Homework 7 Solutions
Chapter 17
1. (a) The smallest the sum can be is 100. This will only happen if
every draw is the lowest number in the box (that's 1).
The largest the sum can be is 1000. This will only happen if
every draw is the highest number in the box (that's 10).
(b) The sum is between 650 and 750 with a chance of about 90%.
average of the box: (1+6+7+9+9+10)/6 = 7
SD of the numbers in the box: [(36+1+4+4+9)/6]^0.5 = [54/6]^0.5 = 3
expected value of sum = 7(100) = 700
SE for sum = 30
So 650 and 750 are about 1 2/3 SEs away from average. If you look on
the normal table (which we are using to approximate this percentage),
you will find that about 90% of the area of the normal curve lies
between -1.65 and 1.65 (closest number to 1 2/3 on the table).
3. In each of the five parts of this question, I have used the
shortcut discussed in section 4 of this chapter.
(a) 1, -2, -2 --- choice (iii)
(1-(-2))[(1/3)(2/3)]^0.5 = (3)[(1/3)(2/3)]^0.5
(b) 15, 15, 16 --- choice (i)
(16-15)[(1/3)(2/3)]^0.5 = [(1/3)(2/3)]^0.5
(c) -1, -1, -1, 1 --- choice (v)
(1-(-1))[(1/4)(3/4)]^0.5 = (2)[(1/4)(3/4)]^0.5
(d) 0, 0, 0, 1 --- choice (iv)
(1-0)[(1/4)(3/4)]^0.5 = [(1/4)(3/4)]^0.5
(e) 0, 0, 2 --- choice (ii)
(2-0)[(1/3)(2/3)]^0.5 = 2[(1/3)(2/3)]^0.5
4. The number of aces in 180 rolls of a die is like the sum of 180
draws from a box with 1 ticket marked "1" and 5 tickets marked "0".
expected number of 1's: (1/6)(180) = 30
SE: (180)^0.5 (1-0) [(1/6)(5/6)]^0.5 = [(180)(5)/36]^0.5 = 5
Both 15 and 45 are 3 SD's away from expected value of 30. We can then
expect about 99.73% of people to get counts in the range 15 to 45.
6. observed value for the sum of the draws: 45(1) + 23(2) + 32(3) = 187
observed value for the number of 3's: 32
observed value for the number of 1's: 45
expected value for the sum of the draws: [(1+1+2+3)/4][100] = 175
expected value for the number of 3's: (1/4)(100) = 25
expected value for the number of 1's: (2/4)(100) = 50
chance error in the sum of the draws: 187 - 175 = 12
standard error for the number of 1's: 10(1-0)[((1/2)(1/2))^0.5] = 5
8. A coin tossed 100 times.
(a) The difference "number of heads - number of tails" is like the
sum of 100 draws from box (ii) because you add 1 to the sum when you
throw heads (with probability 1/2) and you lose 1 when you throw
tails.
(b) expected value for this difference is 0
(c) standard error for this difference is
(100^0.5)(1-(-1))[(1/2)(1/2)]^0.5 = 10(2)(1/2) = 10
Chapter 18
1. 20 and 25
5. (i) is the probability histogram for the sum. It's like the
normal curve.
(ii) is the probability histogram for the product. It's like
figure 10, p. 323.
(iii) is the histogram for the numbers drawn. It's like a
histogram for the numbers in the box.
6. No, the COIN program did not pass. The expected number of heads
is 500,000. The standard error for the total number of heads is 500.
So having 2,015 more heads than expected is more than 4 SE's away from
what was expected. It is possible to get this many heads with a fair
coin, but it is extrememly unlikely. The program should be retested,
and if it continues in this way, we will know that it is not
performing properly.
Chapter 20
1. The completed table for the coin-tossing game:
# of heads % of heads
# of tosses exp. value SE exp. value SE
----------------------------------------------------------
100 50 5 50% 5%
2,500 1,250 25 50% 1%
10,000 5,000 50 50% 0.5%
1,000,000 500,000 500 50% 0.05%
5. So, we can think of a box that has an average of 150 lbs. and a SD
of about 35 lbs. We make 50 draws from the box and add them up to
represent the combined weight of the 50 people. (Remember, the
elevator was designed for 4 tons = 8000 lbs.)
expected sum = 50(150) = 7500
SE sum = [(50)^0.5][35] is about 247.49
z-score = [8000-7500]/[(35)(50)^0.5] is about 2.02 (but the closest
number on the table is 2.00)
So, chance of passing 4 tons is about: (100-95.45)/2 = 2.275%
Chapter 21
4. 36.1% knew about Chaucer
95.2% knew about Edison
(a) SE for the percentage knowing about Chaucer:
[(0.361)(0.639)/6000]^0.5 * 100% = 0.6%
95% CI:
36.1% - 2(0.6%) = 34.9%
36.1% + 2(0.6%) = 37.3%
(b) SE for the percentage knowing about Edison:
[(0.952)(0.048)/6000]^0.5 * 100 % = 0.3%
95% CI:
95.2% - 2(0.3%) = 94.6%
95.2% + 2(0.3%) = 95.8%
5. False: It is extremely unlikely that the percentage of the sample
will be exactly equal to the population sample. In general, though,
it is true that as the sample size gets larger, the sample percentage
is likely to draw closer and closer to the population percentage.
12. Figure (ii) shows the histogram for the numbers drawn. It only
has 3 bars, because only 3 numbers can be drawn from this box.
Figure (iii) is the histogram for the sum, becuase it resembles
the normal curve.
Figure (i) is irrelevant.
Chapter 23
2. (a) True. SE for sum/# draws = [(500^0.5)(2.3)]/500 is about
0.10.
(b) True. 68% CI for average of box: 71.3 +/- 0.10
(c) False. No, the confidence interval of 68% just means that if
you build a confidence interval like this for many different samples
from the box, about 68% of them will include the true average. (See
pg. 417.)
8. (a) True. 68% CI: 3700 +/- 325 = (3375, 4025)
(b) True. See Sec 21.3.
(c) False. Data don't follow the normal curve, but 68% may be
about right. Need the data to tell.
(d) False. 325 is not the SD.
(e) False. We're looking at averages.
10. (a) True. SE for the average: 1.75/(625^0.5) = 1.75/25 = 0.07
(b) False. We know what the average household size is in the
sample; this confidence interval would be for
(c) True. 95% CI for average: 2.30 +/- 2(0.07) = (2.16, 2.44)
(d) False. Confuses SD with SE. (Also, household must have
whole number of persons.)
(e) False. We're estimating the average. (If followed the
normal curve, there would be houses with negative numbers of
occupants.)
(f) True.
11. average of box: 3
SD of box: [(4+1+0+1+4)/5]^0.5 = 2^0.5 is about 1.41
So, the range is much larger than we expect. We expect to see a mean
of 3 with a SD of about (2^0.5)/5 = 0.28. That means that 95% of the
histogram's area would be between 3 - 2(0.28) = 2.44 and 3 + 2(0.28) =
3.56.