In opposing the establishment of a new bank in a community anexisting bank state

Question

In opposing the establishment of a new bank in a community anexisting bank stated that the area to be served by the proposedbank contained 75 percent of its accounts. An economic consulting firm randomly sampled the households to determine the sample proportion, p, which had accounts at the existing bank. The sample size was 225.

Find the probability that p is: .

A) Less than 67 percent. =

B) More than 76 percent. =

C) Within 0.05 of the true proportion stated as 0.75.

Please show your calculations!

Explanation / Answer

mean=0.75

Standard Dev=sqrt(P*Q/n)=0.0289

Normal Distribution =Z=X-mean/sd~N(0.1)

P(X<0.69)=0.69-0.75/0.0289=-2.7681

=P(Z<-2.7681)=0.0189

B. P(Z>0.76)=0.76-0.75/0.0289

=0.346

P(Z>0.346)=0.3647

C.P(bwn 0.05 and 0.75)=0.5

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