STAT 152 PRACTICE EXERCISES FOR EXAM 1 SUMMER 2013 The following problems are a good practice for the first midterm. Your homework problems are also great for midterm practice. The problems (and their number) on the exam, may be different from those on this practice sheet or homeworks. Exam will cover all material up to and including section 6-3 (Applications of normal distributions). Each multiple choice problem is worth 1 point. You must circle your answers clearly on the exam paper. In order for your exam to be graded, you must write your name in BLOCK CAPITAL LETTERS, LAST NAME FIRST in the NAME field, your R number (if you have it with you). You may use a calculator, the classroom computer, and one piece of 8.5x11in paper with handwritten notes on both sides. No phones, pda’s laptop’s or other electronic devises are allowed. Your phones must be switched off until you leave the room after the exam. GOOD LUCK (1) Which of the following variables is qualitative? (a) weight of a cat in kg (e) none of those in (a) -(d)

(b) marital status

(c) family income in $

(d) mountain top elevation in m

(2) Suppose you missed a STAT 152 exam. The professor said that she can assign you a grade which was a measure of center for the class grades. The histogram of class grades is skewed to the left. Which would you chose: the mean or the median? Of course you want the larger one! (a) mean

(b) median

(3) Compute the minimum, maximum, and the three quartiles (5-number summary) of the weights, in grams, of the orange M& Ms. Also compute the mean and standard deviation of the weights of the orange M & Ms. Use the data on weights of 100 M& Ms given in Data Set 13 in Appendix B. (I) minimum weight: ......................... maximum weight: .................. (II) First quartile of weights: ....................... Median ........................ Third quartile: .................... (III) Mean ....................... and standard deviation of weights:....................... (4) The following data represent grades on a quiz in a Math class: 5, 2, 1, 2, 7, 9, 10, 11, 8, 4, 7, 6. Please find the following: (I) which percentile is score of 9; (a) 58th

(b) 83rd

(c) 90th

(d) 75th

(e) other than given in (a) -(d)

(II) 90th percentile of the scores (a) 11

(b) 10

(c) 9

(d) 8

(e) other than given in (a) -(d) 1

2

(5) Which is relatively better: a score of 85 on a psychology test or a score of 45 of an economics test? Scores on the psychology test have a mean of 90 and standard deviation of 10. Scores on the economics test have a mean of 55 and standard deviation of 5. (a) score on the psychology test is better; (b) score on the economics test is better (6) The probability that Burt can solve the New York Times puzzle in an hour is 0.4. The probability that Mary can do that is 0.6. Burt and Mary work on the puzzle independently. Find the probability that I) both of them can solve the puzzle in an hour; (a) 0.18

(b) 0.24

(c) 1

(d) 0.4

(e) other than given in (a) -(d)

II) neither can solve the puzzle in an hour; (a) 0.36

(b) 0.3

(c) 0.24

(d) 1

(e) other than given in (a) -(d)

III) only Mary can solve the puzzle in an hour; (a) 0.28

(b) 0.36

(c) 1

(d) 0.24

(e) other than given in (a) -(d)

IV) Mary or Burt can solve the puzzle in an hour; (a) 0.24

(b) 0.36

(c) 0.76

(d) 1

(e) other than given in (a) -(d)

(7) Your winnings on a lottery is a random variable with the following probability distribution: x 100 200 1,000 0 P(X=x) 0.4 0.3 ? 0.2 I) Find the probability that you win $1,000. (a) 0.05

(b) 0.15

(c) 0.01

(d) 0.1

(e) other than given in (a) -(d)

II) What is the probability that you will win at least $200? (a) 0.7

(b) 0.4

(c) 0.3

(d) 0.2

(e) other than given in (a) -(d)

III) What is your expected winning? (a) 100

(b) 150

(c) 200

(d) 325

(e) other than given in (a) -(d)

3

(8) The data below represent the results for a test for a certain disease. Assume one individual from the group is randomly selected. Truth (Disease present) Yes No Test Positive 126 31 Negative 12 131 (a) Find the probability of finding someone who tests negative given that the individual had the disease. (b) Find the probability of finding someone who tests positive AND has the disease. (c) Find the probability of finding someone who tests positive OR has the disease. (9) The TV show 60 Minutes has been successful for many years. That show recently had a share of 20%, meaning that among the TV sets in use, 20% were tuned to 60 Minutes. Assume that an advertiser wants to verify that 20% share value by conducting its own survey, and a pilot survey begins with 10 households having TV sets in use at the time of a 60 Minutes broadcast. (a) Find the probability that none of the households are tuned to 60 Minutes. (b) Find the probability that at least one household is tuned to 60 Minutes. (c) Find the probability that at most one household is tuned to 60 Minutes? (d) How many households should they expect in the sample of 10 to be tuned to 60 Minutes? (10) A physical fitness association is including the mile run in its secondary-school fitness test for boys. The time for this event for boys in secondary school is normally distributed with mean 500 seconds and a standard deviation of 45 seconds. (Hint: Draw appropriate graphs of the normal curve for this problem.) (I) What percentage of the boys finish the mile run in less than 400 seconds? (a) 1.32%

(b) 8.68%

(c) 98.68%

(d) 2.43%

(e) other than given in (a) -(d)

(II) What percentage of the boys take between 400 and 590 seconds to finish the mile run? (a) 1.8%

(b) 3.6%

(c) 96.4%

(d) 0%

(e) other than given in (a) -(d)

(III) If the association wants to designate the fastest 1% as “excellent,” what time (in sec) should the association set for this criterion? a) 605.3

(b) 650.2

(c) 640.56

(d) 604.85

(e) other than given in (a) -(d)

4

SOLUTIONS TO THE PRACTICE EXERCISES FOR EXAM 1 (1) Marital status is the only categorical variable on the list. (2) Since the histogram of the grades is skewed to the left, the median is larger than the mean. I would choose the median grade. (3) Weights of M&Ms. (I) minimum weight: ....0.735g................. maximum weight: ....0.977g........... (II) First quartile of weights: ....0.84g............... Median ...0.863g.............. Third quartile: ...0.881g............... (III) Mean ...0.8578g............... and standard deviation of weights:......0.0501g.............. (4) Percentiles of quiz scores. There are 12 scores total. (I) There are 9 scores smaller than the score of 9. Thus, 9 is the following percentile: (9/12)100 = 75th. (II) 90th percentile of the scores? L = (90/100)12 = 10.8 rounded UP L = 11. 11th score from the smallest one up is 10. So, 90th percentile of the scores is 10. (5) To find relative standing of the exam scores, we need z-scores for each exam score. For psychology exam 45−55 zp = 85−90 = −2. Since the z-score for the psychology exam zp is larger 10 = −0.5. For economics exam ze = 5 than the z-score ze for the economics exam, the score on the psychology exam is relatively better than the score on the economics exam. (6) Let B event that Burt can solve the puzzle in an hour; M=event that Mary can do that. P(B)=0.4, P(M)=0.6, B, M independent. I) both can solve the puzzle; P(B and M)=P(B)· P(M)= (0.4) · (0.6) =0.24. II) neither can solve the puzzle; P(not B and not M) = P(not B) · P(not M)= (1-0.4)(1-0.6)=0.24 III) only Mary can solve the puzzle; P(not B and M) = P(not B) · P(M)= (1-0.4)(0.6)=0.36 IV) Mary or Burt can solve the puzzle; P( B or M) = P( B) + P( M)-P(B and M)= 0.4 +0.6 -(0.4)(0.6)=0.76

5

(7) Winnings on a lottery X: I) Find the probability that you win $1,000. P(X=1000)=1-(0.4+0.3+0.2)= 1- 0.9=0.1 II) What is the probability that you will win at least $200? P(X ≥ 200)=0.3+0.1=0.4 III) What is your expected winning? EX=(100)(0.4)+(200)(0.3)+(1000)(0.1)=200 (8) (a) The total number of people in the group surveyed = 300. Number of people who had the disease (disease present) = 126+12=138. Number of people who had the disease and tested negative =12. We need conditional probability.

P (Test negative| Disease present) =

P(test negative and disease present) 12/300 12 = = ' 0.087 P(disease present) 138/300 138

(b) Find the probability of finding someone who tests positive AND has the disease.

P (Test positive AND Disease present) =

126 = 0.42 300

(c) Find the probability of finding someone who tests positive OR has the disease.

P (Test positive OR Disease present) = using Addition rule = = P (Test positive) + P ( Disease present) − P (Test positive AND Disease present = 126 + 31 126 + 12 126 = + − = 0.56 300 300 300 (9) The 60 Minutes popularity question. Let X = number of households tuned to 60 Minutes out of 10 surveyed. Then X has a Binomial distribution with n = 10 and p = 0.2, that is X ∼ Bin(10, 0.2). (a) Find the probability that none of the households are tuned to 60 Minutes. P(X=0) =0.107 (Tables) or = 0.107374 (MINITAB)

6

(b) Find the probability that at least one household is tuned to 60 Minutes. P(X ≥ 1) = 1- P(X < 1)= 1 - P(X = 0)= 1 - 0.107 = 0.893 (c) Find the probability that at most one household is tuned to 60 Minutes? P(X ≤ 1) = P(X = 0 or X = 1) = 0.107 + 0.268. (d) How many households should they expect in the sample of 10 to be tuned to 60 Minutes? The expected value of X is 10(0.2) = 2 households. (10) Times for the mile run. T=time a boy in secondary school takes to do a mile run. T ∼ N (500, 45). (I) Percent that finish in less than 400 sec.

T − 500 400 − 500 < ) = P (Z < −2.22) = 0.0132 (used standarization and Tables) 45 45 Using MINTAB get P(T < 400)= 0.0131341. P (T < 400) = P (

(II) Percent that finish between 400 and 590 sec.

P (400 < T < 590) = P (−2.22 < Z < 2) = 0.9772 − 0.0132 = 0.964 (used standardization and Tables)

(III) Find 99th percentile of the run times. 99th percentile of the standard normal distribution (table) is 2.33. T − 500 45 solving for T, we get T= 604.85, so the 99th percentile of the run times in 604.85 sec. This was done using the Tables. Using MINITAB we get the 99th percentile of the run times equal to 604.686. 2.33 =