Midterm Review (Solutions)-Edited.pdf

Midterm Review (Solutions)-Edited.pdf - STAT 1112 MIDTERM...

This preview shows page 1 out of 9 pages.

Unformatted text preview: STAT 1112 MIDTERM REVIEW QUESTIONS (SOLUTIONS) 1. For the following sample 3.6 5.9 4.0 7.1 8.2 7.1 7.1 8.0 a. Find the mean, median, and mode. Solution: The sample mean is equal to x = (3.6 + 5.9 + + 8.0 ) / 8 = 6.375 ≈ 6.4 . To find the median, we first arrange the days in order (from smallest to largest): 3.6 4.0 5.9 7.1 7.1 7.1 8.0 8.2 ↑ ↑ x 4 = 7.1 x5 = 7.1 x + x5 7.1 + 7.1 median = ~ = x= 4 = 7.1 2 2 mode = 7.1 (occurs three times) b. Find the range and the standard deviation. Solution: x x−x (x − x )2 3.6 − 2.8 7.84 5.9 − 0.5 0.25 4.0 − 2.4 5.76 7.1 0.7 0.49 8.2 1.8 3.24 7.1 0.7 0.49 7.1 0.7 0.49 8.0 1.6 2.56 Totals: The standard deviation is s = 21.12 21.12 = 8 −1 21.12 = 1.74 7 The range is equal to x max − x min = 8.2 − 3.6 = 4.6 Midterm Review Questions by Prof. A. Gokhman STAT 1112 c. Find P25 , P60 , and P75 . Solution: L25 = (n + 1) P 100 = (8 + 1) 25 100 = 2.25 th P25 = 4.0 + 0.25 (5.9 − 4.0 ) = 4.475 ≈ 4.5 L60 = (n + 1) P 100 = (8 + 1) 60 100 = 5.4 th P60 = 7.1 + 0.4 (7.1 − 7.1) = 7.1 L75 = (n + 1) P 100 = (8 + 1) 75 100 = 6.75 th P75 = 7.1 + 0.75 (8.0 − 7.1) = 7.775 ≈ 7.8 d. Construct the box-plot. Solution: -5 -4 -3 -2 -1 0 1 2 4 5 2. Let P( A) = 0.5 , P(B ) = 0.4 , and P( A and B ) = 0.2 . Use Venn diagrams. a. Find P( A or B ) . Solution : P ( A or B ) = P ( A) + P (B ) − P( A and B ) = 0.5 + 0.4 − 0.2 = 0.7 2 Midterm Review Questions by Prof. A. Gokhman STAT 1112 3 b. Find P(A and B ) . Solution : P (A and B ) = P ( A and not B ) = P ( A or B ) − P (B ) = 0.7 − 0.4 = 0.3 Alternatively : P (A and B ) = P( A) − P ( A and B ) = = 0.5 − 0.2 = 0.3 c. Calculate P(A and B ) . Solution : P (A and B ) = 1 − P ( A or B ) = = 1 − 0.7 = 0.3 3. The problem uses a collection containing responses by students of Toronto Pushkin High School to a 2011 survey. The question was asked: “How do you feel about your school?” The following table summarizes the results. Male (M) Female (F) Total I hate school (A) 17 37 54 Don’t like very much (B) 35 64 99 Like a bit (C) 103 138 241 Like quite a bit (D) 177 169 346 Like very much (E) 180 109 289 Total 512 517 1029 a. Find P( A) . Solution : P( A) = 54 = 0.052 1029 Midterm Review Questions by Prof. A. Gokhman STAT 1112 b. Find P(B and F ) . Solution : P(B and F ) = 64 = 0.062 1029 c. Find P(C or M ) . Solution : P(C or M ) = P(C ) + P(M ) − P(C and M ) = = 241 512 103 + − = 1029 1029 1029 650 = 0.632 1029 Alternatively : 17 + 35 + 103 + 177 + 180 + 138 650 == = 0.632 1029 1029 d. Calculate P(C or D ) . Solution : Since C and D are mutually exclusive, P(C or D ) = P(C ) + P(D ) = 241 346 587 + = = 0.570 1029 1029 1029 ( ) ( ) e. Calculate P F B . P (F and B ) 64 1029 64 Solution : P F B = = = 0.646 = 99 1029 99 P (B ) Alternatively : 64 = 0.646 P F B = 99 ( ) ( ) f. Calculate P A M . = P ( A and M ) = 17 1029 = 17 = 0.033 Solution : P A M 512 1029 512 P (M ) Alternatively : 17 = 0.033 P A M = 512 4 Midterm Review Questions by Prof. A. Gokhman STAT 1112 5 4. Suppose that a certain disease is present in 10% of the population, and that there is a screening test designed to detect this disease if present. The test does not always work perfectly. In 5% of the cases the test is negative when the disease is present, and in 3% of the cases the test is positive when the disease is absent. Draw a probability tree. What is the probability that a randomly selected patient still has the disease if the test is negative? Solution: The probability tree is ( P D P (D and − ) 0.005 0.005 ) = = = = 0.0057 − 0.005 + 0.873 0.878 P( − ) 5. The paired data below consists of sizes of the households and weights (in kilograms) of discarded paper Household size 2 3 3 7 4 2 1 5 Paper 1.09 3.43 4.33 5.00 3.96 3.16 3.10 5.18 The Pearson Linear Correlation Coefficient r = 0.736 and the regression equation is yˆ = 1.973 + 0.499 x . a. Find the value of the coefficient of determination and interpret it. Solution: The coefficient of determination is r2 = (0.736)2 = 0.542 It means that 54.2% of the total variation in weights of discarded papers can be explained by the variation in the sizes of households; 45.8% of the total variation in weights of discarded papers is attributed to other factors. b. Predict the amount of discarded paper for a family of 6. Solution : yˆ = 1.973 + 0.499 × 6 = 1.973 + 2.994 = 4.967 Answer : about 5 kg Midterm Review Questions by Prof. A. Gokhman STAT 1112 6 6. In a survey, the Canadian Automobile Association (CAA) found that 10% of its members bought their cars at a used-car lot. a. If 25 CAA members are selected at random, what is the probability that no more than 4 of them bought their cars at a used-car lot? Solution : Using the table : n = 25 , k = 4 , p = 10% = 0.10 P(x ≤ 4 ) = 0.9020 b. If 25 CAA members are selected at random, what is the probability that exactly 4 of them bought their cars at a used-car lot? Solution : Using the table : n = 25 , k = 4 , p = 10% = 0.10 P( x = 4) = P(x ≤ 4 ) − P( x ≤ 3) = 0.9020 − 0.7636 = 0.1384 c. If 25 CAA members are selected at random, what is the probability that more than 4 of them bought their cars at a used-car lot? Solution : Using the table : n = 25 , k = 4 , p = 10% = 0.10 P( x > 4) = 1 − P( x ≤ 3) = 1 − 0.7636 = 0.2364 d. If 25 CAA members are selected at random, what is the probability that less than 4 of them bought their cars at a used-car lot? Solution : Using the table : n = 25 , k = 4 , p = 10% = 0.10 P( x < 4) = P(x ≤ 3) = 0.7636 7. A used-car dealership has found that the length of time before a major repair is required on the cars it sells is normally distributed, with a mean equal to 10 months and a standard deviation of 3 months. a. What is the probability that a randomly selected car will need repair in 9 months or earlier? Midterm Review Questions by Prof. A. Gokhman STAT 1112 7 b. What is the probability that a randomly selected car will not need any repair for at least a year? 12 − 10 = 0.67 3 P( x ≥ 12 ) = 1 − P( x ≤ 12) = 1 − P( z ≤ 0.67) = 1 − 0.7486 = 0.2514 Solution : z = c. What is the probability that for a randomly selected car the length of time before a major repair needed will fall between 12 and 15 months? 15 − 10 12 − 10 = 1.67 = 0.67 and z2 = Solution : z1 = 3 3 P(12 ≤ x ≤ 15) = P ( x ≤ 15) − P( x ≤ 12) = P( z ≤ 1.67) − P( z ≤ 0.67) = = 0.9525 − 0.7486 = 0.2039 Midterm Review Questions by Prof. A. Gokhman STAT 1112 8 d. If a dealer wants only 5% of the cars to fail before the end of the guarantee period, for how many months should the cars be guaranteed? Solution : x = µ + zσ = 10 + (− 1.645)3 = 5.065 Answer : about 5 months e. The dealer randomly selected 20 cars. What is the probability that the average time before major repair exceeds 9 months? (− 1) = −1.49 x−µ 9 − 10 Solution : z = = = σ n 3 20 0.6708 P( x > 9 ) = 1 − P( x < 9) = 1 − P( z < −1.49) = 1 − 0.0681 = 0.9319 Midterm Review Questions by Prof. A. Gokhman STAT 1112 9 DEFINITIONS: 1. 2. 3. 4. 5. 6. 7. 8. Types of data: qualitative, quantitative; discrete, continuous. Levels of measurement: nominal, ordinal, interval. Methods of sampling: random, convenience, systematic, cluster, stratified. Bar graph, pie-chart, frequency table and frequency histogram, ogive, scatter diagram. Measures of centre: mean, median, mode. Measures of variation: range, variance, standard deviation. Measures of position: percentiles. Rules of probability: complementary probability, addition rule, multiplication rule, conditional probability, probability trees. 9. Binomial distribution. 10. Correlation and regression. Coefficient of determination. 11. Normal distribution. Z-score. ...
View Full Document

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern