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AP STAT: MORE Ch. 16 practice!NAME: _______________________________There are 4 runners on the New High School team. The team is planning to participate in a race in which each runner runs a mile. The team time is the sum of the individual times for the 4 runners. Assume that the individual times of the 4 runners are all independent of each other. The individual times, in minutes, of the runners in similar races are approximately normally distributed with the following means and standard deviations.The distribution of possible team times is approximately normal. What are the mean and standard deviation of this distribution?μ1+2+3+4=4.9+4.7+4.5+4.8=18.9 minutes σ1+2+3+4=0.152+0.162+0.142+0.152=0.3003 minutesSuppose the team’s best time to date is 18.4 minutes. What is the probability that the team will beat its own best time in the next race?X = 1+2+3+4X~N(18.9, 0.3003)P(X<18.4) = normalcdf(-E99, 18.4, 18.9, 0.3003) = 0.04796For an upcoming concert, each customer may purchase up to 3 child tickets and 3 adult tickets. Let C be the number of child tickets purchased by a single customer. The probability distribution of the number of child tickets purchased by a single customer is given in the table pute the mean and the standard deviation of C.μC=1.1 tickets σC=1.22 tickets Suppose the mean and the standard deviation for the number of adult tickets purchased by a single customer are 2 and 1.2, respectively. Assume that the numbers of child tickets and adult tickets purchased are independent random variables. Compute the mean and the standard deviation of the total number of adult and child tickets purchased by a single customer.A = adult ticketsμA=2 tickets σA=1.2 tickets μA+C=1.1+2=3.1 tickets σA+C=1.222+1.22=1.7113 tickets Suppose each child ticket costs $15 and each adult ticket costs $25. Compute the mean and the standard deviation of the total amount spent per purchase.μ25A+15C=(25*1.1)+(15*2)=$57.5 σ25A+15C=(25*1.22)2+(15*1.2)2=$35.415 The depth from the surface of Earth to a refracting layer beneath the surface can be estimated using methods developed by seismologists. One method is based on the time required for vibrations to travel from a distant explosion to a receiving point. The depth measurement (M) is the sum of the true depth (D) and the random measurement error (E). That is, M = D + E. The measurement error (E) is assumed to be normally distributed with mean 0 feet and standard deviation 1.5 feet.μE=0 σE=1.5 If the true depth at a certain point is 2 feet, what is the probability that the depth measurement will be negative?M = D+ED = 2 ft. μM=0+2=2 feet σM=1.5 feet M~N(2, 1.5)P(M < 0) = normalcdf -E99, 0, 2, 1.5) = 0.0912Suppose three independent depth measurements are taken at the point where the true depth is 2 feet. What is the probability that at least one of these measurements will be negative?P(Neg) = 0.0912n = 3B(3, 0.0912)P(N > 1) = 1-binomcdf(3, 0.0912, 0) = 0.2494What is the probability that the total of four independent depth measurements taken at the point where the true depth is 2 feet will be negative?M~N(2, 1.5)μM+M+M+M=2*4=8 feet σM+M+M+M=1.52+1.52+1.52+1.52= 3 feet M+M+M+M ~N(8, 3)P(M+M+M+M < 0) = normalcdf(-E99, 0, 8, 3) = 0.00383The weights of red delicious apples are approximately normally distributed with a mean of 9 ounces and a standard deviation of 0.75 ounce. An online gift store sells gift boxes containing 5 red delicious apples. At the time of packaging, 5 red delicious apples are randomly selected and packaged in a box.μA=9 oz σA=0.75 oz A~N(9, 0.75)Describe the distribution of the total weight of the 5 randomly selected apples.μA+A+A+A+A= 45 oz.Total = A+A+A+A+ATotal~N(45, 1.677)σA+A+A+A+A= 1.677 oz.What is the probability that the total weight of the 5 randomly selected apples will be less than 42 ounces?N(45, 1.677)P(total < 42) = normalcdf(-E99, 42, 45, 1.677) = 0.0368The combined weight of the packing material and box in which the apples will be shipped is always 10 ounces. Let W represent the weight of a complete packaged gift box, which consists of the packing material, box, and 5 randomly selected apples. What are the mean and the standard deviation of W ?μW=μtotal+10=45+10= 55 oz σW=σtotal+10 =1.677 oz The company must pay to ship the gift boxes as well. They are charged a flat fee of $3 per box, and then $0.35 per oz. What are the mean and standard deviation of the shipping charge for the complete packaged gift boxes? Cost = 3 + (0.35*W)μcost=3+0.35* 55=$22.25 σcost=0.35*1.677=$0.587 Assuming the cost of the complete packaged gift box follows a normal model, what is the probability that the cost of the complete packaged gift boxes exceeds $24?Cost ~N(22.25, 0.587)P(Cost > 24) = normalcdf(24,E99,22.25, 0.587 = 0.0014 ................
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