SOLUTION: I just need someone to review my work. Question: If the cholesterol level of men in the community is normally distributed with a mean of 200 and a standard deviation of 25, what

Algebra ->  Probability-and-statistics -> SOLUTION: I just need someone to review my work. Question: If the cholesterol level of men in the community is normally distributed with a mean of 200 and a standard deviation of 25, what       Log On


   

Question 655116: I just need someone to review my work.
Question:
If the cholesterol level of men in the community is normally distributed with a mean of 200 and a standard deviation of 25, what is the probability that a randomly selected sample of of 49 men will have a mean between 190 and 205? (hint: this question pertains to the sampling distribution of the mean).
(190-200)/(25/√49)= -10/3.57= -2.80
(205-200)/(25//√49)= 5/3.57= 1.4
P(X-Bar > 190) = 0.50 –.4974= .0026
0.26% of sample means (from repeated samples of n=49 men) will have cholesterol mean value greater than 190.

P(X-Bar <205) = 0.50 + .4192 = .9192
91.92% of sample means (from repeated samples of n=49) will have a cholesterol level less than 205.


Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

RE TY, Yes, NORMSDIST(-2.8) is .00256  and NORMSDIST(1.4) is .91924

 .91924

-.00256  is  .91668  Or = .9167   Or 91.67%

(Ans Correct, as listed, rounded to 4 decimal points, calculation  Using 5)

As to: .9192 - .0026 (Using 4 decimal points for calculation...)= 91.66% 



t = (190-200)/(25/√49)= -10/3.57= -2.80 Yes

t =  (205-200)/(25//√49)= 5/3.57= 1.4   Yes

P(180< x    <205) = P( -2.80  ≤ t  ≤ 1.4) = .91924 - .00256 = .9167