1. The data in the following table represent the number of miles per gallon achieved on the highway for compact cars for the model year 2005.

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1 Millersville University Name Answer Key Department of Mathematics MATH 130, Elements of Statistics I, Test 2 March 5, 2010, 10:00AM-10:50AM Please answer the following questions. Your answers will be evaluated on their correctness, completeness, and use of statistical concepts we have covered. Please show all work and write out your work neatly. Answers without supporting work will receive no credit. The point values of the problems are listed in parentheses. 1. The data in the following table represent the number of miles per gallon achieved on the highway for compact cars for the model year (a) (15 points) Find the five-number summary for these data. minimum: 18 Q 1 : The index for the 25th percentile is i = (63 + 1) = 16, thus Q 1 = 26. M: The index for the 50th percentile (median) is i = 50 (63 + 1) = 32, thus 100 M = 30. Q 3 : The index for the 75th percentile is i = (63 + 1) = 48, thus Q 3 = 32. maximum: 46

2 (b) (5 points) Construct a box plot for the data IQR: Q 3 Q 1 = = 6 lower fence: Q 1 1.5(IQR) = (6) = 17 upper fence: Q (IQR) = (6) = 41

3 2. (3 points each) The following table represents the medal tallies of the top eight countries in the 2010 Winter Olympics in Vancouver. Country Gold Silver Bronze Total United States Germany Canada Norway Austria Russian Federation Korea China Total (a) If a medal is randomly selected from the top eight countries, what is the probability it is gold? P(gold) = (b) If a medal is randomly selected from the top eight countries, what is the probability it was won by Korea? P(Korea) = (c) If a medal is randomly selected from the top eight countries, what is the probability it was gold and was won by Korea? P(gold and Korea) =

4 (d) If a medal is randomly selected from the top eight countries, what is the probability it was gold or was won by Korea? P(gold or Korea) = P(gold) + P(Korea) P(gold and Korea) = = (e) If a bronze medal is randomly selected from the top eight countries, what is the probability it was won by Norway? P(Norway bronze) = 6 50 = (f) If a medal that was won by Norway is randomly selected, what is the probability that it is silver? P(silver Norway) =

5 3. (4 points each) The following data represent (in thousands) the number of 5- to 9-yearold females in the United States in X (age) Freq. Rel. Freq (a) Verify that the relative frequency column is a probability distribution for the ages. We note that for each value of X, the relative frequency (which in this case is interpreted as the probability) is a number between 0 and 1. Also the sum of the probabilities is 1. P(X) = = 1 (b) Find the mean of the random variable X. µ X = (X P(X)) = (5)(0.1929) + (6)(0.1956) + (7)(0.2003) + (8)(0.2036) + (9)(0.2076) = (c) Find the standard deviation of the random variable X. σ 2 X = (X 2 P(X)) µ 2 X = (5 2 )(0.1929) + (6 2 )(0.1956) + (7 2 )(0.2003) + (8 2 )(0.2036) + (9 2 )(0.2076) (7.0374) 2 = = σ X =

6 4. (6 points) The Pennsylvania Mega Millions lottery requires a player to select (without repetition) five numbers from 1 to 56, and then one number (called the Mega Ball) from 1 to 46. The Mega Ball number picked can be the same as one of the first five numbers picked. In how many different ways can the Mega Millions lottery ticket numbers be selected? number of selections = ( 56 C 5 )( 46 C 1 ) = ( )(46) = 175, 711, (4 points each) According to the Higher Education Research Institute, 55% of college freshmen in 4-year colleges and universities are female. Suppose 15 freshmen are randomly selected and the number of females in the sample is observed. (a) Explain why the number of females in samples of 15 freshmen is a binomial random variable. Since the samples are of size 15, there are a fixed number of trials per sample. Since the freshmen are randomly sampled, the gender of one freshman in the sample does not affect the gender of another freshman in the sample. Thus the trials are independent. There are only two outcomes per trial, either a freshman is male or female. The probability of being female on a trial is fixed at 55%. (b) Find the probability that exactly 10 of the freshmen are female. According to the binomial probability formula using n = 15, p = 0.55, x = 10, P(X = 10) = ( 15 C 10 )(0.55) 10 (1 0.55) = The same result can be found in Table II. (c) Find the probability that less than 8 of the freshmen are female. Since the binomial random variable is a discrete random variable according to Table III. P(X < 8) = P(X 7) =

7 (d) Find the probability that between 7 and 12 freshmen, inclusive, are female. P(7 X 12) = P(X 12) P(X 6) = according to Table III. 6. (4 points each) Before a criminal trial can begin, a jury of twelve impartial people must be selected. The pool of potential jurors, contains 20 men and 16 women. (a) How many different 12 member juries can be formed from this jury pool? Since there are = 36 people in the jury pool, the number of 12 member juries which can be formed is 36C 12 = 1, 251, 677, 700. (b) What is the probability of getting an all male jury of 12? P(12 male) = 20 C 12 36C 12 = 125, 970 1, 251, 677, (c) What is the probability of getting an all female jury of 12? P(12 female) = 16 C 12 36C 12 = , 251, 677, 700

8 (d) What is the probability of getting a jury of 12 consisting of 6 men and 6 women? P(6 male and 6 female) = ( 20C 6 )( 16 C 6 ) 36C 12 = (38760)(8008) 1, 251, 677, (4 points) Lunesta is a drug for people with insomnia. One of the most common side effects of the drug is headache which occurs in 21% of people taking it. (a) If 150 users of Lunesta are randomly selected, what is the expected number of users who would experience headache as a side effect? Since the number of users experiencing headache as a side effect of Lunesta is a binomial random variable with n = 150 and p = 0.21, then µ X = np = (150)(0.21) = (b) If 150 users of Lunesta are randomly selected, what is the standard deviation in the number of users who would experience headache as a side effect? σ X = np(1 p) = 150(0.21)(1 0.21) 5.0 (c) Would it be unusual to observe 45 reports of headache out of 150 users of Lunesta? Since np(1 p) we may apply the Empirical Rule to this situation. The typical number of reports of headache out of 150 would be within two standard deviations of the mean, thus between the numbers µ X 2σ X = (5.0) = 21.5 and µ X + 2σ X = (5.0) = Since 45 is larger than 41.5 then this would be an unusual number of reports of headache out of a sample of 150 users.