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A department store has determined that 25% of all their sales are credit sales. A random sample of 75 sales is selected and the proportion of credit sales in the sample is computed.
a) What is the probability that the sample proportion will be greater than 0.34?
my answer is:
n=75 p=0.25
mu=np=18.75
sigma=sqrt(n*p*(1-p)) = 3/75
0.34 * 75=25.5
z=x-mu/sigma = 25.5-18.75/3.75 =1.8
P(z>1.8)=0.0359
the answer i got is 0.0359 but i am not confident, please tell me if i am correct or not. Thanks.
0.34 of 75 is 25.5, so we're looking for the probability that 26 or more of the 75 samples are credit sales (if I understand the question).
I would calculate this as
,
where n=75 and p=0.25.
The last time I did a calculation like this, I used Matlab. I saved the expression I used that time, so I could check your result just by substituting my numbers for yours. This is the result I got:
» p=0.25; k=26; n=75; s=0; for x=k:n; t=p.^x.*(1-p).^(n-x).*factorial(n)/(factorial(x)*factorial(n-x)); s=s+t; end; s
s =
0.0393
So I would say that your approximate answer is pretty close to the exact one.