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AP Statistics 4.11 Parameters for a Binomial Distribution- MCQs - Exam Style Questions

Question

The delivery times for a local food-delivery service are approximately normally distributed with a mean of \(50\) minutes and a standard deviation of \(10\) minutes. Any customer who must wait more than \(60\) minutes for their delivery will receive a coupon for 50 percent off their next order.
If \(30\) customers are randomly selected from all customers who purchased food from the delivery service, then the probability that at least \(2\) of the customers will have to wait more than \(60\) minutes and receive a coupon is closest to which of the following?
(A) \(\displaystyle \binom{30}{2}(0.16)^2(0.84)^{28}\)
(B) \(\displaystyle 1-\Big[\binom{30}{0}(0.84)^{0}(0.16)^{30}+\binom{30}{1}(0.84)^{1}(0.16)^{29}\Big]\)
(C) \(\displaystyle 1-\Big[\binom{30}{0}(0.84)^{0}(0.16)^{30}+\binom{30}{1}(0.84)^{1}(0.16)^{29}+\binom{30}{2}(0.84)^{2}(0.16)^{28}\Big]\)
(D) \(\displaystyle 1-\Big[\binom{30}{0}(0.16)^{0}(0.84)^{30}+\binom{30}{1}(0.16)^{1}(0.84)^{29}\Big]\)
(E) \(\displaystyle 1-\Big[\binom{30}{0}(0.16)^{0}(0.84)^{30}+\binom{30}{1}(0.16)^{0}(0.84)^{29}+\binom{30}{2}(0.16)^{2}(0.84)^{28}\Big]\)
▶️ Answer/Explanation
Single-customer probability of a coupon: \[ p=\Pr(X>60)=\Pr\!\left(Z>\frac{60-50}{10}\right)=\Pr(Z>1)\approx0.1587\approx0.16,\quad q=1-p\approx0.84. \] Let \(Y\sim\mathrm{Bin}(n=30,p\approx0.16)\). Then \[ \Pr(Y\ge2)=1-\Pr(Y=0)-\Pr(Y=1) =1-\Big[(0.84)^{30}+30(0.16)(0.84)^{29}\Big] =1-\Big[\binom{30}{0}(0.16)^0(0.84)^{30}+\binom{30}{1}(0.16)^1(0.84)^{29}\Big]. \] ✅ Answer: (D)
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