

I went like this:

Mu = np

5.5 = 9p

p = 0.611

q would then be:

q = 1-p

q = 1-0.611

q = 0.389

right?

So then:

P[X=5] = (9C5) (0.611)^5 (0.389)^9-5

= 0.2457?

The answer choices I have are:

(A) 0.1056 (B) 0.3203 (C) 0.7539 (D) 0.0667 (E) 0.4364 (F) 0.8944 (G) 0.5636 (H) 0.2461 (I) 0.9333 (J) 0.6797

...and none of them are what I came up with. (H) 0.2461 is close, but I tried again without rounding anything and its still the same answer. =/ One thing I'm worried about is that I didn't use the standard deviation at all. It's mentioned right in the question, so it should be used in the calculation somewhere, shouldn't it? D:

Am I even close to the right direction? :( Any help would be appreciated.

HDL cholesterol levels of males aged 20-29 are Normally distributed with a mean of 5.5 and a standard deviation of 1.2 For the nine male employees of a certian company, what is the probability that exactly five of them have HDL cholesterol levels exceeding 5.5?

Since it uses "exactly 5", I went to the binomial distribution and tried to use formulas I know to fill in the various parts of the binomial.

I went like this:

Mu = np

5.5 = 9p

p = 0.611

q would then be:

q = 1-p

q = 1-0.611

q = 0.389

right?

So then:

P[X=5] = (9C5) (0.611)^5 (0.389)^9-5

= 0.2457?

The answer choices I have are:

(A) 0.1056 (B) 0.3203 (C) 0.7539 (D) 0.0667 (E) 0.4364 (F) 0.8944 (G) 0.5636 (H) 0.2461 (I) 0.9333 (J) 0.6797

...and none of them are what I came up with. (H) 0.2461 is close, but I tried again without rounding anything and its still the same answer. =/ One thing I'm worried about is that I didn't use the standard deviation at all. It's mentioned right in the question, so it should be used in the calculation somewhere, shouldn't it? D:

Am I even close to the right direction? :( Any help would be appreciated.