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89 . Applying the law of large numbers, which sample mean would expect to be closer to the population mean, a sample of size ten or a sample of size 100?
Use this information for the next three questions. A manufacturer makes screws with a mean diameter of 0.15 cm (centimeters) and a range of 0.10 cm to 0.20 cm; within that range, the distribution is uniform.
90 . If X = the diameter of one screw, what is the distribution of X ?
91 . Suppose you repeatedly draw samples of size 100 and calculate their mean. Applying the central limit theorem, what is the distribution of these sample means?
92 . Suppose you repeatedly draw samples of 60 and calculate their sum. Applying the central limit theorem, what is the distribution of these sample sums?
1 . The domain of X = {English, Mathematics,….], i.e., a list of all the majors offered at the university, plus “undeclared.”
2 . The domain of Y = {0, 1, 2, …}, i.e., the integers from 0 to the upper limit of classes allowed by the university.
3 . The domain of Z = any amount of money from 0 upwards.
4 . Because they can take any value within their domain, and their value for any particular case is not known until the survey is completed.
5 . No, because the domain of Z includes only positive numbers (you can’t spend a negative amount of money). Possibly the value –7 is a data entry error, or a special code to indicated that the student did not answer the question.
6 . The probabilities must sum to 1.0, and the probabilities of each event must be between 0 and 1, inclusive.
7 . Let X = the number of books checked out by a patron.
8 . P ( x >2) = 0.10 + 0.05 = 0.15
9 . P ( x ≥ 0) = 1 – 0.20 = 0.80
10 . P ( x ≤ 3) = 1 – 0.05 = 0.95
11 . The probabilities would sum to 1.10, and the total probability in a distribution must always equal 1.0.
12 . $\overline{x}$ = 0(0.20) + 1(0.45) + 2(0.20) + 3(0.10) + 4(0.05) = 1.35
13 .
x | P ( x ) | x P ( x ) |
---|---|---|
30 | 0.33 | 9.90 |
40 | 0.33 | 13.20 |
60 | 0.33 | 19.80 |
14 . $\overline{x}$ = 9.90 + 13.20 + 19.80 = 42.90
15 .
P (
x = 30) = 0.33
P (
x = 40) = 0.33
P (
x = 60) = 0.33
16 .
x | P ( x ) | xP ( x ) | ( x – μ ) ^{2} P ( x ) |
---|---|---|---|
30 | 0.33 | 9.90 | (30 – 42.90) ^{2} (0.33) = 54.91 |
40 | 0.33 | 13.20 | (40 – 42.90) ^{2} (0.33) = 2.78 |
60 | 0.33 | 19.90 | (60 – 42.90) ^{2} (0.33) = 96.49 |
17 . ${\sigma}_{x}=\sqrt{54.91+2.78+96.49}=12.42$
18 . q = 1 – 0.65 = 0.35
19 .
20 . No, because there are not a fixed number of trials
21 . X ~ B (100, 0.65)
22 . μ = np = 100(0.65) = 65
23 . ${\sigma}_{x}=\sqrt{npq}=\sqrt{100(0.65)(0.35)}=4.77$
24 . X = Joe gets a hit in one at-bat (in one occasion of his coming to bat)
25 . X ~ B (20, 0.4)
26 . μ = np = 20(0.4) = 8
27 . ${\sigma}_{x}=\sqrt{npq}=\sqrt{20(0.40)(0.60)}=2.19$
28 .
29 . T T T T H
30 . The domain of X = {1, 2, 3, 4, 5, ….n}. Because you are drawing with replacement, there is no upper bound to the number of draws that may be necessary.
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