SOLUTION: What is the formula to find the following; The probability that the mean of a sample of 50 babies' weights lies between 3330 and 3510, with a standard deviation of 493.5 (answer

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Question 440553: What is the formula to find the following; The probability that the mean of a sample of 50 babies' weights lies between 3330 and 3510, with a standard deviation of 493.5 (answer given is .803, how do I get that????) Thanks!

Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
What is the formula to find the following; The probability that the mean of a sample of 50 babies' weights lies between 3330 and 3510, with a standard deviation of 493.5 (answer given is .803, how do I get that????) Thanks!


We are given a confidence interval and are asked to find 
the level of confidence, which is the probability that the
true mean lies between the two given endpoints of the 
confidence interval, 3330 and 3510.

A confidence interval is given by the formula

(xbar-z%5Balpha%5D%2Asigma%2Fsqrt%28n%29, xbar%2Bz%5Balpha%5D%2Asigma%2Fsqrt%28n%29)

We know that xbar has to be the average of those endpoints 
of the confidence interval, so we average 3330 and 3510 
and get xbar = 3420

(3420-z%5Balpha%5D%2A493.5%2Fsqrt%2850%29, 3420%2Bz%5Balpha%5D%2A493.5%2Fsqrt%2850%29)

we are given that the confidence interval is

(3330, 3510)

So we can set 

3420-z%5Balpha%5D%2A493.5%2Fsqrt%2850%29=3330 and 3420%2Bz%5Balpha%5D%2A493.5%2Fsqrt%2850%29=3330 

and solve either one for z%5Balpha%5D. We only need to do
one because we'll get the same answer, since the sample mean
is the midpoint of the given confidence inveral.

3420-z%5Balpha%5D%2A493.5%2Fsqrt%2850%29=3330

-z%5Balpha%5D%2A493.5%2Fsqrt%2850%29=-90

z%5Balpha%5D%2A493.5%2Fsqrt%2850%29=90

z%5Balpha%5D%2A69.7914393=90

z%5Balpha%5D=90%2F69.7914393

z%5Balpha%5D=90%2F69.7914393

z%5Balpha%5D=1.28955644

We round that z-score to 1.29.

Next we look up that z-score in a normal table
and read .4015 area between the lower bound of
the confidence interval 3330 and the sample mean
of 3420.

There is also .4015 area between the sample mean
of 3420 and the upper bound of the given confidence
interval of 3510.

So the answer is .4015 + .4015 = .803

Edwin