A recent poll has suggested that 61 % of Canadians will be spending money - decorations, halloween treats, etc. - to celebrate Halloween this year. 23 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 16 of these Canadians indicate they will be spending money to celebrate Halloween. P(X = 16) = (use four decimals in your answer) Part (b) Compute the probability that between 7 and 16 of these Canadians, inclusive, indicate they will be spending money to celebrate Halloween. P(7 ≤ x ≤ 16) = (use four decimals in your answer) Part (c) How many of the 23-Canadians randomly chosen would you expect to indicate they will be spending money to celebrates Halloween? Compute the standard deviation as well.. E(X) = x= SD(X)=ox= (use two decimals in your answer) (use two decimals in your answer) hility that the 14-th Canadian random chosen is the 7-th to say they will be spending money to celebrate

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A: X~N( μ , ?)μ=195     , ?=20Z-score =( x - μ )/?

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A:

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A: Givenp=0.41n=12Let "x" be no. of U.S. adults who favor the use of unmanned drones by police…

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A: Consider X denotes adults favor the use of unmanned drones by police agencies which follows binomial…

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A: Binomial Distribution : The binomial distribution is a discrete probability distribution with two…

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A:

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A: Given data, p=46%=0.46 n=80 P(X=43)=?

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A: From the given information,Binomial parameters, n = 12 and P=34% .034.Here we can use binomial…

Q: 45% percent of consumers prefer to purchase electronics online. You randomly select 8 consumers.…

A:

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A:

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A:

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A: This is a binomial distribution problem because there are only two possible outcomes (an adult is…

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A:

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