1. Five cards are drawn from an ordinary deck without replacement. Find the probability of getting a. All red cards π(πππ πππ πππππ ) = 26 25 24 23 22 7,893,600 253 × × × × = ππ 52 51 50 49 48 311,875,200 9996 b. All diamonds π(πππ ππππππππ ) = 13 12 11 10 9 154,440 33 × × × × = ππ 52 51 50 49 48 311,875,200 66640 c. All aces π(πππ πππ) = 4 3 2 1 24 1 × × × = ππ 52 51 50 49 311,875,200 12994800 2. For the given data below. 102 168 168 116 179 186 113 170 117 132 160 108 128 163 171 117 187 173 156 185 161 182 158 163 183 163 168 171 167 182 a. Construct a grouped frequency distribution. Use six classes. Range = Highest Value - Lowest Value = 187 - 102 = 85 Class Limits 102 - 116 116 - 129 130 - 143 144 - 157 158 - 171 172 - 187 Freque ncy 4 3 1 1 13 8 Cumulative Frequency 4 7 8 9 22 30 b. Draw a histogram and frequency polygon for the frequency distribution obtained from the data. Class Limits 102 - 116 116 - 129 130 - 143 144 - 157 158 - 171 172 - 187 Class Midpoint 109 123 137 144 165 179 Frequen cy 4 3 1 1 13 8 FREQUENCY 14 12 10 8 6 4 2 0 13 8 4 ΠΠΠΠΠΠΠΠ ΠΠ‘Π 109 3 123 1 1 137 144 165 CLASS MIDPOINT 179 14 12 10 8 6 4 2 0 109 123 137 144 165 179 ΠΠΠΠΠΠΠΠ ΠΠ‘Π c. Find the mean, median, mode, mid-range, range, variance and standard deviation for the data. 102 116 128 158 163 167 168 171 182 185 108 117 132 160 163 168 170 173 182 186 113 117 156 161 163 168 171 179 183 187 ππ’π ππ π‘ππππ mean = π 4967 mean = 30 mean = 156.5666667 Median = 165 Mode = 168 Standard Deviation, σ: 25.85 Count, N: 30 Sum, Σx: 4697 Mean, μ: 156.56666666667 Variance, σ2: 668.44555555556 π΄(π₯π − π)2 σ2 = π σ2 = (102 − 156.56666666667)2 + ...+ (182 − 156.56666666667)2 20053.366666667 σ2 = 30 σ2 = 668.44555555556 σ= √668.44555555556 SD = 25.854314060821 30 3. On the daily run of an express bus, the average number of passengers is 48. The standard deviation is 3. Assume the data are normally distributed. Find the probability that the bus will have a. Between 36 and 40 passengers b. Fewer than 42 passengers c. More than 48 passengers d. Between 43 and 47 passengers. 4. The data represent the heights in feet and the number of stories of the tallest buildings in Cleveland. a. Draw a scatter plot. y 60 No. of stories 50 40 30 y 20 10 0 0 200 400 600 800 1000 height of Building b. Find the value of r. c. x y 947 708 658 529 450 446 430 420 xy 57 52 46 40 31 28 24 26 53979 36816 30268 21160 13950 12488 10320 10920 x2 y2 896809 3249 501264 2704 432964 2116 279841 1600 202500 961 198916 784 184900 576 176400 676 419 5007 9 ∑ n π= π= π= π= π= π= 32 13408 175561 1024 336 203309 3049155 13690 n(∑xy) − (∑x)(∑y) √[n∑x 2 − (∑x) 2 ][n∑y 2 − (∑y) 2 )] 9(203,309) − (5,007)(336) √9(3,049,155) − (5007) 2 ][9(13,690) − (336) 2 )] 1,829,781 − 1,682,352 √27,442,395 − 25,070,049][123,210 − 112,896] 147,429 √(2,372,346)(10,314 ) 147,429 √24,468,376,644 147,429 156,424 π = 0.942497792 or 0.94 d. Test the significance of r at the 5% level and at the 1% level. d. Find the equation of the regression line and draw the line on the scatter plot, but only if r is significant. y 70 y = 0,0621x + 2,7601 R² = 0,8883 No. of stories 60 50 40 30 y 20 ΠΠΈΠ½Π΅ΠΉΠ½Π°Ρ (y) 10 0 0 200 400 600 800 height of Building 1000 e. Describe the nature of the relationship if one exists. There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero. f. Predict the number of stories in a 500 – feet building. Y = 0.0621x + 2.7601 Y = 0.0621(500)+2.7601 Y = 31.05+2.7601 Y = 33.8101
