Mathematics 375 – Probability Theory Solutions for Problem Set 1 September 9, 2011

1.4. (a) Using bins [2, 3), [3, 4), etc. the histogram would show boxes of heights

10/40, 12/40, 7/40, 5/40, 2/40, 3/40,

then three boxes with height zero, and one final box of height 1/40.

(b) The proportion is 18/40 = 9/20 or about 45%

(c) The probability is the proportion 29/40, or 72.5%.

1.7. (a) and (b) This histogram has a sort of “two-hump” shape with a lower portion in the center.

(c) The best explanation is the difference between average heights of males and females. We saw a similar pattern in the heights data for our class.

1.8. (a) Using six bins [10.1, 12.1), [12.1, 14.1), …, [20.1, 22.1) the histogram would show boxes of heights

7/26, 7/26, 4/26, 2/26, 5/26, 1/26.

(b) The lower portion in the center (on the interval [16.1, 18.1)) and the two humps in the histogram on either side of the central depression is not like many other distributions (not like the normal distribution or “bell-shaped curve”). One possibility is that the pottery found at the first two sites (Llanederyn and Caldicott) has some definitely different composition from the pottery found at the other two sites (Island Thorns, Ashley Rails). Some reasonable explanations: the clay used at the two groups of sites could have had different chemical compositions, or there might have been some different sort of glaze used on the pottery from the second two sites that contained more aluminum oxide. Note: The fact that there are more data values from the first two sites just says the total area on the left side of the histogram is larger than the total area on the right. That fact does not account for the gap between the two subsets of data values that you can see if you include enough bins in the histogram, though.

1.11. We will show the desired equality by starting on the left. Begin by squaring out:

(y_{i} − \overline{y})^{2} = y_{i}^{2} − 2y_{i}\overline{y} + \overline{y}^{2}.

Then and using a, b, c

s^{2} = \frac{1}{n − 1} ∑_{i=1}^{n}(y_{i} − \overline{y})^{2}

= \frac{1}{n − 1} ∑_{i=1}^{n}(y_{i}^{2} − 2y_{i}\overline{y} + \overline{y}^{2})

= \frac{1}{n − 1} (∑_{i=1}^{n} y_{i}^{2} − 2\overline{y} ∑_{i=1}^{n} y_{i} + n\overline{y}^{2}) ,

where we note that \overline{y} is constant, so ∑_{i=1} \overline{y} = n\overline{y}.

Then using the definition of \overline{y}:

= \frac{1}{n − 1} (∑_{i=1}^{n} y_{i}^{2} − 2n\overline{y}^{2} + n\overline{y}^{2})

= \frac{1}{n − 1} (∑_{i=1}^{n} y_{i}^{2} − n\overline{y}^{2})

= \frac{1}{n − 1} (∑_{i=1}^{n} y_{i}^{2} − \frac{1}{n} (∑_{i=1}^{n} y_{i})^{2}) .

1.12. Using the formula of 1.11, we have ∑_{i=1}^{6} y_{i} = 14, and ∑_{i=1}^{6} y_{i}^{2} = 40. So s^{2} = \frac{1}{5}(40− \frac{1}{6}(14)^{2}) = 1.47. Then s ≐ \sqrt{1.47} ≐ 1.21.

1.15. (a) \overline{y} = 4.39, s = 1.87. (b) The interval \overline{y} ± s is (2.52, 6.26) There are 36 data points in this interval. 35/40 = 87.5% is much larger than one would expect by the empirical rule. The interval \overline{y} ± 2s is (0.65, 8.13). This contains 39/40 = 97.5% of the data points – again somewhat larger than we would expect. The interval \overline{y} ± 3s is (−1.22, 10.0). This interval again contains 39/40 of the data points. (Note: This data set is somewhat unusual because of the one very large “outlier” value 11.88 at the start. Because that number is so much larger than the others, the value of \overline{y} is pulled up and the the value of s is larger relative to the mean and standard deviation of the remaining data points if the outlier is removed. This explains the relatively poor agreement with the empirical rule in this example.)

1.17. The range of the measurements in 1.4 is 11.88 − 2.61 = 9.27. Since the interval \overline{x} − 2s to \overline{x} + 2s contains the large majority of the data, we can estimate s by taking the range and dividing by 4. This yields an approximation to s: s ≐ 9.27/4 ≐ 2.32. Using the formula from 1.11, the actual s is approximately 1.87. Note that the one very large value has the effect of increasing the range, and making one quarter of the range an overestimate.

1.21. By the empirical rule, the number of measurements larger than \overline{y} − s should be about (\frac{1}{2}(68) + 50)% = 84%. (Note that this includes the “upper tail” of values greater that \overline{y} + 2s. The empirical rule says that tail should account for about \frac{1}{2}(100−95) = 2.5% of the total number of data values.) The fact that the claim is 80% while the actual data suggests 84% is not a contradiction. Think about it from the point of view of a farmer using So no, the manufacturer is just being slightly conservative with the 80% figure.

2.14. Note: this problem is all about being able to interpret the headings on the table properly!

(a) The probability that a single adult needs glasses is the sum of the entries in the first row: 0.58.

(b) The probability that an adult needs glasses but does not use them is 0.14. (c) The probability that a single adult uses glasses whether or not he/she needs them is the sum of the entries in the first column: 0.46.

2.15. (a) The entry with the question mark is 1−(.01+.09+.81) = .09 since these four alternatives are mutually exclusive and should add up to 1. (b) The probability that the company hits gas or oil on at least one drilling is (.01 + .09 + .09) = .19.

2.16. (a) The probability that the randomly selected person has O^{+} blood is \frac{1}{3} = .\overline{3}. (b) The probability the person has type O blood is \frac{1}{3} + \frac{1}{15} = \frac{6}{15} = \frac{2}{5} = .4. (c) The probability the person has type A blood is \frac{1}{3} + \frac{1}{16} = \frac{19}{48} ≐ .396. (d) The probability that the person has neither type A nor type O blood is ≐ 1 − .4 − .396 = .204 (using the answers from (b) and (c)).

2.18. (a) The possible outcomes are S = {HH, HT, T H, T T }. (b) If both coins are fair (balanced), it is reasonable to assign a probability of \frac{1}{2} · \frac{1}{2} = \frac{1}{4} to each sample point. They are equally likely.

(c) A = \{HT, T H\} and B = \{HT, T H, HH\}. (d) P (A) = \frac{1}{2}, P (B) = \frac{3}{4}, P (A ∪ B) = P (B) = \frac{3}{4} since A ⊂ B, P (A ∩ B) = \frac{1}{2} since A ∩ B = A whenever A ⊂ B. P (\overline{A} ∪ B) = 1.

2.19. (a) The sample space is

S = {(V_{1}, V_{1}), (V_{1}, V_{2}), (V_{1}, V_{3}), (V_{2}, V_{1}), (V_{2}, V_{2}), (V_{2}, V_{3}), (V_{3}, V_{1}), (V_{3}, V_{2}), (V_{3}, V_{3})}

(note that the same vendor could be used both days since that is not ruled out). (b) Each of the 9 points in S gets probability \frac{1}{9}. (c) A = \{(V_{1}, V_{1}), (V_{2}, V_{2}), (V_{3}, V_{3})\} and

B = {(V_{2}, V_{1}), (V_{2}, V_{2}), (V_{2}, V_{3}), (V_{1}, V_{2}), (V_{3}, V_{2})}.

Therefore P (A) = \frac{1}{3}, P (B) = \frac{5}{9}, P (A ∪ B) = \frac{7}{9} and P (A ∩ B) = \frac{1}{9}.