,{"id":"item-3","content":"Build the cumulative frequency column."},{"id":"item-4","content":"Find the row containing the middle position or positions."},{"id":"item-5","content":"Read off the value, or the class interval if the data is grouped."}],"order":8,"instruction":"Order the steps for finding the median from a frequency table.","correctOrder":["item-1","item-2","item-3","item-4","item-5"]},{"id":"card-9","type":"content","order":9,"title":"Grouped data and midpoints","blocks":[{"id":"block-1","content":"![Number line of plant height class intervals from 140 to 200 with midpoint markers at 145, 155, 165, 175 and 190, showing the last class is wider](https://assets.mathsgenie.co.uk/lesson-v2-img/inline/e5bf6633-c550-4ff0-ba34-f317637673eb.jpeg)\n\nGrouped tables give intervals such as $140 < h \\le 150$ instead of exact values. To estimate the mean, we pretend every value in a class sits at the midpoint of that interval."},{"id":"block-2","content":"The midpoint is halfway between the class boundaries. For $180 < h \\le 200$, the midpoint is calculated as:\n\n$$\\frac{180+200}{2}=190$$\n\nNote that the midpoint is 190, not 185, because this specific class is wider than the previous ones."},{"id":"block-3","content":"For grouped data, use midpoints $m$ instead of exact values. The formula is:\n\n$$\\text{estimated mean} = \\frac{\\sum (m \\times f)}{\\sum f}$$\n\nThis is only an estimate because the exact values inside each class are unknown to us."}]},{"id":"card-10","type":"fill_blank","order":10,"blanks":[{"id":"group_mean","isMath":true,"correctAnswer":"18"}],"sentence":"For groups $0 < t \\le 10$, $10 < t \\le 20$, and $20 < t \\le 30$ with frequencies 2, 3 and 5, the estimated mean is {{blank:group_mean}} minutes."},{"id":"card-11","type":"content","order":11,"title":"Worked example: estimated mean from grouped data","blocks":[{"id":"block-1","content":"The heights of 60 plants are grouped below. Add a midpoint column and then multiply each midpoint by its frequency.\n\n| Height | Frequency $f$ | Midpoint $m$ | $mf$ |\n|---|---:|---:|---:|\n| $140 < h \\le 150$ | 5 | 145 | 725 |\n| $150 < h \\le 160$ | 11 | 155 | 1705 |\n| $160 < h \\le 170$ | 16 | 165 | 2640 |\n| $170 < h \\le 180$ | 18 | 175 | 3150 |\n| $180 < h \\le 200$ | 10 | 190 | 1900 |"},{"id":"block-2","content":"Now add the two totals. The total frequency is 60, and the total of the $mf$ column is 10120.\n$$\\sum f = 60 \\qquad \\sum mf = 10120$$"},{"id":"block-3","content":"Divide to estimate the mean height.\n$$\\text{estimated mean} = \\frac{10120}{60} = 168.666\\ldots \\approx 168.7 \\, \\text{cm}$$\n\nThis answer is an estimate, not an exact mean. The real plant heights could be anywhere inside each class interval."}]},{"id":"card-12","type":"mcq","order":12,"options":[{"id":"a","text":"10","isCorrect":false},{"id":"b","text":"13","isCorrect":true},{"id":"c","text":"15","isCorrect":false},{"id":"d","text":"26","isCorrect":false}],"question":"A football team scored 54 goals in a season. Goals per game had frequencies 0 goals: 9, 1 goal: 13, 2 goals: $x$, 3 goals: 5, 4 or more goals: 0. What is $x$?","explanation":"The known goals contribute $0 \\times 9 + 1 \\times 13 + 3 \\times 5 = 28$, so the 2-goal games contribute $54 - 28 = 26$ goals. Since each of those games adds 2 goals, $2x = 26$ and $x = 13$; the open-ended row does not matter because its frequency is 0, and the most common wrong answer is D, which confuses 26 goals with 26 games.","correctOptionId":"b"},{"id":"card-13","type":"content","order":13,"title":"Median class and percentages","blocks":[{"id":"block-1","content":"A travel-time table for 100 people is shown below. For grouped data, the exact median is usually unknown, so we find the median class instead.\n\n| Time (minutes) | Frequency | Cumulative frequency |\n|---|---:|---:|\n| $0 < t \\le 10$ | 12 | 12 |\n| $10 < t \\le 20$ | 18 | 30 |\n| $20 < t \\le 30$ | 24 | 54 |\n| $30 < t \\le 40$ | 28 | 82 |\n| $40 < t \\le 50$ | 13 | 95 |\n| $50 < t \\le 60$ | 5 | 100 |"},{"id":"block-2","content":"With 100 people, the middle positio

Averages from Frequency Tables Lesson