1. According to a Adobe Interactive (Dec 2018) survey of U.S. adults, 61% of U.S

Question

1. According to a Adobe Interactive (Dec 2018) survey of U.S. adults, 61% of U.S. adults work during their holidays. In a random sample of 10 U.S. adults, let y represent the number who do not work during holidays. Find the probability that four of the 10 U.S. adults work during their holidays, P(y=4)

2. The supervisor of a soft-drink bottling company believes that when a new beverage-dispensing machine is set to dispense 11.5 oz, it in fact dispenses an amount y at random anywhere between 11 and 12 oz inclusive. Suppose y has a uniform probability distribution. Find P(11.3<y<11.7)

Explanation / Answer

1) n = 10

p = 1 - 0.61 = 0.39

This is a binomial distribution.

P(Y = y) = 10Cy * 0.39y * (1 - 0.39)10-y

P(Y = 4) = 10C4 * 0.394 * 0.616 = 0.2503

2) Y ~ U (11 , 12)

P(11.3 < y < 11.7) = (11.7 - 11.3) / (12 - 11) = 0.4 (ans)

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