SOLUTION: Fifty per cent of students attend the schools in our less privileged township. In a sample of 500, what is the probability that the sample proportion will be between 0.45 and 0.55

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Question 862569: Fifty per cent of students attend the schools in our less privileged township. In a sample of 500, what is the probability that the sample proportion will be between 0.45 and 0.55
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
p = .5 = probability student will attend school in a less privileged township.
q = 1 - .5 = .5 = probability student will attend school in a more privileged township

the population proportion is given as .5.

sample proportion range is between .45 and .55

we assign x1 = .45 and x2 = .55.

n = sample size = 500

s = standard error of the distribution of the sample proportion = sqrt(pq/n) = sqrt(.5^2/500) = .02236 rounded to 5 decimal places.

z1 = (x1-p)/s = (.45-.5)/.02236 = -2.24 rounded to 2 decimal places.

z2 = (x2 - p)/s = (.55-.5)/.02236 = 2.24 rounded to 2 decimal places.

the probability that the sample proportion will be between .45 and .55 is equal to the area under the normal distribution curve to the left of z2 minus the area under the normal distribution curve to the left of z1.

those areas are found in the following z-score table:

http://lilt.ilstu.edu/dasacke/eco148/ztable.htm

this turns out to be .9875 - .0125 = .975.

there is a probability of .975 that the sample mean proportion of a sample size of 500 will be between .45 and .55.

you can also works this as a mean rather than a proportion.

the mean of the population is calculated to be p*n = .5*500 = 250.

the standard error of the distribution of the sample means is assumed to be sqrt (n*p*q) which is equal to sqrt(500*.5*.5) which is equal to sqrt(125) which is equal to 11.18034 rounded to 5 decimal places.

the sample mean proportions are between .45 and .55.

this means that the sample means are between 500*.45 and 500*.55 because the sample mean is equal to n * (p of the sample).

you get:

x1 = 500*.45 = 225.

x2 = 500*.55 = 275.

z is the same formula of (x-m)/s, only now we are dealing with different numbers that should give us the same answer.

z1 = (x1 - m)/s = (225 - 250) / 11.18034 = -2.24 rounded to 2 decimal places.

z2 = (x2 - m)/s = (275 - 250) / 11.18034 = 2.24 rounded to 2 decimal places.

since we got the same z's as before, the area under the normal distribution curve calculated before will apply here as well.

the assumed population proportion is used in the calculation of the standard error.

a good reference discussing sampling distribution of p can be found at the following link:

http://onlinestatbook.com/2/sampling_distributions/samp_dist_p.html