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Mean and standard deviation probability help

  1. Nov 23, 2006 #1

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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.
     
  2. jcsd
  3. Nov 23, 2006 #2

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    oh, i just found another equation for mean and standard deviation of a sample proportion is mu=p and sigma=sqrt(p(1-p)/n)
    which one should i use for this question?
     
  4. Nov 27, 2006 #3
    One big thing:

    The z-table in your book gives you P(z<Z) for a standardized normal random variable Z, not P(z>Z). So, P(z>1.8) = 1 - P(z<1.8) = 1 - phi(1.8). You may have already done that in your calculations but you did not specify that if you did.

    The formula I have for sigma is: sqrt(npq) where q=(1-p).
     
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