6. Cold U? Max = 51.8 F Range = 59.4 F Mean = 33.8 F s = 12.6 F med = 35.6 F IQR = 28.8 F

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1 AP Stat Ch. 6 Practice Worksheet - KEY BOOK PROBLEMS: p. 129 #2-24 even 2. Hotline a) Median = 264 seconds IQR = 138 seconds b) Median = 240 seconds IQR = 138 seconds 4. Hams a) Range = 3.3 lbs. IQR = 0.95 lbs. b) Boxplot: The distribution looks to be skewed to the left. When looking at the boxplot we see a tailing out the left. We see a large difference between the min and Q1 and between Q1 and the Median. Also the mean is less than the median. c) Mean = 96 oz. s = 10.4 oz. Q1 = 89.6 oz. Q3 = oz. Med = 99.2 oz. IQR = 15.2 oz. Range = 52.8 oz. d) Mean = 126 oz. s = 10.4 oz. Q1 = oz. Q3 = oz. Med = oz. IQR = 15.2 oz. Range = 52.8 oz. e) Q1, Median, Q3, IQR would probably stay the same 6. Cold U? Max = 51.8 F Range = 59.4 F Mean = 33.8 F s = 12.6 F med = 35.6 F IQR = 28.8 F 8. Checkup The boy s height is 1.88 standard deviations below the mean height for all American 2-yearolds. 10. Mensa x = 16 x = 140 An IQ of 140 would be needed to be considered a genius 12. Placement Exam French Mathematics z F = = 1.25 z M = = She did better on the Mathematics Test. 14. Combining scores again Reginald Sara Test z = = 0 z = = Test z = = 1 z = = Combined As seen from the combined score, Sara has done better overall when compared to how the other students did on both exams even though her mean raw score is lower.

2 16. MP3s a) The standard deviations would let me know how consistent the battery lives are and also allow me to figure out the chance they will last certain times b) DuraTunes RockReady z D = = 1.5 z R = = Comparing the standard scores it can be seen that the standard score for RockReady is further from the mean. It is unlikely that it would last less than 8 hours. RockReady is more likely to last more than 8 hours than DuraTunes. c) DuraTunes RockReady z D = = 2.5 z R = = Comparing the standard scores it can be seen that the standard score for RockReady is further from the mean. It is unlikely that it would last up to 16 hours. Neither are likely to last 16 hours but DuraTunes has a better chance than RockReady. 18. Car speeds a) z = = It would be standard deviations below the mean speed b) 34 MPH 10 MPH z34 = = z10 = = A car driving 10 mph would be more unusual since it has a standard score that is further away from the mean. 20. Car speeds again a) Mean = 3.84 mph; s = 3.56 mph b) Mean = 6.18 kph; s = 5.73 kph 22. Caught speeding Mean = $180 max = $230 s = $24 IQR = $ Rock concerts Most likely it would be 2,000

3 1. Suppose a Normal model describes the fuel efficiency of cars currently registered in your state. The mean is 24 mpg, with a standard deviation of 6 mpg. a. Sketch the Normal model. Be sure to include the rule. N(24, 6) b. What percent of all cars get less than 15 mpg? P(x < 15) = 6.68% c. What percent of all cars get between 20 and 30 mpg? P(20 < x < 30) = 58.89% d. What percent of cars get more than 40 mpg? P(x > 40) = 0.38% e. Describe the fuel efficiency of the worst 20% of all cars. P(x < A) = 20% A = mph f. What gas mileage represents the third quartile? P(x < Q 3 ) = 75% Q 3 = invnorm(0.75,24,6) = mph g. What would be the IQR? P(x < Q 1 ) = 25% Q 1 = invnorm(0.25, 24,6) = mph IQR = = 8.1 mph h. Describe the fuel efficiency of the worst 5% of all cars. P(x < A) = 5% A = invnorm(0.05,24,6) = mph i. What percentage of cars get under 20 mpg? P(x < 20) = 25.25%

4 j. An ecology group is lobbying for a national goal calling for no more than 10% of all cars to be under 20 mpg. If the standard deviation does not change what average fuel efficiency must be attained? invnorm(0.10) = μ = 6 μ = mph k. Car manufacturers argue that they cannot raise the average that much they believe they can only get to 26 mpg. What standard deviation would allow them to meet the only 10% under 20 mpg goal? invnorm(0.10) = = = 4.68 mph l. What change in fuel economy of cars would achieve that standard deviation bring about? What are the advantages and disadvantages? 2. The life expectancy of a particular brand of light bulb is normally distributed with a mean of 1500 hours and a standard deviation of 75 hours. a. What is the probability that a light bulb will last less than 1410 hours? N(1500, 75) P(x < 1410) = 11.51% b. What is the probability that a light bulb will last more than 1550 hours? P(x > 1550) = 25.25% c. What is the probability that a light bulb will last between 1563 and 1648 hours? P(1563 < x < 1648) = 17.62% d. 15% of the time a light bulb will last more than how many hours? P(x < A) = 85% A = invnorm(0.85,1500,75) = hours 3. Given a normal distribution with a standard deviation of 10, what is the mean (μ) if 21% of the values are below 50? invnorm(0.21) = μ = 10 x = 58.06

5 4. Given a normal distribution with 80% of the values above 125 and 90% of the values above 110, what are the mean and standard deviation? invnorm(0.20) = invnorm(0.10) = μ = 110 μ = 125 μ 110 μ = = μ 110 μ = (125 μ) = (110 μ) μ = μ 0.44μ = μ = = = A water fountain is designed to dispense a volume of 12.2 oz. with a standard deviation of 0.5 oz. N(12.2, 0.5) a. What percentage of cups end up with at least 12 oz.? P(X > 12) = P(z > -0.4) =65.5% b. 75% of the cups contain more than how much water? P(x < A) = 25% invnorm(0.25,12.2,0.5) = oz c. Find the IQR for the amount of water dispensed. Q 1 = oz P(x < Q 3 ) = 75% Q 3 = invnorm(0.75,12.2,0.5) = oz IQR = = 0.68 oz d. Find the 90 th percentile for the amount of water dispensed. P(x < A) = 90% A = invnorm(0.90,12.2,0.5) = oz

6 6. A tire manufacturer believes that the treadlife of its snow tires can be described by a Normal model with a mean of 32,000 miles and a standard deviation of 2500 miles. N(32000, 2500) a. If you buy a set of these tires, would it be reasonable for you to hope they ll last 40,000 miles? z = = ,000 miles is over 3 standard deviations above the mean so it is highly unlikely that they would last that long. P(X > 40000) = P(z > 3.2) = x 10-4 b. In planning a marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer does not want to take too big a risk. If the dealer is willing to give refunds to no more than 1 of every 25 customers, for what mileage can he guarantee these tires to last? 1/25 = 4% P(x < A) = 4% A = invnorm(0.04,32000,2500) = 27, miles c. The manufacturer has located a new process to increase the treadlife of the tires. What new mean would the treadlife be if the tire dealer can keep the same money back guarantee but now only have to refund 1% of tires? invnorm(0.01) = μ z = = 2500 μ = miles d. The manufacturer can t actually increase the mean treadlife but he can reduce the standard deviation. What new standard deviation would be needed for the dealer to keep the same guarantee but only refund 1% of tires? What would this mean in the context? z = = = 1, miles