(1 point) e-Marketer conducted a survey of Internet shoppers and found 13% said

Question

(1 point) e-Marketer conducted a survey of Internet shoppers and found 13% said they shop more "online" now than they did last year. If a sample of 75 Internet shoppers is randomly selected (Input answers to four decimal places): (a) What is the probability that exactly 16 individuals in the sample will say they do more online shopping now than they did a year ago? (b) What is the probability that the number of people in the sample who are doing more shopping online will be more than 4 but at most 17? (c) What is the probability that at least 5 people in the sample will say they are doing more online shopping?

Explanation / Answer

p = 0.13
n = 75

(A)
P(X = 16) = 75C16 * (0.13)^16 * (0.87)^59 = 0.01537

(B)
P(4 < X <= 17) = P(X <= 17) - P(X <= 3)
Using excel binomial funtion we can compute these probabilties as
P(X <= 17) = 0.9933 (excel formula: =BINOM.DIST(17,75,0.13,TRUE))

P(X <= 3) = 0.0087 (excel formula: =BINOM.DIST(3,75,0.13,TRUE))

P(4 < X <= 17) = 0.9933 - 0.0087 = 0.9846

(c)
P(X >= 5) = 1 - P(X <= 4)

Using excel binomial funtion we can compute these probabilties as
P(X <= 4) = 0.0264 (excel formula: =BINOM.DIST(4,75,0.13,TRUE))

P(X >= 5) = 1 - 0.0264 = 0.9736

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