The table below shows the marks obtained by 278 candidates in an examination. a) State the modal class of the distribution. b) Construct a cumulative frequency table and draw a cumulative frequency cu

Welcome back Faith — missed you this week. Here are the solutions to your question: a) State the modal class of the distribution. Step 1: Identify the highest frequency in the "Number of candidates" column. The highest frequency is 81. Step 2: Determine the class interval corresponding to this frequency. The class interval for a frequency of 81 is 41-50. The modal class is 41-50. b) Construct a cumulative frequency table and draw a cumulative frequency curve. Step 1: Create a cumulative frequency table by adding the frequencies sequentially. Also, determine the upper class boundaries for plotting the curve. |c|c|c|c| Marks & Upper Class Boundary & Number of candidates (f) & Cumulative Frequency (cf) \\ 1-10 & 10.5 & 2 & 2 \\ 11-20 & 20.5 & 6 & 2+6=8 \\ 21-30 & 30.5 & 30 & 8+30=38 \\ 31-40 & 40.5 & 58 & 38+58=96 \\ 41-50 & 50.5 & 81 & 96+81=177 \\ 51-60 & 60.5 & 65 & 177+65=242 \\ 61-70 & 70.5 & 28 & 242+28=270 \\ 71-80 & 80.5 & 8 & 270+8=278 \\ Step 2: To draw the cumulative frequency curve (ogive), plot the upper class boundaries against their corresponding cumulative frequencies. The points to plot are: (10.5, 2), (20.5, 8), (30.5, 38), (40.5, 96), (50.5, 177), (60.5, 242), (70.5, 270), (80.5, 278). Start the curve from (0.5, 0) and connect the plotted points with a smooth curve. c) Use your curve to estimate: i) the median of the distribution Step 1: Calculate the position of the median. The total number of candidates (N) is 278. The median position is (N)/(2) = (278)/(2) = 139. Step 2: Locate the median on the cumulative frequency curve (or by interpolation). Find the mark corresponding to a cumulative frequency of 139. The 139th value falls within the 41-50 class (cumulative frequency 177). Using interpolation: Median = L + ((N)/(2) - cf_bf_m) × c Where L = 40.5 (lower boundary of median class), cf_b = 96 (cumulative frequency before median class), f_m = 81 (frequency of median class), c = 10 (class width). Median = 40.5 + ((139 - 96)/(81)) × 10 Median = 40.5 + ((43)/(81)) × 10 Median = 40.5 + 5.3086 Median ≈ 45.81 ii) the 40th percentile. Step 1: Calculate the position of the 40th percentile. The position is 0.40 × N = 0.40 × 278 = 111.2. Step 2: Locate the 40th percentile on the cumulative frequency curve (or by interpolation). Find the mark corresponding to a cumulative frequency of 111.2. The 111.2th value falls within the 41-50 class (cumulative frequency 177). Using interpolation: 40th Percentile = L + (Position - cf_bf_p) × c Where L = 40.5, cf_b = 96, f_p = 81, c = 10. 40th Percentile = 40.5 + ((111.2 - 96)/(81)) × 10 40th Percentile = 40.5 + ((15.2)/(81)) × 10 40th Percentile = 40.5 + 1.8765 40th Percentile ≈ 42.38 d) What is the probability that a candidate chosen at random scored a mark of at least 60.5%? Step 1: Identify the number of candidates who scored at least 60.5%. A mark of at least 60.5% means marks 60.5. This includes candidates in the 61-70 class and the 71-80 class. Number of candidates with mark 60.5 = (frequency of 61-70) + (frequency of 71-80) Number of candidates = 28 + 8 = 36. Step 2: Calculate the probability. The total number of candidates is 278. P(mark 60.5\%) = Number of candidates with mark 60.5Total number of candidates P(mark 60.5\%) = (36)/(278) P(mark 60.5\%) = (18)/(139) P(mark 60.5\%) ≈ 0.1295 Send me the next one 📸