n value: \n$$\\bar{x} = 1.5$$"},{"type":"text","order":3,"title":"State the final answer","content":"The mean number of pets per family is **$1.5$**.\n\n*Note: If you are showing your working on an exam to guarantee full marks, always write down your calculations:*\n$$\\text{Mean} = \\frac{\\sum (\\text{Pets} \\times \\text{Frequency})}{\\sum \\text{Frequency}} = \\frac{90}{60} = 1.5$$"}]}],"generatedAt":"2026-06-12T13:05:35.732Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"59f50585-fc8a-47fe-bced-b6bcd915fcbd","revisionId":"1e279396-7c0c-468b-a569-65108ca9cc62","questionNumber":null,"questionType":"LA","specification":"The table shows information about the age, $a$ years, of 40 members of a local chess club.\n\n| Age ($a$ years) | Frequency |\n| :--- | :--- |\n| $10 < a \\leq 20$ | 8 |\n| $20 < a \\leq 30$ | 15 |\n| $30 < a \\leq 40$ | 11 |\n| $40 < a \\leq 60$ | 6 |","difficulty":3,"difficultyBand":"hard","totalMarks":4,"sourceTitle":null,"calculatorUse":"non_calc","parts":[{"id":"fa07b99a-636b-4fd4-b766-0073868924db","label":"a","content":"Find the class interval that contains the median age.","markscheme":"- Total frequency is 40. Median is between 20th and 21st member. Cumulative frequencies are 8, 23. Both 20th and 21st members are in $20 < a \\leq 30$. B1","marks":1,"order":0,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Identify the correct interval $20 < a \\leq 30$","code":"B","acceptable":["20 < a <= 30","20-30"]}],"finalAnswers":[],"generatedAt":"2026-05-06T14:00:47.070Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"cumulative_frequency","label":"Use cumulative frequency","steps":[{"type":"text","order":0,"title":"Locate the middle members","content":"There are $40$ members, so the median lies between the $20$th and $21$st members when the ages are placed in order. We only need to identify which class contains these members; we do not need to estimate the median age.","highlights":[{"text":"40 members","location":"specification"}]},{"type":"text","order":1,"title":"Build cumulative frequencies","content":"The cumulative frequencies are $8$ and then $8+15=23$. This means the second class contains members $9$ to $23$, so it contains both the $20$th and $21$st members.\n\nTherefore both middle members, and hence the median age, are in $20M1\n- $\\sum f x = (8 \\times 15) + (15 \\times 25) + (11 \\times 35) + (6 \\times 50) = 120 + 375 + 385 + 300 = 1180$. Calculation $1180 / 40$ M1\n- Final answer $29.5$ A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct midpoints (15, 25, 35, 50) identified.","code":"M"},{"id":"p2","marks":1,"description":"Use of sum of (frequency x midpoint) divided by 40.","code":"M"},{"id":"p3","marks":1,"description":"Correct calculation of mean (29.5).","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:00:47.070Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"grouped_mean","label":"Use class midpoints","steps":[{"type":"text","order":0,"title":"Find each midpoint","content":"Because the exact ages are unknown, represent each group by its midpoint, found using\n\n$$\\text{midpoint}=\\frac{\\text{lower boundary}+\\text{upper boundary}}{2}.$$ \n\nThis gives midpoints $15$, $25$, $35$ and $50$. Finding all four correctly earns the first **M1** mark.","highlights":[{"text":"Work out an estimate for the mean age of the members.","location":"specification"}]},{"type":"text","order":1,"title":"Multiply by frequency","content":"Multiply each midpoint, $x$, by

Number of Pets	Frequency
0	15
1	20
2	12
3	8
4	3
5	2

Number of Pets	Frequency
0	15
1	20
2	12
3	8
4	3
5	2

Averages from Frequency Tables

Question 10

Averages from Frequency Tables Questions

Averages from Frequency Tables

Question 10

Averages from Frequency Tables Questions