Step 1: Find the total number of goals (Goals × Frequency). 0 × 2 = 0 1 × 3 = 3 2 × 4 = 8 3 × 1 = 3 Total goals = 0 + 3 + 8 + 3 = 14 Step 2: Find the total frequency (number of matches). Total matches = 2 + 3 + 4 + 1 = 10 Step 3: Calculate the mean. Mean = 14 ÷ 10 = 1.4 The mean number of goals per match is 1.4.Step 1: The data is already in order. 12, 13, 13, 14, 14, 14, 15, 15, 16 Step 2: Find the median. There are 9 values, so the median is the 5th value. Median = 14 Step 3: Find the mode. 14 appears 3 times (the most). Mode = 14 Step 4: Comparison. Both median and mode are 14. Either measure represents the typical age well since they are equal. The mode is useful because it shows the most common age.Step 1: Use the mean formula to find the total. Mean = Sum ÷ Count 12 = Sum ÷ 5 Sum = 12 × 5 = 60 Step 2: Find the sum of the known numbers. 8 + 10 + 15 + 13 = 46 Step 3: Find the missing number. Missing number = Total sum - Sum of known numbers Missing number = 60 - 46 = 14 Therefore, the fifth number is 14. Check: (8 + 10 + 15 + 13 + 14) ÷ 5 = 60 ÷ 5 = 12 ✓Finding the Mean: Sum = 45 + 52 + 38 + 67 + 52 + 48 + 52 = 354 Mean = 354 ÷ 7 = 50.57 (to 2 decimal places) Finding the Median: Ordered data: 38, 45, 48, 52, 52, 52, 67 Middle value (4th value) = 52 Median = 52 Finding the Mode: 52 appears 3 times (the most) Mode = 52 Recommendation: I would use the mode (52) or median (52) for advertising. Both represent the typical day well. The mean (50.57) is affected by the low value of 38. Using 52 customers per day is both accurate and easy to communicate.Original Data (12 students): Mean: Sum = 45 + 67 + 72 + 72 + 75 + 78 + 80 + 82 + 85 + 88 + 90 + 92 = 926 Mean = 926 ÷ 12 = 77.17 (to 2 dp) Median: Data is ordered. Middle values are 6th and 7th (78 and 80). Median = (78 + 80) ÷ 2 = 79 Without the 45 score (11 students): New sum = 926 - 45 = 881 New mean = 881 ÷ 11 = 80.09 (to 2 dp) Comment: Removing the outlier (45) increased the mean from 77.17 to 80.09. The mean increased by about 3 marks. This shows that the mean is sensitive to extreme values (outliers). The median (79) would change less if we removed this outlier.Dot Plot Description: Draw a number line from 12 to 30 with intervals of 3. Place dots above each value for each occurrence: 12: • (1 dot) 15: • (1 dot) 18: •• (2 dots) 21: • (1 dot) 24: ••• (3 dots) 27: • (1 dot) 30: • (1 dot) Calculations: Mean: Sum = 12 + 15 + 18 + 18 + 21 + 24 + 24 + 24 + 27 + 30 = 213 Mean = 213 ÷ 10 = 21.3 cm Median: Data is ordered. Middle values are 5th and 6th (21 and 24). Median = (21 + 24) ÷ 2 = 22.5 cm Mode: 24 appears 3 times (the most). Mode = 24 cmStep 1: Find the total number of students. Total = 5 + 12 + 8 + 4 + 1 = 30 students Step 2: Calculate the mean. Total siblings = (0×5) + (1×12) + (2×8) + (3×4) + (4×1) = 0 + 12 + 16 + 12 + 4 = 44 Mean = 44 ÷ 30 = 1.47 siblings (to 2 dp) Step 3: Find the median. Total frequency = 30, so median is average of 15th and 16th values. Cumulative frequencies: 0 siblings (5), 1 sibling (5+12=17) The 15th and 16th values both fall in the "1 sibling" category. Median = 1 sibling Step 4: Find the mode. The highest frequency is 12 (for 1 sibling). Mode = 1 sibling Summary: Mean = 1.47, Median = 1, Mode = 1
