Midterm Review (Solutions)-Edited.pdf

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

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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. ...
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