Solutions for Friday, 4-7-2014 AP Statistics Final Review: ANSWER KEY 6a. The shape is single peaked and approximately symmetric. 6b. Bin Freq Rel. Freq. Cum. Rel. Freq. 10-<15 1 0.03333333 0.03333333 15-<20 1 0.03333333 0.06666667 20-<25 1 0.03333333 0.1 25-<30 3 0.1 0.2 30-<35 7 0.23333333 0.43333333 35-<40 7 0.23333333 0.66666667 40-<45 4 0.13333333 0.8 45-<50 3 0.1 0.9 50-<55 2 0.06666667 0.96666667 55-<60 0 0 0.96666667 60-<65 0 0 0.96666667 65-<70 1 0.03333333 1 Ogive of Times for Test 1 0.9 0.8
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6c. In the histogram, most of the data is between 25 and 55, so this is where the ogive is most steep (where the most data is being added to the cumulative relative frequency). There is less data below 25 and above 55, so the ogive is relatively flat here. It is perfectly flat from 55 to 65 where there is no data at all. 6d. Tracing over from .75, we estimate Q3 = 43 and tracing over from .25, we estimate Q1 to be 32. So, the IQR is approximately 11. 7a.
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7b. Since the data is strongly skewed to the right, I would expect the mean to be greater than the median since the mean is pulled in the direction of the skew. 7c. I would use a relative frequency histogram anytime I wanted to know the percent in a category and especially when I am comparing two distributions with different sample sizes. 8a. range = 20 – 6 = 14, IQR = 16.5 – 11 = 5.5 8b. mean = 13.44(3) + 10 = 50.32 s = 3.67(3) = 11.01 median = 13.5(3) + 10 = 50.5 IQR = 5.5(3) = 16.5 8c. Sally’s score was the same or better than 39% of the test takers. 8d. Still in the 39th percentile. 8e. To make the SD = 15, multiply each value by 15/3.67 = 4.09. Multiplying everything by 4.09 will make the mean = 4.09(13.44) = 54.97 so add an additional 25.03 to each score. 9a. min = 1, Q1 = 10, med = 30.5, Q3 = 38, max = 50
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9b. With the stemplot you were able to see the double-peaked shape, which is not evident in the boxplot. However, in the boxplot it is easy to see where the median is as well as measure the interquartile range (length of the box). 10. Let x = distance the golf ball travels ~ N(250,15)
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220 235 250 265 280 231 − 250 P(x < 231) = P(z < ) = P(z < -1.27) = .1020 15 OR P(x < 231) = normalcdf(-999,231,250,15) = .1026 P(x > 300) = normalcdf(300,999,250,15) = .0004 (can also do with z-scores) P(240 < x < 260) = normalcdf(240,260,250,15) = .4950 (can also do with z-scores) (draw picture with area of .75 to left of boundary) boundary = invnorm(.75,250,15) = 260.1 yards
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x − 250 . Solving for x = 260.05 yards 15 f) (draw picture with area of .10 to left of boundary so .90 is to the right of the boundary. Boundary = invnorm(.10, 250, 15) = 230.8 yards (also can do with z-scores)
From table: z = .67 =
