Question (1 point) A recent poll has suggested that 63 % of Canadians will be spending money - decorations, halloween treats, etc. - to celebrate Halloween this year. 22 Canadians are randomly chosen, and the number that will be spending money to celebrate Halloween is to be counted. This count is represented by the random variable X. Part (a) Compute the probability that 15 of these Canadians indicate they will be spending money to celebrate Halloween. P(X = 15) = !! (use four decimals in your answer) Part (b) Compute the probability that between 7 and 15 of these Canadians, inclusive, indicate they will be spending money to celebrate Halloween. P(7 < X < 15) = (use four decimals in your answer) Part (c) How many of the 22-Canadians randomly chosen would you expect to indicate they will be spending money to celebrate Halloween? Compute the standard deviation as well. E(X) = uy = (use two decimals in your answer) SD(X) = 0x = || (use two decimals in your answer) Part (d) Compute the probability that the 10-th Canadian random chosen is the 9-th to say they will be spending money to celebrate Halloween. . (use four decimals in your answer)
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Transcribed Image Text: (1 point) A recent poll has suggested that 63 % of Canadians will be spending money - decorations, halloween treats, etc. - to celebrate Halloween this year. 22 Canadians are randomly chosen, and the number that will be spending money to celebrate Halloween is to be counted. This count is represented by the random variable X. Part (a) Compute the probability that 15 of these Canadians indicate they will be spending money to celebrate Halloween. P(X = 15) = !! (use four decimals in your answer) Part (b) Compute the probability that between 7 and 15 of these Canadians, inclusive, indicate they will be spending money to celebrate Halloween. P(7 < X < 15) = (use four decimals in your answer) Part (c) How many of the 22-Canadians randomly chosen would you expect to indicate they will be spending money to celebrate Halloween? Compute the standard deviation as well. E(X) = uy = (use two decimals in your answer) SD(X) = 0x = || (use two decimals in your answer) Part (d) Compute the probability that the 10-th Canadian random chosen is the 9-th to say they will be spending money to celebrate Halloween. . (use four decimals in your answer)
More Transcribed Image Text: (1 point) A recent poll has suggested that 63 % of Canadians will be spending money - decorations, halloween treats, etc. - to celebrate Halloween this year. 22 Canadians are randomly chosen, and the number that will be spending money to celebrate Halloween is to be counted. This count is represented by the random variable X. Part (a) Compute the probability that 15 of these Canadians indicate they will be spending money to celebrate Halloween. P(X = 15) = !! (use four decimals in your answer) Part (b) Compute the probability that between 7 and 15 of these Canadians, inclusive, indicate they will be spending money to celebrate Halloween. P(7 < X < 15) = (use four decimals in your answer) Part (c) How many of the 22-Canadians randomly chosen would you expect to indicate they will be spending money to celebrate Halloween? Compute the standard deviation as well. E(X) = uy = (use two decimals in your answer) SD(X) = 0x = || (use two decimals in your answer) Part (d) Compute the probability that the 10-th Canadian random chosen is the 9-th to say they will be spending money to celebrate Halloween. . (use four decimals in your answer)
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(c) {:[E(x)=np=22 xx0.63=13.86],[v(x)=npq=22 xx0.63 xx0.37=5.1282],[" S.D.(x) "=sqrt(V(x))=sqrt5.1282=2.2646]:}{:[p=63%=(63)/(100)=0.63=>q=1-0.63],[n=22]:}By binomial distribution,P(x=x)=^(n)c_(x)*p^(x)*q^(n-x)(a){:[P(x=15)=^(22)C_(15)*(0.63)^(15)*(0.37)^(7)],[=1 ... See the full answer