This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — here's the solution for question 12: a) Construct a grouped frequency distribution table using the class interval 20-29, 30-39, 40-49... Step 1: Define class intervals and tally the given marks. The marks are: 29 52 37 79 63 48 65 46 54 32 51 66 74 24 35 61 58 39 49 53 45 58 56 38 59 28 50 49 67 56 37 45 79 68 43 66 26 47 55 71 The class intervals are: 20-29 30-39 40-49 50-59 60-69 70-79 Step 2: Count the frequency for each class interval. • 20-29: 29, 24, 28, 26 (4 marks) • 30-39: 37, 32, 35, 39, 38, 37 (6 marks) • 40-49: 48, 46, 49, 45, 49, 43, 45, 47 (8 marks) • 50-59: 52, 54, 51, 53, 58, 58, 56, 59, 50, 56, 55 (11 marks) • 60-69: 63, 65, 66, 61, 67, 68, 66 (7 marks) • 70-79: 79, 74, 79, 71 (4 marks) Step 3: Construct the frequency distribution table. |c|c|c| Class Interval & Tally & Frequency (f) \\ 20-29 & |||| & 4 \\ 30-39 & ||||| | & 6 \\ 40-49 & ||||| ||| & 8 \\ 50-59 & ||||| ||||| | & 11 \\ 60-69 & ||||| || & 7 \\ 70-79 & |||| & 4 \\ Total & & 40 \\ b) Using an assumed mean of 54.5, calculate, correct to the nearest whole number, the: (i) mean mark; (ii) standard deviation of the distribution. Step 1: Create an extended frequency table to calculate class marks (x), deviations (d), fd, and fd^2. The assumed mean A = 54.5. The class mark x is the midpoint of each interval. For example, for 20-29, x = (20+29)/(2) = 24.5. The deviation d = x - A. |c|c|c|c|c|c|c| Class Interval & f & x & d = x - 54.5 & fd & d^2 & fd^2 \\ 20-29 & 4 & 24.5 & -30 & -120 & 900 & 3600 \\ 30-39 & 6 & 34.5 & -20 & -120 & 400 & 2400 \\ 40-49 & 8 & 44.5 & -10 & -80 & 100 & 800 \\ 50-59 & 11 & 54.5 & 0 & 0 & 0 & 0 \\ 60-69 & 7 & 64.5 & 10 & 70 & 100 & 700 \\ 70-79 & 4 & 74.5 & 20 & 80 & 400 & 1600 \\ Total & f = 40 & & & fd = -170 & & fd^2 = 9100 \\ (i) Calculate the mean mark. The formula for the mean using an assumed mean is x = A + ( fd)/( f). x = 54.5 + (-170)/(40) x = 54.5 - 4.25 x = 50.25 Correct to the nearest whole number, the mean mark is 50. (ii) Calculate the standard deviation of the distribution. The formula for standard deviation using an assumed mean is = sqrt(( fd^2)/( f) - (( fd)/( f))^2). = sqrt((9100)/(40) - ((-170)/(40))^2) = sqrt(227.5 - (-4.25)^2) = sqrt(227.5 - 18.0625) = sqrt(209.4375) ≈ 14.472 Correct to the nearest whole number, the standard deviation is 14. What's next?