0.2 Practice tests (1-4) and final exams  (Page 13/36)

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?

Practice test 2 solutions

Probability distribution function (pdf) for a discrete random variable

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

Mean or expected value and standard deviation

13 .

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 .

17 . ${\sigma }_x=\sqrt{54.91+2.78+96.49}=12.42$

Binomial distribution

18 . q = 1 – 0.65 = 0.35

19 .

1. There are a fixed number of trials.
2. There are only two possible outcomes, and they add up to 1.
3. The trials are independent and conducted under identical conditions.

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 . ${\sigma }_x=\sqrt{npq}=\sqrt{20(0.40)(0.60)}=2.19$

4.4: geometric distribution

28 .

1. A series of Bernoulli trials are conducted until one is a success, and then the experiment stops.
2. At least one trial is conducted, but there is no upper limit to the number of trials.
3. The probability of success or failure is the same for each trial.

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.