[{"data":1,"prerenderedAt":991},["ShallowReactive",2],{"layer:data-handling:deepen":3},{"layer":4,"contentHash":967,"dependencyHashes":968,"approval":984,"releaseId":990},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":962,"reviewStatus":963,"authoring":964},1,"data-handling","en","deepen","Why averages work, and which one to trust","Balance points, proofs, grouped data, combined groups and the art of choosing an average","Prove the mean is a balance point and how it reacts to shifts and scaling. Combine groups correctly, handle grouped data with class intervals, read double bar graphs, and choose between mean, median and mode with outliers, cricket averages and average speeds. Plus a history of statistics in India.",[13,14,15,16,17],"Explain why deviations from the mean always add to zero, and use sum = mean × count in proofs.","Prove how adding a constant or multiplying by a constant changes the mean, median, mode and range.","Find a combined (weighted) mean and avoid averaging averages.","Group data into class intervals, find the modal class and estimate the mean from midpoints; read double bar graphs.","Choose and justify the right average for a situation, including outliers, cricket batting averages and average speed.",55,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Go deeper",{"label":26,"value":27},"Reading time","≈ 55 minutes",{"label":29,"value":30},"Prior knowledge","All four measures; outliers (Investigate)",{"label":32,"value":33},"Chapters","10",{"label":35,"value":36},"Labs","Salary lab, innings lab, which-average sort, match, sprint",{"label":38,"value":39},"Stretch","Class 8 grouped data, olympiad-style traps",[41,45,51,54,87,98,104,117,122,125,135,154,159,170,180,185,188,193,202,213,223,228,231,235,276,286,291,313,318,321,377,386,391,396,399,428,461,521,526,531,534,545,550,576,587,592,595,599,604,607,652,656,661,697,736,790,909,924,928,934,938,943,948],{"id":42,"type":43,"markdown":44},"intro","prose","You can now calculate every average and predict how it behaves. This layer asks **why**. Why does sum ÷ count give a fair share? Why must deviations from the mean cancel out? Why is averaging two class averages usually wrong? Why is a cricketer's \"average\" not the mean of their scores? And when a newspaper says \"the average Indian…\", which average should it be using?\n\nThe arguments here are the kind mathematicians call **proofs**: reasons that work for *every* data set, not just the examples we try.\n\nThe class marks, salaries and journeys used as examples here are invented; the rainfall figures and Bradman's record are real, and the chapters say where they come from.",{"id":46,"type":47,"title":48,"eyebrow":49,"navLabel":50},"ch01","chapter","The mean is a balance point","Chapter 01","1 Balance point",{"id":52,"type":43,"markdown":53},"balance","Put a ruler on your finger and place identical coins at the positions 3, 4, 5, 6 and 7 cm. Where must your finger go for the ruler to balance? At **5 cm**, the mean. Move the coin at 7 to 12 and the balance point shifts right, to 6 cm, which is the new mean (3 + 4 + 5 + 6 + 12 = 30, and 30 ÷ 5 = 6).\n\nThis is not a coincidence. The mean is exactly the point where the \"pull\" of the values on each side cancels out. To see why, measure each value's **deviation**: how far it is from the mean, with a sign (+ above, − below).",{"id":55,"type":56,"caption":57,"columns":58,"rows":62},"table-dev","table","Deviations from the mean for 3, 7, 4, 6, 5 (mean 5)",[59,60,61],"Value x","Deviation x − 5","Meaning",[63,67,71,75,79,83],[64,65,66],"3","−2","2 below",[68,69,70],"7","+2","2 above",[72,73,74],"4","−1","1 below",[76,77,78],"6","+1","1 above",[80,81,82],"5","0","on the mean",[84,85,86],"**Sum**","**0**","below and above cancel",{"id":88,"type":89,"title":90,"problem":91,"steps":92},"we-proof-dev","worked_example","Proof: deviations from the mean always add to 0","Show that for **any** data set, the deviations from the mean add up to zero.",[93,94,95,96,97],"Say there are n values with sum S. The mean is M = S ÷ n, so n × M = S.","The deviations are (x₁ − M), (x₂ − M), …, (xₙ − M).","Add them: (x₁ + x₂ + … + xₙ) − (M + M + … + M) = S − n × M.","But n × M = S, so the total is S − S = **0**. ✓","This is exactly the levelling picture from Discover: the amount given away by values above the mean equals the amount received by values below it. Nothing is created or lost, only moved.",{"id":99,"type":100,"variant":101,"title":102,"markdown":103},"balance-aha","callout","aha","Two definitions, one idea","\"Share everything equally\" and \"find the balance point\" sound different, but the proof shows they are the same thing. That is why the mean feels so natural: it is where the surplus above exactly pays for the shortfall below.",{"id":105,"type":106,"itemId":107,"prompt":108,"check":109,"hints":113,"feedback":114},"pr-dev","practice","data-handling.deepen-deviation-missing","The deviations of five values from their mean are −4, +1, +3, −2 and d. What is d?",{"kind":110,"answer":111,"tolerance":112},"number",2,0,[],{"correct":115,"incorrect":116},"Right: deviations must add to 0, so −4 + 1 + 3 − 2 + d = 0, giving d = 2.","Deviations from the mean always add up to zero. Add the four known ones (−2) and choose d to cancel it.",{"id":118,"type":47,"title":119,"eyebrow":120,"navLabel":121},"ch02","Properties of the mean, proved","Chapter 02","2 Properties",{"id":123,"type":43,"markdown":124},"props","Investigate showed *by experiment* that adding a constant shifts the mean and multiplying scales it. Now we can **prove** it using nothing but sum = mean × count.",{"id":126,"type":89,"title":127,"problem":128,"steps":129},"we-proof-add","Proof: add k to every value → the mean increases by k","n values have mean M. Each value is increased by k. Show the new mean is M + k.",[130,131,132,133,134],"Old sum = n × M.","Adding k to each of the n values adds k a total of n times: new sum = n × M + n × k.","New mean = (n × M + n × k) ÷ n = M + k. ✓","Similarly, multiplying every value by k multiplies the sum by k, so the new mean is k × M.","For the median and mode, the argument is about **position**: adding k (or multiplying by a positive k) keeps the values in the same order, so the middle value and the most common value are changed in exactly the same way.",{"id":136,"type":137,"items":138},"formulas-props","formulas",[139,142,145,148,151],{"expression":140,"caption":141},"sum = mean × count","The master key: switch between an average and a total.",{"expression":143,"caption":144},"Σ(x − mean) = 0","Deviations from the mean always cancel out.",{"expression":146,"caption":147},"mean(x + k) = mean(x) + k","Shifting every value shifts the mean; the range is unchanged.",{"expression":149,"caption":150},"mean(k × x) = k × mean(x)","Scaling every value scales the mean; the range is scaled too.",{"expression":152,"caption":153},"combined mean = total ÷ total count","Never just average the averages when group sizes differ.",{"id":155,"type":100,"variant":156,"title":157,"markdown":158},"neg-k","careful","Multiplying by a negative number","If every value is multiplied by −1, the mean becomes −(mean) as the proof says, but the **order reverses**: the largest becomes the smallest. The median still becomes −(median), and the range stays the same size (it is always a positive distance). For school data you rarely do this, but it is a good test of whether you understand the argument rather than the rule.",{"id":160,"type":106,"itemId":161,"prompt":162,"check":163,"hints":166,"feedback":167},"pr-props","data-handling.deepen-convert-temp","The mean of a week's temperatures in Bengaluru was 25 °C. To convert °C to °F you multiply by 1.8 and add 32. What is the mean in °F?",{"kind":110,"answer":164,"tolerance":112,"unit":165},77,"°F",[],{"correct":168,"incorrect":169},"Yes: multiply the mean by 1.8 (45) and add 32, giving 77 °F. The same rule that converts each value converts the mean.","Apply the same conversion to the mean: 25 × 1.8 + 32.",{"id":171,"type":106,"itemId":172,"prompt":173,"check":174,"hints":176,"feedback":177},"pr-props-range","data-handling.deepen-convert-range","The same week's temperatures had a range of 10 °C. What is the range in °F?",{"kind":110,"answer":175,"tolerance":112,"unit":165},18,[],{"correct":178,"incorrect":179},"Correct: the multiplication stretches the range (10 × 1.8 = 18), but adding 32 shifts every value equally and does not change the range.","Multiplying scales the range; adding a constant does not change it.",{"id":181,"type":47,"title":182,"eyebrow":183,"navLabel":184},"ch03","Combining groups: the weighted mean","Chapter 03","3 Combined means",{"id":186,"type":43,"markdown":187},"weighted","Section A of Class 7 has 30 students with a mean mark of 62. Section B has 20 students with a mean of 72. What is the mean of all 50 students?\n\nIt is tempting to say (62 + 72) ÷ 2 = 67. **That is wrong.** Section A has more students, so its average should count for more. Go back to totals:\n\n- Section A total = 30 × 62 = 1,860.\n- Section B total = 20 × 72 = 1,440.\n- All 50 students: (1,860 + 1,440) ÷ 50 = 3,300 ÷ 50 = **66**.\n\nThe combined mean, 66, is closer to Section A's 62 because Section A is bigger. This is called a **weighted mean**: each group's mean is weighted by its size.",{"id":189,"type":100,"variant":190,"title":191,"markdown":192},"avg-avg","misconception","\"The average of averages is the overall average\"","Only when the groups are the **same size**. With different sizes, always multiply each mean by its count, add the totals and divide by the total count. News reports of state-wise averages make this mistake surprisingly often: the average of 28 states' literacy rates is not India's literacy rate, because Uttar Pradesh has far more people than Goa.",{"id":194,"type":89,"title":195,"problem":196,"steps":197},"we-weighted-marks","A weighted report card","A school counts the final exam as 60% of a subject grade, the half-yearly as 30% and projects as 10%. Meera scored 80 in the final, 70 in the half-yearly and 95 in projects. What is her grade?",[198,199,200,201],"Weights 60, 30 and 10 add up to 100.","Weighted sum = 80 × 60 + 70 × 30 + 95 × 10 = 4,800 + 2,100 + 950 = 7,850.","Divide by the total weight: 7,850 ÷ 100 = **78.5**.","Compare with the plain mean (80 + 70 + 95) ÷ 3 ≈ 81.7. Her excellent project counts for less in the weighted version, because projects carry only 10% of the weight.",{"id":203,"type":106,"itemId":204,"prompt":205,"check":206,"hints":208,"feedback":210},"pr-weighted","data-handling.deepen-combined-mean","A bus carries 40 passengers with a mean age of 30 and a second bus carries 10 passengers with a mean age of 55. What is the mean age of all 50 passengers?",{"kind":110,"answer":207,"tolerance":112},35,[209],"Total ages: 40 × 30 and 10 × 55.",{"correct":211,"incorrect":212},"Right: 40 × 30 = 1,200 and 10 × 55 = 550; 1,750 ÷ 50 = 35. Much closer to 30 than 55, because the first bus is four times fuller.","Find each bus's total age, add, and divide by 50, not 2.",{"id":214,"type":89,"title":215,"problem":216,"steps":217},"we-speed","The average-speed trap","A family drives 120 km to a wedding at 60 km\u002Fh and returns the same 120 km at 40 km\u002Fh because of traffic. What is the average speed for the whole trip?",[218,219,220,221,222],"The tempting answer is (60 + 40) ÷ 2 = 50 km\u002Fh. But the two speeds were kept for **different amounts of time**.","Time out = 120 ÷ 60 = 2 hours. Time back = 120 ÷ 40 = 3 hours.","Total distance = 240 km; total time = 5 hours.","Average speed = 240 ÷ 5 = **48 km\u002Fh**.","The slow leg lasted longer, so it gets more weight. Average speed is always total distance ÷ total time: a weighted mean in disguise.",{"id":224,"type":47,"title":225,"eyebrow":226,"navLabel":227},"ch04","Grouped data and class intervals","Chapter 04","4 Grouped data",{"id":229,"type":43,"markdown":230},"grouped-intro","When data has many different values, a frequency table with one row per value becomes long and bumpy. Here is an invented but realistic set of marks (out of 50) for 40 students in a Class 8 test:\n\n*8, 22, 47, 18, 13, 49, 40, 39, 22, 30, 37, 48, 30, 27, 32, 33, 39, 31, 36, 21, 17, 43, 17, 47, 27, 21, 22, 39, 29, 28, 39, 45, 38, 28, 30, 19, 7, 33, 20, 24.*\n\nAlmost every mark appears only once or twice, so a mark-by-mark table tells us little. Instead we **group** the marks into **class intervals** of equal width: 0–10, 10–20, 20–30, 30–40, 40–50.",{"id":232,"type":100,"variant":156,"title":233,"markdown":234},"boundary","Where does a mark of 20 go?","Intervals like 10–20 and 20–30 share the boundary 20. The usual convention (used in NCERT) is that a class interval **includes its lower limit and excludes its upper limit**: 20 goes in 20–30, not 10–20. So 10–20 means 10 ≤ mark \u003C 20. The difference between the limits (here 10) is the **class width** or **class size**.",{"id":236,"type":56,"caption":237,"columns":238,"rows":244},"table-grouped","Test marks of 40 students grouped into class intervals (lower limit included)",[239,240,241,242,243],"Marks","Tally","Frequency f","Midpoint x","f × x",[245,249,254,260,266,271],[246,247,248,80,33],"0–10","||","2",[250,251,80,252,253],"10–20","卌","15","75",[255,256,257,258,259],"20–30","卌 卌 ||","12","25","300",[261,262,263,264,265],"30–40","卌 卌 ||||","14","35","490",[267,268,68,269,270],"40–50","卌 ||","45","315",[272,273,274,273,275],"**Total**","—","**40**","**1,190**",{"id":277,"type":89,"title":278,"problem":279,"steps":280},"we-grouped","Summarising grouped data","Use the grouped table to find the modal class, the class containing the median, and an estimate of the mean. Compare the estimate with the exact mean.",[281,282,283,284,285],"**Modal class:** the interval with the highest frequency, **30–40** (14 students). With grouped data we can name the modal *class* but not an exact mode.","**Median class:** 40 values, so the median lies between the 20th and 21st. Running totals: 2, 7, 19, 33, 40. The 20th and 21st students both fall in **30–40**.","**Estimated mean:** we no longer know each mark, so assume every student in an interval scored its **midpoint** (5, 15, 25, 35, 45).","Sum of f × midpoint = 10 + 75 + 300 + 490 + 315 = 1,190. Estimated mean = 1,190 ÷ 40 = **29.75**.","**Exact mean** from the raw marks: 1195 ÷ 40 = **29.875**. The estimate is off by only 0.125. Grouping loses a little detail but keeps the big picture.",{"id":287,"type":100,"variant":288,"title":289,"markdown":290},"histogram","observation","Histograms: bar graphs for grouped data","Grouped continuous data is drawn as a **histogram**: bars for class intervals standing side by side with **no gaps**, because the intervals join up (the end of 20–30 is the start of 30–40). In an ordinary bar graph the gaps say \"these categories are separate\"; in a histogram the missing gaps say \"this is one continuous scale\".",{"id":292,"type":106,"itemId":293,"prompt":294,"check":295,"hints":309,"feedback":310},"pr-grouped","data-handling.deepen-grouped-interval","Using the convention \"lower limit included, upper limit excluded\", in which class interval does a mark of 40 belong?",{"kind":296,"options":297,"correct":308},"choice",[298,300,302,305],{"id":299,"label":261},"a",{"id":301,"label":267},"b",{"id":303,"label":304},"c","both",{"id":306,"label":307},"d","neither",[301],[],{"correct":311,"incorrect":312},"Yes: 40 is the lower limit of 40–50, so it belongs there; 30–40 stops just below 40.","Each interval includes its lower limit but not its upper limit.",{"id":314,"type":47,"title":315,"eyebrow":316,"navLabel":317},"ch05","Double bar graphs: comparing side by side","Chapter 05","5 Double bar graphs",{"id":319,"type":43,"markdown":320},"double-intro","A **double bar graph** puts two related data sets on the same axes, with a pair of bars for each category and a **legend** saying which bar is which. It is the natural way to compare two groups, two years or two places.\n\nIndia has two monsoons. Mumbai, on the west coast, is soaked by the **south-west monsoon** from June to September. Chennai, on the east coast, gets its heaviest rain from the **north-east monsoon** in October to December. A double bar graph makes the contrast leap out.",{"id":322,"type":56,"caption":323,"columns":324,"rows":329},"table-double","Average monthly rainfall: Mumbai (Santacruz) and Chennai (Nungambakkam), IMD normals 1991–2020, rounded, mm. ▇ ≈ 50 mm",[325,326,327,328],"Month","Mumbai (M)","Chennai (C)","Wetter city",[330,335,338,341,344,348,353,357,361,365,369,373],[331,332,333,334],"Jan","M 0","C 16","Chennai",[336,332,337,334],"Feb","C 6",[339,332,340,334],"Mar","C 2",[342,332,343,334],"Apr","C 14",[345,346,347,334],"May","M 7","C ▇ 43",[349,350,351,352],"Jun","M ▇▇▇▇▇▇▇▇▇▇▇ 526","C ▇ 59","Mumbai",[354,355,356,352],"Jul","M ▇▇▇▇▇▇▇▇▇▇▇▇▇▇▇▇▇▇ 920","C ▇▇ 102",[358,359,360,352],"Aug","M ▇▇▇▇▇▇▇▇▇▇▇ 561","C ▇▇▇ 133",[362,363,364,352],"Sep","M ▇▇▇▇▇▇▇▇ 384","C ▇▇▇ 146",[366,367,368,334],"Oct","M ▇▇ 91","C ▇▇▇▇▇▇ 300",[370,371,372,334],"Nov","M 11","C ▇▇▇▇▇▇▇ 374",[374,375,376,334],"Dec","M 2","C ▇▇▇▇ 182",{"id":378,"type":89,"title":379,"problem":380,"steps":381},"we-double","Reading the double bar graph","Using the table: (a) In how many months is Chennai wetter than Mumbai? (b) Which city has more rain in the year, and by how much? (c) Which city's rainfall is more concentrated in its wettest three months?",[382,383,384,385],"(a) Chennai is wetter in Jan, Feb, Mar, Apr, May, Oct, Nov and Dec: **8 months** (in March by only 2 mm).","(b) Mumbai ≈ 2,502 mm; Chennai ≈ 1,377 mm. Mumbai gets about **1,125 mm more**.","(c) Mumbai's wettest three (Jun, Jul, Aug): 526 + 920 + 561 = 2,007 mm, about 80% of its year. Chennai's (Oct, Nov, Dec): 300 + 374 + 182 = 856 mm, about 62%. Mumbai's rain is far more concentrated.","Notice that the same story could not be seen from two separate graphs so easily. Putting the bars side by side is what makes comparison quick.",{"id":387,"type":100,"variant":388,"title":389,"markdown":390},"double-limit","model_limit","What these rainfall numbers are","These are the India Meteorological Department's long-term averages (normals) for 1991–2020 at two particular stations, Mumbai (Santacruz) and Chennai (Nungambakkam), rounded to the nearest millimetre. A different station in the same city gives different numbers — Mumbai (Colaba), on the tip of the island, averages about 2,213 mm a year rather than 2,502 — and any one year can differ hugely from the normal: Chennai's 2015 floods came from an extraordinary November and early December.",{"id":392,"type":47,"title":393,"eyebrow":394,"navLabel":395},"ch06","Choosing the right average","Chapter 06","6 Which average?",{"id":397,"type":43,"markdown":398},"billionaire","Nine people sit in a tea stall. Each earns about ₹25,000 a month. Their mean, median and mode income are all ₹25,000. Then a billionaire walks in who earns ₹100 crore a month.\n\n- Total income in the room: 9 × 25,000 + 1,00,00,00,000 = ₹1,00,02,25,000.\n- **Mean** income: ₹1,00,02,25,000 ÷ 10 = **₹10,00,22,500**, about ₹10 crore a month each!\n- **Median** income: the 10 values in order are nine 25,000s and one 100 crore; the middle two (5th and 6th) are both 25,000, so the median is **₹25,000**.\n- **Mode:** **₹25,000**.\n\nThe mean now describes nobody in the room. Nine people earn far less, and one earns ten times more. This is why reports on incomes, house prices and wealth use the **median**: it tells you about a typical person, not about how much money is in the room.",{"id":400,"type":56,"caption":401,"columns":402,"rows":407},"table-which","Which average to use",[403,404,405,406],"Average","Best when…","Weakness","Example",[408,413,418,423],[409,410,411,412],"Mean","Data is numerical and fairly symmetric, or the total matters","Dragged by outliers","Mean daily electricity use to size a solar panel",[414,415,416,417],"Median","Data is skewed or has outliers; you want a typical member","Ignores how big the extremes are","Typical income, house price, waiting time",[419,420,421,422],"Mode","Data is categorical, or only actual values make sense","May not exist or may not be unique","Most popular flavour, shoe size to stock",[424,425,426,427],"(Range)","Not an average: describes spread","Uses only two values","Temperature swing in a day",{"id":429,"type":430,"component":431,"componentVersion":5,"config":432,"objective":459,"textAlternative":460},"lab-salaries","interactive","data-lab",{"datasets":433,"valueRange":444,"step":445,"challenges":446},[434],{"label":435,"unit":436,"values":437},"Monthly pay at a small workshop (₹ thousand), owner included","₹ thousand",[438,438,439,439,439,440,440,441,442,443],20,30,40,50,60,400,{"min":112,"max":443},10,[447,450,453,456],{"measure":448,"target":440,"prompt":449},"mean","The owner cuts their own pay. What pay makes the mean ₹40 thousand?",{"measure":451,"target":440,"prompt":452},"median","Give pay rises so the median becomes ₹40 thousand. What is the fewest workers you must change?",{"measure":454,"target":438,"prompt":455},"mode","Make ₹20 thousand the only mode.",{"measure":457,"target":441,"prompt":458},"range","Make the pay gap (range) only ₹50 thousand by changing one person.","See how one very high salary makes the mean misleading while the median still describes a typical worker.","A dot plot of 10 monthly pays (₹ thousand): 20, 20, 30, 30, 30, 40, 40, 50, 60 and the owner's 400. Sum 720, so the mean is **72**, higher than 9 of the 10 people. Median = (30 + 40) ÷ 2 = **35**. Mode **30**. Range 400 − 20 = **380**.\n\nChallenges:\n\n1. **Mean 40:** the total must be 400, so the owner's pay must fall by 320, to **80**.\n2. **Median 40:** the 5th and 6th ordered pays must average 40. Raising one ₹30 thousand worker to 40 works: ordered 20, 20, 30, 30, 40, 40, 40, … so the 5th and 6th are both 40.\n3. **Mode 20 only:** 20 must appear more than 30 does; move two ₹30 thousand workers to ₹20 thousand (20 appears 4 times, 30 once).\n4. **Range 50:** change the owner's 400 to 70 (70 − 20 = 50).\n\nThe owner's pay controls the mean and range, but has no effect on the median or mode.",{"id":462,"type":430,"component":463,"componentVersion":5,"config":464,"objective":519,"textAlternative":520},"lab-which-sort","sort-game",{"prompt":465,"bins":466,"items":470,"seconds":112},"Which average best describes each situation?",[467,468,469],{"id":448,"label":409},{"id":451,"label":414},{"id":454,"label":419},[471,475,479,483,487,491,495,499,503,507,511,515],{"id":472,"label":473,"bin":454,"why":474},"a1","Most popular ice-cream flavour at a stall","Flavours are categories; only the mode works.",{"id":476,"label":477,"bin":451,"why":478},"a2","Typical house price in a city with a few huge mansions","Mansions would drag the mean far above a typical house.",{"id":480,"label":481,"bin":454,"why":482},"a3","Which shoe size a shop should stock most of","The shop needs the most common real size, not 5.3.",{"id":484,"label":485,"bin":448,"why":486},"a4","Average daily rainfall, to plan how much a reservoir will collect","The total matters: mean × days gives the total rain.",{"id":488,"label":489,"bin":451,"why":490},"a5","Typical income in a village where one person is a crorepati","One huge income would distort the mean.",{"id":492,"label":493,"bin":448,"why":494},"a6","Class test marks that are close together with no outliers","With symmetric data the mean uses every mark and works well.",{"id":496,"label":497,"bin":454,"why":498},"a7","The usual number of passengers in an auto-rickshaw","Only whole numbers of passengers make sense; the most common one is useful.",{"id":500,"label":501,"bin":451,"why":502},"a8","Typical waiting time at a clinic where one patient waited 5 hours","The 5-hour wait is an outlier; the median shows the usual wait.",{"id":504,"label":505,"bin":448,"why":506},"a9","Mean daily electricity use, to size a rooftop solar system","The panel must supply the total energy, so the mean (total ÷ days) matters.",{"id":508,"label":509,"bin":454,"why":510},"a10","Most common blood group among donors","Blood groups are categories.",{"id":512,"label":513,"bin":451,"why":514},"a11","Typical finishing time in a race where one runner fell and walked","The fallen runner is an outlier.",{"id":516,"label":517,"bin":448,"why":518},"a12","Runs per match for a steady batter over a season","With no extreme innings, the mean summarises all the matches.","Choose the most suitable average for twelve real situations and say why.","A sorting game with three bins and 12 situations.\n\n- **Mean** (the total matters, or data is symmetric): rainfall for a reservoir; class marks close together; daily electricity use to size a solar system; runs per match for a steady batter.\n- **Median** (outliers or skew): house prices with mansions; village income with one crorepati; clinic waiting times with one 5-hour wait; race times with one fallen runner.\n- **Mode** (categories or real values only): ice-cream flavour; shoe size to stock; passengers per auto-rickshaw; blood group.\n\nThree questions help: Is the data categorical? (mode) Are there outliers or a long tail? (median) Does the total matter? (mean)",{"id":522,"type":100,"variant":523,"title":524,"markdown":525},"which-nuance","nuance","There is not always one right answer","Sometimes two averages are both reasonable, and the honest thing is to report both. A company might say \"the mean salary is ₹72 thousand\" and be telling the truth, while a worker says \"the median salary is ₹35 thousand\" and also tells the truth. Which one is quoted often depends on what the speaker wants you to believe. A good reader of data asks: *which average, and why that one?*",{"id":527,"type":47,"title":528,"eyebrow":529,"navLabel":530},"ch07","Cricket averages are special","Chapter 07","7 Batting average",{"id":532,"type":43,"markdown":533},"bat-avg","In cricket, a batter's **batting average** is not the mean runs per innings. It is\n\n**batting average = total runs ÷ number of times out.**\n\nInnings in which the batter was **not out** (marked with an asterisk, like 45*) add their runs to the total but do not add to the count of dismissals. The idea: a not-out innings is unfinished, so it would be unfair to treat it as a completed score. The side effect: batters who are often not out, like tail-enders or finishers, can have a batting average higher than any \"fair share\" of their innings.",{"id":535,"type":89,"title":536,"problem":537,"steps":538},"we-bat-avg","Three averages for one batter","In 7 innings a batter scored 12, 45*, 0, 78, 30*, 5, 60 (* = not out). Find the batting average, the mean runs per innings and the median innings.",[539,540,541,542,543,544],"Total runs = 12 + 45 + 0 + 78 + 30 + 5 + 60 = 230.","Times out = 7 innings − 2 not outs = 5.","**Batting average** = 230 ÷ 5 = **46**.","**Mean runs per innings** = 230 ÷ 7 ≈ **32.86**.","**Median innings:** ordered 0, 5, 12, 30, 45, 60, 78 → **30**.","Three honest numbers: 46, 32.86 and 30. The batting average is highest because the not-outs are left out of the divisor. For judging how many runs this batter usually makes, the median innings (30) tells yet another story.",{"id":546,"type":100,"variant":547,"title":548,"markdown":549},"bradman","example","The most famous average in sport","Sir Donald Bradman of Australia scored **6,996** runs in **80** Test innings, of which **10** were not out, so he was dismissed 70 times. His batting average is 6,996 ÷ 70 ≈ **99.94**. In his final innings, at The Oval in 1948, he needed just **4 runs** to finish with an average of exactly 100, since 7,000 ÷ 70 = 100; Eric Hollies bowled him for 0 with the second ball he faced. No one else with more than 20 Test innings has finished with an average above 62.",{"id":551,"type":430,"component":431,"componentVersion":5,"config":552,"objective":574,"textAlternative":575},"lab-innings",{"datasets":553,"valueRange":563,"step":558,"challenges":564},[554],{"label":555,"unit":556,"values":557},"A batter's runs in 10 innings (all out)","runs",[112,558,445,559,560,439,440,441,561,562],5,15,25,75,150,{"min":112,"max":562},[565,567,569,571],{"measure":448,"target":439,"prompt":566},"Replace the century (150) with a score that makes the mean 30.",{"measure":451,"target":440,"prompt":568},"Change the fewest innings you can so the median is 40.",{"measure":448,"target":441,"prompt":570},"Make the mean 50 by changing a single innings.",{"measure":457,"target":572,"prompt":573},100,"Make the range 100.","Compare a batter's mean and median innings, and see how one big century inflates the mean.","A dot plot of 10 innings: 0, 5, 10, 15, 25, 30, 40, 50, 75, 150. Total 400, mean **40**, median (25 + 30) ÷ 2 = **27.5**, no mode, range **150**.\n\nThe one century pulls the mean 12.5 runs above the median. In half of the innings the batter made 25 or fewer.\n\nChallenges:\n\n1. **Mean 30:** the total must be 300, so change 150 to **50**.\n2. **Median 40:** the 5th and 6th ordered scores must average 40; e.g. change 25 → 40 and 30 → 40.\n3. **Mean 50 with one change:** the total must be 500, 100 more; change any innings by +100 within the 0–150 scale, e.g. 0 → 100 or 50 → 150.\n4. **Range 100:** change 150 to 100.",{"id":577,"type":106,"itemId":578,"prompt":579,"check":580,"hints":582,"feedback":584},"pr-bat","data-handling.deepen-batting-average","A batter has scored 540 runs in 15 innings, 3 of them not out. What is the batting average?",{"kind":110,"answer":581,"tolerance":112},45,[583],"How many times was the batter actually dismissed?",{"correct":585,"incorrect":586},"Right: times out = 15 − 3 = 12, and 540 ÷ 12 = 45.","Divide total runs by the number of times out, not the number of innings.",{"id":588,"type":47,"title":589,"eyebrow":590,"navLabel":591},"ch08","Measuring spread more robustly","Chapter 08","8 Beyond the range",{"id":593,"type":43,"markdown":594},"iqr","The range has the same weakness as the mean: one outlier can wreck it. In the ten values 12, 14, 15, 15, 16, 17, 18, 19, 20, 48, the range is 48 − 12 = 36, but nine of the ten values lie within just 8 of each other.\n\nA sturdier idea, used from Class 9 onwards, is the **interquartile range (IQR)**. Put the data in order and split it into two halves. The median of the lower half is the **lower quartile**, the median of the upper half is the **upper quartile**. The IQR is their difference, and it measures the spread of the **middle half** of the data, ignoring the extremes.\n\nHere the lower half is 12, 14, 15, 15, 16 (median 15) and the upper half is 17, 18, 19, 20, 48 (median 19). IQR = 19 − 15 = **4**, a far better description of how spread out most values are.",{"id":596,"type":100,"variant":523,"title":597,"markdown":598},"iqr-nuance","Different books, slightly different quartiles","There are several slightly different rules for finding quartiles (for example, whether the median itself is included in each half when the count is odd), so calculators and books sometimes give slightly different answers. The idea is what matters here: **quartiles cut the ordered data into four equal parts, and the IQR is the spread of the middle two parts.**",{"id":600,"type":47,"title":601,"eyebrow":602,"navLabel":603},"ch09","A short history of counting people and things","Chapter 09","9 History",{"id":605,"type":43,"markdown":606},"history","Collecting data is as old as governments. Rulers needed to know how many people, fields and animals they had, to raise taxes and armies and to plan for famine. The word **statistics** itself comes from *state*: it began as \"facts about the state\".\n\nIndia has a long record. The *Arthashastra*, a text on statecraft traditionally linked with Kautilya (Chanakya) more than 2,000 years ago, describes officials who kept records of households, land and livestock. Under Akbar, Raja Todar Mal's revenue system measured land and recorded crop yields, and Abul Fazl's *Ain-i-Akbari* is packed with tables of data about the empire. Modern India has one of the largest statistical systems in the world, and one of its founders, **P. C. Mahalanobis**, is remembered on 29 June, his birthday, as **National Statistics Day**.",{"id":608,"type":609,"title":610,"items":611},"timeline-history","timeline","From clay tablets to the Census of India",[612,616,620,624,628,632,636,640,644,648],{"time":613,"title":614,"text":615},"c. 300 BCE","Arthashastra","A text on statecraft describes keeping records of population, land and cattle for taxation.",{"time":617,"title":618,"text":619},"1590s","Ain-i-Akbari","Abul Fazl's record of Akbar's empire includes detailed revenue and crop data from Todar Mal's land surveys.",{"time":621,"title":622,"text":623},"1786","The bar graph","William Playfair, a Scottish engineer, publishes some of the first bar charts; he later popularises the pie chart (1801).",{"time":625,"title":626,"text":627},"1858","Nightingale's diagrams","Florence Nightingale uses coloured diagrams of army death data to show that most soldiers died of disease, and wins hospital reform.",{"time":629,"title":630,"text":631},"1872","First census in India","The first census across much of British India is completed in 1872, counted area by area over several years rather than everywhere on one date.",{"time":633,"title":634,"text":635},"1881","First synchronous census","The first census counting everyone as at the same date; since then India has held a census about every ten years.",{"time":637,"title":638,"text":639},"1931","Indian Statistical Institute","P. C. Mahalanobis founds the ISI in Kolkata, which becomes a world centre for statistics and sample surveys.",{"time":641,"title":642,"text":643},"1950","National Sample Survey","India begins large nationwide sample surveys of households to measure spending, jobs and living conditions.",{"time":645,"title":646,"text":647},"2011","Census 2011","The 15th census counts about 121 crore people (1,21,08,54,977).",{"time":649,"title":650,"text":651},"2026–27","Census 2027","The census due in 2021 was postponed. India's 16th census is its first digital one: houses are listed between April and September 2026, people are counted in February 2027, and the reference moment is midnight on 1 March 2027.",{"id":653,"type":100,"variant":523,"title":654,"markdown":655},"hist-nuance","Old data, careful reading","Ancient and medieval records are precious, but they were collected for taxes, not for science, and people had reasons to under-report their fields and cattle. Historians treat such data with the same questions you should ask of any secondary data. Who collected it, why, and what might be missing?",{"id":657,"type":47,"title":658,"eyebrow":659,"navLabel":660},"ch10","Round-up","Chapter 10","10 Round-up",{"id":662,"type":430,"component":663,"componentVersion":5,"config":664,"objective":695,"textAlternative":696},"lab-props-match","match-pairs",{"prompt":665,"mode":666,"pairs":667},"Match each situation to its effect or value.","connect",[668,671,674,677,680,683,686,689,692],{"a":669,"b":670},"Add 10 to every value","Mean rises by 10; range unchanged",{"a":672,"b":673},"Multiply every value by 3","Mean and range both × 3",{"a":675,"b":676},"Add a new value equal to the mean","Mean unchanged",{"a":678,"b":679},"Sum of deviations from the mean","Always 0",{"a":681,"b":682},"Mean × number of values","Sum of the values",{"a":684,"b":685},"Remove a value above the mean","Mean goes down",{"a":687,"b":688},"30 students at 62 and 20 at 72","Combined mean 66",{"a":690,"b":691},"120 km at 60 km\u002Fh, back at 40 km\u002Fh","Average speed 48 km\u002Fh",{"a":693,"b":694},"6,996 runs, 70 dismissals","Batting average ≈ 99.94","Connect the properties of the mean to their effects, and the Deepen examples to their answers.","Nine pairs:\n\n- Add 10 to every value → mean rises by 10; range unchanged.\n- Multiply every value by 3 → mean and range both multiplied by 3.\n- Add a new value equal to the mean → mean unchanged.\n- Sum of deviations from the mean → always 0.\n- Mean × number of values → the sum.\n- Remove a value above the mean → mean goes down.\n- 30 students with mean 62 and 20 with mean 72 → combined mean (1,860 + 1,440) ÷ 50 = 66.\n- 120 km at 60 km\u002Fh and back at 40 km\u002Fh → 240 km in 5 hours = 48 km\u002Fh.\n- 6,996 runs and 70 dismissals → batting average ≈ 99.94 (Bradman).",{"id":698,"type":430,"component":699,"componentVersion":5,"config":700,"objective":734,"textAlternative":735},"lab-deepen-sprint","arith-sprint",{"operations":701,"ranges":704,"rounds":445,"secondsTotal":112,"estimateFirst":708,"wordProblems":709},[702,703],"×","÷",{"a":705,"b":706},{"min":111,"max":438},{"min":111,"max":707},12,false,[710,713,717,719,723,726,729,732],{"prompt":711,"answer":712,"operation":702},"The mean of 8 numbers is 15. What is their sum?",120,{"prompt":714,"answer":715,"operation":716},"The mean of 5 numbers is 12. Four of them add up to 47. What is the fifth?",13,"-",{"prompt":718,"answer":581,"operation":703,"unit":556},"A batter has 630 runs and has been out 14 times. What is the batting average?",{"prompt":720,"answer":721,"operation":703,"unit":722},"20 girls have a mean height of 140 cm and 30 boys have a mean of 145 cm. What is the mean height of all 50 children?",143,"cm",{"prompt":724,"answer":439,"operation":725},"Every value in a data set with mean 23 is increased by 7. What is the new mean?","+",{"prompt":727,"answer":440,"operation":703,"unit":728},"A car goes 60 km at 30 km\u002Fh and then 60 km at 60 km\u002Fh. What is its average speed for the 120 km?","km\u002Fh",{"prompt":730,"answer":731,"operation":703},"In grouped data, 4 students are in 10–20 and 6 in 20–30. Using midpoints, estimate the mean.",21,{"prompt":733,"answer":581,"operation":703},"The mean of 10 numbers is 50. One number, 95, is removed. What is the mean of the other 9?","Rapid practice with totals, missing values, combined means and batting averages.","A sprint of multiplication and division warm-ups, then eight word problems:\n\n- Mean 15 of 8 numbers → sum 8 × 15 = **120**.\n- Mean 12 of 5 numbers → total 60; fifth = 60 − 47 = **13**.\n- 630 runs, 14 dismissals → **45**.\n- 20 × 140 + 30 × 145 = 2,800 + 4,350 = 7,150; ÷ 50 = **143 cm**.\n- Mean 23, add 7 to each → **30**.\n- 60 km at 30 km\u002Fh (2 h) + 60 km at 60 km\u002Fh (1 h) → 120 ÷ 3 = **40 km\u002Fh**.\n- Midpoints 15 and 25: (4 × 15 + 6 × 25) ÷ 10 = 210 ÷ 10 = **21**.\n- Total 500 − 95 = 405; ÷ 9 = **45**.",{"id":737,"type":738,"title":739,"terms":740},"glossary-deepen","glossary","Words from this layer",[741,745,748,752,755,759,762,766,769,771,774,777,781,784,787],{"term":742,"meaning":743,"example":744},"deviation","How far a value is from the mean, with a sign: positive above, negative below.","7 − 5 = +2.",{"term":746,"meaning":747},"balance point","The point where the values' deviations cancel out; the mean.",{"term":749,"meaning":750,"example":751},"weighted mean","A mean in which some values or groups count more than others, according to their weights (such as group sizes).","Combined mean of two sections.",{"term":753,"meaning":754},"class interval","A range of values used as one group in grouped data, such as 20–30.",{"term":756,"meaning":757,"example":758},"class width","The difference between the upper and lower limits of a class interval.","30 − 20 = 10.",{"term":760,"meaning":761},"lower limit \u002F upper limit","The smallest and largest boundaries of a class interval; usually the lower is included and the upper excluded.",{"term":763,"meaning":764,"example":765},"midpoint (class mark)","The middle of a class interval, (lower + upper) ÷ 2, used to estimate the mean of grouped data.","(20 + 30) ÷ 2 = 25.",{"term":767,"meaning":768},"modal class","The class interval with the highest frequency.",{"term":287,"meaning":770},"A bar graph for grouped continuous data, with bars touching because the intervals join up.",{"term":772,"meaning":773},"double bar graph","A bar graph with pairs of bars for comparing two data sets, with a legend.",{"term":775,"meaning":776},"legend","The key on a graph that says what each colour or bar style stands for.",{"term":778,"meaning":779,"example":780},"batting average","In cricket, total runs divided by the number of times the batter was out.","6,996 ÷ 70 ≈ 99.94.",{"term":782,"meaning":783},"quartiles","The values that cut ordered data into four equal parts; the lower and upper quartiles are the medians of the two halves.",{"term":785,"meaning":786},"interquartile range (IQR)","Upper quartile − lower quartile: the spread of the middle half of the data.",{"term":788,"meaning":789},"statistics","The science of collecting, organising, summarising and interpreting data; the word comes from \"state\".",{"id":791,"type":792,"title":793,"questions":794},"quiz-deepen","quiz","Reasoning check",[795,806,819,832,844,855,865,874,885,896],{"itemId":796,"prompt":797,"options":798,"correct":301,"why":805},"data-handling.deepen-q-dev","The deviations from the mean of a data set are 3, −5, 1, x. What is x?",[799,800,802,803],{"id":299,"label":73},{"id":301,"label":801},"1",{"id":303,"label":81},{"id":306,"label":804},"9","Deviations sum to 0: 3 − 5 + 1 + x = 0, so x = 1.",{"itemId":807,"prompt":808,"options":809,"correct":301,"why":818},"data-handling.deepen-q-combined","Class A (10 students) has mean 80; Class B (40 students) has mean 60. The combined mean is…",[810,812,814,816],{"id":299,"label":811},"70",{"id":301,"label":813},"64",{"id":303,"label":815},"66",{"id":306,"label":817},"76","(800 + 2,400) ÷ 50 = 3,200 ÷ 50 = 64.",{"itemId":820,"prompt":821,"options":822,"correct":301,"why":831},"data-handling.deepen-q-speed","Going to school at 10 km\u002Fh and returning the same distance at 15 km\u002Fh, the average speed is…",[823,825,827,829],{"id":299,"label":824},"12.5 km\u002Fh",{"id":301,"label":826},"12 km\u002Fh",{"id":303,"label":828},"13 km\u002Fh",{"id":306,"label":830},"25 km\u002Fh","For 30 km each way: 3 h + 2 h = 5 h for 60 km, so 12 km\u002Fh.",{"itemId":833,"prompt":834,"options":835,"correct":301,"why":843},"data-handling.deepen-q-bat","A batter scores 300 runs in 10 innings, 4 of them not out. Batting average?",[836,838,840,841],{"id":299,"label":837},"30",{"id":301,"label":839},"50",{"id":303,"label":253},{"id":306,"label":842},"42.9","Times out = 6; 300 ÷ 6 = 50.",{"itemId":845,"prompt":846,"options":847,"correct":301,"why":854},"data-handling.deepen-q-interval","In grouped data with intervals 0–5, 5–10, 10–15, where is the value 10?",[848,850,852,853],{"id":299,"label":849},"5–10",{"id":301,"label":851},"10–15",{"id":303,"label":304},{"id":306,"label":307},"Lower limit included, upper limit excluded.",{"itemId":856,"prompt":857,"options":858,"correct":299,"why":864},"data-handling.deepen-q-midpoint","Estimated mean: 3 values in 0–10 and 7 values in 10–20?",[859,860,861,862],{"id":299,"label":257},{"id":301,"label":252},{"id":303,"label":33},{"id":306,"label":863},"13","(3 × 5 + 7 × 15) ÷ 10 = (15 + 105) ÷ 10 = 12.",{"itemId":866,"prompt":867,"options":868,"correct":301,"why":873},"data-handling.deepen-q-income","For describing a typical family income in a city, the best average is usually…",[869,870,871,872],{"id":299,"label":448},{"id":301,"label":451},{"id":303,"label":454},{"id":306,"label":457},"Incomes are skewed by a few very high earners; the median resists them.",{"itemId":875,"prompt":876,"options":877,"correct":301,"why":884},"data-handling.deepen-q-scale","Every value is multiplied by 5 and then 2 is added. The range was 6. It becomes…",[878,879,880,882],{"id":299,"label":76},{"id":301,"label":837},{"id":303,"label":881},"32",{"id":306,"label":883},"8","Multiplying scales the range to 30; adding 2 does not change it.",{"itemId":886,"prompt":887,"options":888,"correct":301,"why":895},"data-handling.deepen-q-iqr","Which measure of spread ignores the most extreme values?",[889,891,893,894],{"id":299,"label":890},"Range",{"id":301,"label":892},"Interquartile range",{"id":303,"label":409},{"id":306,"label":419},"The IQR only uses the middle half of the ordered data.",{"itemId":897,"prompt":898,"options":899,"correct":306,"why":908},"data-handling.deepen-q-avg-avg","When is the mean of two group means equal to the overall mean?",[900,902,904,906],{"id":299,"label":901},"Always",{"id":301,"label":903},"Never",{"id":303,"label":905},"Only when each group has exactly 10 members",{"id":306,"label":907},"When the groups are the same size (or the two means are equal)","With equal sizes each group gets equal weight; and if the two means are equal, any weighting gives that same value. Group size 10 is not special.",{"id":910,"type":911,"title":912,"points":913},"cheat-deepen","summary","Cheat sheet",[914,915,916,917,918,919,920,921,922,923],"**Deviations** from the mean always add to 0: the mean is the **balance point**. Proof: Σx − n × mean = S − S = 0.","**sum = mean × count** is the master key for proofs, missing values and combined groups.","Add k to all values → mean, median, mode + k, range same. Multiply all by k → all × k (range by the size of k).","**Combined mean** = (n₁ × mean₁ + n₂ × mean₂) ÷ (n₁ + n₂). Never average averages of unequal groups.","**Average speed** = total distance ÷ total time, not the mean of the speeds.","**Grouped data.** Lower limit included, upper excluded. Modal class = highest frequency. Estimated mean uses midpoints.","**Histogram:** bars touch. **Double bar graph:** paired bars plus a legend.","**Which average?** Categories → mode. Outliers or skew → median. Total matters or symmetric → mean.","**Batting average** = runs ÷ times out (not innings). Bradman: 6,996 ÷ 70 ≈ 99.94.","**IQR** = upper quartile − lower quartile: a spread measure that resists outliers.",{"id":925,"type":926,"prompt":927},"reflect-deepen","reflection","A politician says \"the average income in our district has risen by 20%\". Write three questions you would ask before believing this means most families are better off.",{"id":929,"type":930,"conceptId":931,"relation":932,"explanation":933},"conn-ops","connection","order-of-operations","helps_understand","Weighted means like (30 × 62 + 20 × 72) ÷ 50 need the correct order of operations.",{"id":935,"type":930,"conceptId":936,"relation":932,"explanation":937},"conn-props","properties-of-numbers","The proof that deviations sum to zero uses the distributive property: n × M = M + M + … + M.",{"id":939,"type":930,"conceptId":940,"relation":941,"explanation":942},"conn-angles","angles","related_to","Pie charts, in the next layer, turn frequencies into angles at the centre of a circle.",{"id":944,"type":930,"conceptId":945,"relation":946,"explanation":947},"conn-elec","electricity","applied_in","Sizing a rooftop solar system uses the mean daily electricity use, because the total energy is what matters.",{"id":949,"type":950,"sourceIds":951},"sources-deepen","sources",[952,953,954,955,956,957,958,959,960,961],"data-handling-ncert-class7","data-handling-ncert-class8","data-handling-mathsisfun-histograms","data-handling-khan-summarizing","data-handling-wiki-bradman","data-handling-wiki-mahalanobis","data-handling-census-india","data-handling-census-2027","data-handling-imd","data-handling-imd-normals",[952,953,954,955,956,957,958,959,960,961],"needs_review",{"generatedBy":965,"notes":966},"claude-code","Draft generated with Python-checked statistics; pending owner review.","e8b649d1db44ae0cf29b8c244a5e105a176047d40e4c3c6f1181d326939991bf",{"logic:practice":969,"component:data-lab@1":970,"component:sort-game@1":971,"component:match-pairs@1":972,"component:arith-sprint@1":973,"source:data-handling-census-2027":974,"source:data-handling-census-india":975,"source:data-handling-imd":976,"source:data-handling-imd-normals":977,"source:data-handling-khan-summarizing":978,"source:data-handling-mathsisfun-histograms":979,"source:data-handling-ncert-class7":980,"source:data-handling-ncert-class8":981,"source:data-handling-wiki-bradman":982,"source:data-handling-wiki-mahalanobis":983},"3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","466896cc37735f48db03875fe9c9ce42fc8bcb7e5f937c9779d70513703b91bd","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","c0c63ed40e1bca5ba6d446d43886b887a3a7709e5cd6679419f64fd3ba62afe6","b97af6e5e0ca6802ee13a84af02594d9b371855310cc090e74f2b102ed4403ee","934b2e0d1b6222e6346f51e64ffc3e6c8d98db3be7d9a3c8d4cb3da843e8becc","834b555363edd0d5f3a42f5870d2acb5a33c3af5e40ecb43649732edd4cdce2c","f8d2cede5dff9df165f0ad49c28625d281417b1abac9405104cd7e3ff5c50f88","d3f02b5bb1750887d4c5e469441199469eba3f40ba38f08a0f15e6dc123789da","9e8960bd9772301ed39d3d4df6fc1c2c06a84501ac0c1bc78dc4b5aec44347a6","73c4938c33c83e313b3e69ea6d4b3fcd121cfeab65eda7e07e1df7901cfaea2c","9c5530663fa1281b78426a548e1488607d718897d812d85bf6a79a1ac289ffb9","72303bbcacfb7c6c2edf5594818d8cc402de91d19c7f16248a63102dd96ba0de","8b262d93bdc3277658143b72d0f370f2357f46f259c578ba699a36937d3759c6",{"state":985,"reviewer":986,"selfReview":987,"reviewedAt":988,"method":989},"approved","The library owner",true,"2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899598093]