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8.4.12 A survey of 1020 workers in a certain year found that 65% of the respondents...

8.4.12

A survey of 1020 workers in a certain year found that 65% of the respondents spend a total of $40 or less on lunch each week. If 10 of the workers who participated in the survey were chosen at random, what is the probability that at most 3 of them spend a total of $40 or less on lunch each week? (Round your answer to three decimal places.)

math   Statistics-And-Probability   
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Answer #1

n = 10, p = 0.65 This is Binomial distribution

P(X at most 3) = P(0) + P(1) + P(2) + P(3)

Hence, answer is 0.026

Please comment if any doubt. Thank you.

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