SOLUTION: Please help...i am taking an online stats class and am so confused on this problem..I have been working on it for so long Im getting a headache...thank you very much
Assume that
Algebra ->
Probability-and-statistics
-> SOLUTION: Please help...i am taking an online stats class and am so confused on this problem..I have been working on it for so long Im getting a headache...thank you very much
Assume that
Log On
Question 490443: Please help...i am taking an online stats class and am so confused on this problem..I have been working on it for so long Im getting a headache...thank you very much
Assume that the population of heights of female college students is approximately normally distributed with mean µ of 64.64 inches and standard deviation ð of 6.02 inches. A random sample of 98 heights is obtained. Show all work.
(A) Find the mean and standard error of the x distribution
(B) Find p(x>63.75)
You can put this solution on YOUR website! Assume that the population of heights of female college students is approximately normally distributed with mean µ of 64.64 inches and standard deviation ð of 6.02 inches. A random sample of 98 heights is obtained. Show all work.
(A) Find the mean and standard error of the x-bar distribution.
mean of the sample means = 64.64
-----------------
(B) Find p(x>63.75
If you really mean x and not x-bar you get:
z(63.75) = (63.75-64.64)/6.02 = -0.1478
P(x > 63.75) = P(z > -0.1478) = 0.5587
--------------
But, if you mean P(x-bar > 63.75) you get:
t(63.75) = (63.75-64.64)/[6.02/sqrt(98)] = -1.4635
---
P(xbar > 63.75) = P(t > -1.4635 when df = 97) = 0.9267
============================
Cheers,
Stan H.