Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 1 September 10, 2009

1.4. (a) Using bins [2, 3), [3, 4), etc. the histogram should 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) The distribution seems to have two “humps” with a small gap between them.

(b) The lower portion in the center (on the interval from 68 to 69) is not like many other distributions.

(c) One possibility is that the students are from two distinct subpopulations (males and females) whose average heights are different. The lower “hump” would then be composed primarily of females, and the upper “hump” would be composed primarily of males.

1.9. For all parts here, we use the “empirical rule:” (a) This is \overline{x} ± s, so the fraction is about .68, or 68%.

(b) This is \overline{x} ± 2s, so the fraction is about .95, or 95%.

(c) This is \overline{x} − s to \overline{x} + 2s. By symmetry the fraction is about (.5)(.68) + (.5)(.95) = .815, or 81.5%

(d) This is < \overline{x} − 3s and > \overline{x} + 3s, so very few (approximately zero).

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.

1.17. 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 yield 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 probably has the effect of increasing the range, and making this an overestimate.

2.1. We have A = {F F }, B = {MM}, C = {F M, MF, MM}, A ∩ B = ∅, A ∪ B = {F F, MM}, A ∩ C = ∅, A ∪ C = {F F, F M, MF, MM}, B ∩ C = {MM}, B ∪ C = {F M, MF, MM}, and C ∩ \overline{B} = \{F M, MF \}.

2.3.

2.6. (a) We have

A = {(i, 2), (i, 4), (i, 6) | 1 ≤ i ≤ 6}
B = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6),
(5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3),

= (3, 4), (3, 5), (3, 6), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) = (6, 1), (6, 3), (6, 5)}

(b) For A, see above. \overline{C} = \{(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)\} (that is, both entries even). Then

A ∩ B = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

(both entries must be even). For A ∩ \overline{B}, we must have the first entry odd:

A ∩ \overline{B} = \{(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)\}.

For \overline{A} ∪ B, the second entry must be odd, or the sum must be even, so

\overline{A} ∪ B = \{(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (2, 2), (4, 2), (6, 2), (1, 3), (2, 3),
(3, 3), (4, 3), (5, 3), (6, 3), (2, 4), (4, 4), (6, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)
(2, 6), (4, 6), (6, 6)}.

For \overline{A} ∩ C, \overline{A} ∩ C = \overline{A}, since \overline{A} ⊂ C. This is the set

\overline{A} = \{(i, 1), (i, 3), (i, 5) | 1 ≤ i ≤ 6\}.

2.7. Call the male candidates M1, M2, M3, and the females W 1, W 2. The sample space contains the ten points

S = {M1M2, M1M3, M1W 1, M1W 2, M2M3, M2W 1, M2W 2, M3W 1, M3W 2, W 1W 2}.

Then

A = {M1M2, M1M3, M2M3},
B = {M1W 1, M1W 2, M2W 1, M2W 2, M3W 1, M3W 2, W 1W 2}.

So \overline{B} = A, A ∪ B = S, A ∩ B = ∅, and A ∩ \overline{B} = A.

2.8. (a) There were 36 + 9 − 3 = 42 students who were undergraduates, living off campus, or both. (We must subtract the three undergraduates living off campus since they were counted in both totals.)

(b) There were 36 − 3 = 33 undergraduates living on campus.

(c) There were 60 − 36 = 24 graduate students. 9 students total lived off-campus, three of which were undergraduates, so there were 6 graduate students living off campus. Hence there were 24 − 6 = 18 graduate students living on campus.