percentage of sample mean

K

kahkah

New member
Joined
Jan 18, 2012
Messages
14
A random sample of size 81 is drawn from a population with a standard deviationof 12. If only 18% of the time a sample mean greater than 300 is obtained, whatis the mean of the population?


I have not done this as a percentage yet, so am not sure how to approach this, can someone please help me?
 
Yes, you have done this, probably many times. You just have a different piece of information missing.

Z = ((Sample Mean)-(Population Mean))/(Standard Deviation of the Sample Mean)

Fill in the blanks with the given information an calcualte whatever is left.
 
answer

kahkah said:
A random sample of size 81 is drawn from a population with a standard deviationof 12. If only 18% of the time a sample mean greater than 300 is obtained, whatis the mean of the population?

ANSWER: _
n = 81, s = 12, Prob(X > 300) = .18
.5000 – .1800 = .3200

From Z Table, z.3200 = 0.92
Solving for µ:

Z formula = ((sample mean - population mean))÷(standar deviation÷square root of n or 81) (√81=9)

(0.92x12/9) = 1.2267

1.2267 = 300 – m
µ = 300 – 1.2267 = 298.77
Click to expand...
 
You must log in or register to reply here.
Top