Stat 1001 Winter 1998 Geyer Homework 2 Problem 3.1 66 inches and 72 inches. Problem 3.2 % per year 0.0 0.5 1.0 1.5 0 20 40 60 80 Age (years) (a) Age 1. (b) More 31-year olds (c) More people age 35{44, but only because this age range is wider. density is higher in the 30{34 range, but that's not what was asked. The (d) About 50% just from looking at the histogram. A more precise estimate using the actual numbers in the table would be 52%. Problem 3.4 (a) 25%. (b) 99%. 1
(c) 140{150. The area of the bar over 140{150 is more than that of the one bar over 135{140 even though the latter is higher. (d) 135{140. The density is higher there. (e) The width of the bar is 5mm(130, 125) so the area is 5 mm times 2.1 percent per mm = 10.5 percent. (f) As far as we can tell from the histogram, 102{103. The two intervals we are comparing have the same width and the density is higher over 102{103. (g) Somewhere between 115 and 120 mm. Problem 3.5 This is like part (e) of Question 3.4. The width of the interval in question is $10,000 and the height is 1%per $1,000 so the area is width times height equals 10%. Problem 3.11 The top 15 scores are all above 90 and the rest are all below 84. There is no other ve-point gap with no scores. That such a gap occurs separating those who get the jobs from those who don't is highly suspicious. Problem 4.1 (a) The sum of the numbers in the list is 300 and there are 6 numbers, so the average is 300=6 = 50. The deviations from average are,9,,2, 0, 0, 4, 7. The squared deviations are 81, 4, 0, 0, 16, 49. The sum of the squared deviations is 150. The average squared deviation is 150=6 = 25 and the SD is p 25 = 5. (b) 0.5 SD is 2.5. The numbers within 2.5 of average are 48 and the two 50s. 1.5 SD is 7.5. The numbers with 7.5 of average are 48, 50, 50, 54, 57 (all except 41). Problem 4.4 Average would be higher for income. Billionaires don't have any inuence on the median, but they do bring up the average. Not so clear for years of schooling. Most people have either 12 or 16 years of education completed. Figure 5 of Chapter 3 shows a histogram. It looks like this is a \long left hand tail" distribution so the average is smaller than the median (see Figure 7 of Chapter 4). 2
Problem 4.7 (a) This multiplies the averages and the SDs by 2.2. The men have average weight 145.2 pounds and SD 19.8, and the women have average weight 121 pounds and SD 19.8 pounds. Since the original numbers do not seem to have had three-gure accuracy, one could also round these to average weight 145 and SD 20 for men and average weight 120 and SD 20 for women. (b) 57 kg is 1 SD below average and 75 kg is 1 SD above average. The box on p. 68 in the book says about 68% of the data will be in this range. (c) Bigger. Putting two groups together always increases the spread unless both groups have the same histogram. The reason is that the \within group" spread stays the same but there is also now \between group" spread. The gure shows what happens. The higher curve is the histogram sketch for men and women. 20 40 60 80 100 kilograms The lower curves are the histogram sketches for men and women separately. Problem 4.8 (a) No. The combined average (147) is larger than the average for boys alone (146). The the average for girls must be larger (about 148). (b) About half way between the averages at age 9 and 11, that is 141. Problem 4.10 (a) The best guess is the average 163. 3
(b) One SD $8. The box onp.68says about 32% of the data will be more than one SD from the average. This is close to 1 in 3 (33.3%). Problem 5.1 (a) The normal table says 78.87% within 1.25 SDs of average. There are 25 numbers on the list, 78.87% of 25 is 19.7, so about 20 numbers should be within 1.25 SDs of average. (b) The actual numbers corresponding to 1.25 SDs above and below average are 50, 1:25 10 = 37:5 and 50 + 1:25 10 = 62:5. By actual count 18ofthe 25 numbers in the list are between 37.5 and 62.5. That's 72%. Problem 5.3 (a) 600 is 600, 466 = 134 or 134=100 = 1:22 SDs above the average in 1967. From the normal curve tail area table about 10.94 percent of the area is in the tail past 1.22 (somewhere between 11.51 and 10.56). (b) 600 is 600, 423 = 177 or 177=110 = 1:61 SDs above the average in 1994. From the normal curve tail area table about 5.4 percent of the area is in the tail past 1.61 (somewhere between 5.48 and 4.95). Problem 5.6 No. There should be about 16% of the scores more than one SD aboveaverage if the distribution of the scores follows the normal curve, but for the LSAT, there are no scores more than one SD above average (because that's the maximum possible score). Problem 5.7 (a) 350 is 150 points below average or 1.5 SD. From the normal curve tail area table about 6.68 percent is in the tail below,1:5. So one would say the student at the 7th percentile (rounding o to an even percent). (b) This problem involves a \backwards" table look-up. In the normal curve tail area table we see z Area 0.65 25.78 0.70 24.20 Thus the z corresponding to a 25% tail area is about 0.675 (half way between 0.65 and 0.70). Thus in standard units the 25th percentile is,0:675 and the 75th percentile is 0:675. Now we must convert standard units to original units. 0.675 SD above average is 500 + 0:675 100 = 567:5. 4
Problem 5.8 (a) True. This shifts the histogram 7 units to the right, which shifts the balance point 7 units to the right. (b) False. Shifting the histogram doesn't change the spread. (c) True. Doubling all the numbers doesn't change the shape of the histogram, it just doubles all the numbers on the horizontal axis. This also doubles the number corresponding to the balance point. (d) True. spread. Doubling all the numbers on the horizontal axis also doubles the (e) True. Changing the sign of all the numbers does change the shape of the histogram (unless it is symmetric). We get the mirror image histogram. This also changes the sign of the balance point. 0 2 4 6 8 10-10 -8-6 -4-2 0 (f) False. The mirror image histogram has the same spread. Problem 5.9 (a) False. The average can be very far from the median if the distribution has one long tail or if there are outliers. See Figure 7 of Chapter 4. (b) False. This would be true if it said median rather than average. If a list has 100 numbers, one of which is 101 and the other 99 are all 1, then the average is 2 and 99% are below average. (c) False. With a large, representative sample the sample histogram will be quite close to the population histogram, but not all populations follow the normal curve. 5
(d) False. If one list follows the normal curve, then there would be 68% between 40 and 60 (within one SD of average). If the other list has half the numbers equal to 40 and the other half equal to 60, then the average is 50 and the SD is 10. The fraction between 40 and 60 is anything between 0 and 100 percent, depending on how you interpret \between" when the numbers are on the boundary (and by adding or subtracting very small amounts from some of the numbers, the fraction between 40 and 60 can be made anything between 0 and 100 percent with no problems of interpretation). Problem 5.10 This is a (somewhat disguised) question about what long-tailed distributions do to the median and the mean. Since $150,000 is almost the top of the income distribution and $35,000 is the average, the question can be rephrased as: what percentage of incomes are above average? Income is the canonical example of a distribution with a long right tail. From Figure 7 of Chapter 4 we see that the average is bigger than the median. 50% are above the median, so less than 50% are above average. The only possible answer of the choices given is 40%. 6