NCERT Exemplar Solutions Class 10 Maths Chapter 13 Statistics

Reading the formula symbol by symbol. The quickest way to answer this is to remember that the whole assumed-mean machinery replaces each class by one representative value, and that value is the class mark.

• Locate the symbol: the letter d_i appears in the source formula only inside the definition d_i=x_i-a, so its meaning is fixed entirely by what x_i stands for.
• Decode x_i: in every grouped-data formula in this chapter x_i is the mid-point of the ith class, which is the class mark, and it is never a class limit and never a frequency.
• Read off the gap: therefore d_i is just the distance of the class mid-point from the assumed mean, so the distractors that name a limit or the frequency are wrong.
• Why it matters: take d_i from a class limit by mistake and every entry in the deviation column shifts, dragging the computed mean off by a fixed amount, so pinning x_i to the mid-point keeps the whole table honest.

Mid-points of the classes; option (C).

Why the mid-point and not a limit. The assumption is chosen so that, on average, the over- and under-estimates inside a class cancel.

• Picture a class: take the interval 20–30, where some observations sit close to 20 and others close to 30, yet we have no record of exactly where each one lands.
• Balance the two sides: placing all of them at the mid-point 25 lets the low values balance the high values, so the error that grouping introduces stays small either way.
• Rule out the limits: centring at an upper or lower limit would push the estimated mean up or down for every class, so the only fair centring point is the class mark, option (B).
• Why it matters: this same balancing idea explains why the mean of ungrouped and grouped data come out close but rarely identical, a contrast the reasoning questions in this chapter test directly.

Centred at the class marks; option (B).

One-line algebra. The identity is a direct consequence of how x̄ is defined, so no actual data values are needed at all.

• Start from the mean: the definition rearranges to x̄∑ f_i=∑ f_i x_i, which is the only fact this problem relies on.
• Distribute the sum: expanding the bracket gives ∑ f_i x_i-x̄∑ f_i, two pieces built from the same quantities.
• Cancel to zero: those two pieces are equal, so their difference is exactly zero, and the options minus one, one, and two simply cannot arise for any data set.
• Why it matters: this zero-sum property is the algebraic seed of variance and standard deviation in higher classes, where deviations are squared precisely because their plain signed sum is always zero.

0; option (A).

Building u_i from d_i. The step deviation is just the ordinary deviation measured in units of one class width.

• Take the deviation: start with the gap of the class mark from the assumed mean, the familiar d_i=x_i-a.
• Count the steps: express that gap as a number of steps of size h, giving u_i=x_i-ah, and when a is a class mark every u_i becomes a tiny whole number such as -2 or 2.
• Check the formula: substituting d_i=h u_i back into the assumed-mean formula rebuilds the step-deviation formula exactly, so only the form in option (C) is correct.
• Why it matters: dividing by h turns three-digit products into one-digit products, which is the whole reason the step-deviation method is the fastest hand method whenever the class widths are equal.

x_i-ah; option (C).

Reading the graph. The fastest justification is to track what height the two curves share when they meet.

• Opposite directions: both ogives reach the cumulative frequency n at the extremes, but one rises and the other falls as the class value increases across the axis.
• Shared height: they can be equal only at the common height n2, because the two running totals always add up to the full count n at every point.
• Drop the foot: a perpendicular from that crossing point down to the horizontal axis lands on the median value, while the mean and mode are never read off a graph this way.
• Why it matters: this graphical median is examiner-friendly because it finds the median without any formula, and it underpins the popular board long-answer task of drawing both ogives and reading their intersection.

Median; option (B).

Table-first method. Setting up one cumulative column answers both halves of the question at the same time.

• Build the column: the cumulative frequencies run 10,25,37,57,66 with total n=66, so the half-way mark we hunt for is n2=33.
• Place the median class: the value 33 first lands in the 10–15 row, where the cumulative count jumps from 25 to 37, so the median class lower limit is 10.
• Place the modal class: the highest frequency, 20, sits in 15–20, giving a modal class lower limit of 15, and the two lower limits add to 25.
• Why it matters: many MCQs ask only for lower limits, not the full median or mode, so noticing that the cumulative column and the maximum frequency are all you need saves real time in the exam.

25; option (B).

Continuity-correction first. The whole question turns on the single 0.5 adjustment to the class boundaries.

• Spot the gap: the jump from 5 to 6 shows the classes are inclusive, so shift every lower limit down and every upper limit up by half a unit before doing anything else.
• Cumulate the counts: the cumulative column on the given frequencies reads 13,23,38,46,57, and half the total, n2=28.5, falls inside the third class.
• Read the upper limit: that third class after correction is 11.5–17.5, so the upper limit of the median class is 17.5.
• Why it matters: the distractors 17 and 18 are exactly what you would write if you skipped the correction, which is precisely why examiners slip inclusive classes in to test this one habit.

17.5; option (B).

Spotting the biggest jump. The modal class is simply wherever the cumulative total climbs the fastest.

• Difference the totals: subtracting each below row from the next gives the plain class frequencies 3,9,15,30,18,5, which is the table you actually need.
• Find the steepest rise: the largest single jump is 30, between Below 30 and Below 40, showing where most students are packed.
• Name the modal class: that jump of 30 is the frequency of the interval 30–40, so the modal class is 30–40.
• Why it matters: reading a cumulative table as if it were a plain frequency table is a classic slip, but training the eye to hunt for the biggest jump turns this into a five-second question.

30–40; option (C).

Width in disguise. When the two classes coincide, the answer is simply the common class width.

• Build the column: the cumulative frequencies are 4,9,22,42,56,63,67, and half the total, n2=33.5, sits in 125–145, so that is the median class with upper limit 145.
• Find the peak: the maximum frequency 20 also belongs to 125–145, making it the modal class with lower limit 125.
• Take the difference: subtracting the lower limit of the modal class from the upper limit of the median class gives 145-125=20, exactly the width shared by every class.
• Why it matters: recognising that this difference collapses to the class width whenever the two classes agree lets you check the arithmetic instantly, since each class here spans 20 units.

20; option (C).

Cumulating to the boundary. Read the phrase "less than" as a plain signal to cumulate the frequencies.

• Find the boundary: the value 14.6 is the upper end of the fourth class, so every athlete in that class and below it counts toward the total we want.
• Add up to it: summing the frequencies up to and including that class gives 2+4+5+71, which works out to 82 athletes finishing under 14.6 seconds.
• Check the rest: the remaining 48+20=68 athletes finished in 14.6 seconds or more, and the two parts add back to 82+68=150, the full field.
• Why it matters: that self-check against the total head count confirms the cumulative reading is consistent, a quick habit worth keeping on every ogive question.

82; option (C).

Difference of adjacent cumulatives. A more than column behaves just like a less than column read upside down.

• State the rule: the frequency of a class equals its more-than-or-equal total at the lower limit minus the same total at the next higher lower limit, so only two numbers ever matter.
• Bracket the class: for the interval 30–40 the two bracketing entries are the count at ≥ 30 and the count at ≥ 40, namely 51 and 48.
• Subtract: taking 51-48 leaves exactly 3 students in the class 30–40, while 48 and 51 themselves are traps.
• Why it matters: whether the table is below type or more than type, a single-class frequency is always the difference of two neighbouring cumulative entries, and only the direction of subtraction flips.

3; option (A).

Pin down the hidden assumption. The cleanest reasoning answer names exactly which assumption breaks the equality.

• Exact side: the ungrouped median is read straight from the raw ordered list, so it is the genuine middle observation with no approximation at all.
• Estimated side: the grouped median formula interpolates inside the median class, assuming the frequency there is spread perfectly evenly across the interval.
• Compare them: that interpolation is only an approximation, so the two medians need not match, which makes the claim that they are always the same false.
• Why it matters: knowing the grouped formula is really an interpolation explains why board questions say calculate the median of grouped data and never find the exact median, since the result is deliberately an estimate.

The statement is false; the grouped median only estimates the ungrouped one because of the uniform-spread assumption.

Show a cancels out. The strongest justification proves the final mean does not depend on the choice of a at all.

• Substitute the deviation: putting d_i=x_i-a into the formula gives a+∑ f_i(x_i-a)∑ f_i=a+∑ f_i x_i∑ f_i-a, splitting the assumed mean cleanly out of the sum.
• Watch it cancel: the two copies of a cancel against each other, leaving precisely ∑ f_i x_i∑ f_i, which is the ordinary mean x̄.
• Draw the conclusion: since this holds for every value of a, the demand that a must be a class mid-point is simply false.
• Why it matters: watching a cancel reassures students that an unusual choice of assumed mean can never produce a wrong answer, because the choice only affects how tidy the working column looks.

The statement is incorrect; the assumed mean a may be any number, not necessarily a class mid-point.

One counterexample is enough. To disprove an always statement, a single case where the three averages agree is sufficient.

• Pick symmetric data: take any grouped distribution whose frequencies are mirrored about the central class, the easiest kind of counterexample to set up.
• Let symmetry act: that balance puts the mean, the median, and the mode all on the same central class, since each average is pulled to the centre.
• Reach the verdict: the three averages then coincide, which directly contradicts the claim that they are always different, so the statement is false.
• Why it matters: the gap between mean, median, and mode actually measures skewness, and when that gap is zero the data are symmetric, which is exactly why such always-different claims break down for balanced data.

The statement is false; symmetric data make the three measures equal.

Two rules, one class. The disproof is to exhibit data where the two selection rules point at the very same interval.

• Median rule: the median class is the one holding the n2th cumulative position, decided purely by the running total.
• Modal rule: the modal class is the one with the greatest frequency, decided purely by the tallest bar, and nothing forces these two rules apart.
• Make them agree: for a distribution that peaks at its centre, both rules pick the central class, so the median class and modal class coincide and the always-different claim fails.
• Why it matters: seeing that one class can play both roles heads off a common exam slip, where students assume the modal class must sit away from the median class and then misread the table.

The statement is false; the median class and modal class can be the same interval.

Direct method with care on widths. Because the class widths are not all equal, the step-deviation shortcut is awkward, so the plain direct method is cleanest here.

• Class marks: averaging each pair of limits gives 2,4,6,8.5, where the last comes from the wider class as 7+102=8.5 rather than a careless 9.
• Products and totals: the products sum to ∑ f_i x_i = 18+88+162+144.5 = 412.5, and the frequencies sum to ∑ f_i = 75.
• Divide once: dividing the two totals gives x̄=412.575=5.5, the mean of the distribution.
• Why it matters: mixed class widths appear in exemplar data precisely to catch students who memorise one class mark and reuse it, so recomputing x_i for every class is the safe habit.

x̄=5.5.

Assumed-mean cross-check. Taking a=35 shows at once that the data are balanced around the middle class.

• List the deviations: measuring each class mark from the guess 35 gives the tidy values -20,-10,0,10,20.
• Sum them weighted: the weighted total ∑ f_i d_i = 2(-20)+4(-10)+0+6(10)+1(20) collapses to exactly zero once the negatives and positives cancel.
• Recover the mean: feeding that zero into x̄=a+∑ f_i d_i∑ f_i=35+020 leaves 35, in full agreement with the direct method.
• Why it matters: a zero deviation sum is a signal that the data are symmetric about the guess, so the assumed mean is already the true mean, which is the fastest possible finish when it happens.

x̄=35 marks.

Why the mean ignores continuity correction. The correction adds and subtracts the same half unit at the two ends of every class.

• Marks stay put: shifting each class by half a unit at both ends leaves the mid-points untouched, so the class marks remain the clean values 5.5,9.5,13.5,17.5 either way.
• Sum the products: with those marks the products total ∑ f_i x_i = 27.5+38+121.5+175 = 362, while the frequencies total ∑ f_i = 28, the head count of the data.
• Divide for the mean: dividing the two totals gives x̄=36228≈ 12.93, with no correction step anywhere in the working.
• Why it matters: continuity correction does change the boundaries that the median and mode rely on, but it never moves a class mark, so the mean is the one measure you can compute straight from inclusive classes without first making them continuous.

x̄≈ 12.93.

Step-deviation in one pass. The equal width of three turns the whole u_i column into a clean run from -2 to 2.

• Set the anchors: choosing a=23 and h=3 makes the step deviations the small integers -2,-1,0,1,2, which are easy to weight by hand.
• Add the products: the weighted sum comes to ∑ f_i u_i = -2-3+0+9+26 = 30, and the frequencies also total ∑ f_i = 30.
• Finish the formula: substituting gives x̄=23+3(3030)=23+3=26 pages per day.
• Why it matters: because the two totals are equal the correction term is exactly one full width h, so the mean lands one step above a, and spotting that ratio of one confirms the answer without a calculator.

Mean =26 pages per day.

Keep the .5 in the marks. The half-rupee in each class mark is what makes the running totals land on a clean number.

• Halved marks: averaging each pair of inclusive limits gives the class marks 100.5,300.5,500.5,700.5, each ending in a half-rupee that must be carried through.
• Add the products: weighting those marks gives ∑ f_i x_i = 1407+4507.5+7007+4903.5 = 17825 against ∑ f_i = 50 employees in the sample.
• Divide once: dividing the two totals gives x̄=1782550=356.5, the mean daily income.
• Why it matters: inclusive money brackets like 1–200 are common in real survey data, and carrying the half-rupee through the whole table prevents a systematic downward bias in the reported average income.

Mean daily income = Rs 356.5.

Small correction, large a. The assumed mean does the heavy lifting while the step column only nudges it a little.

• Set the anchors: taking a=110 and h=4 gives the usual step deviations -2,-1,0,1,2, centred on the busiest band.
• Add the products: the weighted total is only ∑ f_i u_i = -2 against a head count of ∑ f_i = 100, so the correction will be small.
• Finish the formula: substituting gives x̄=110+4(-2100)=110-0.08=109.92 seats.
• Why it matters: when data are nearly symmetric about the central class the step sum is tiny and the mean settles very close to a, which is a handy estimate-first check before you trust the final decimal.

Mean =109.92 seats.

Decode the step back to kg. The correction of -1.6 kg is just the real-unit version of the abstract step sum.

• Set the anchors: with a=125 and h=10 the step deviations are the familiar integers -2,-1,0,1,2 across the five weight bands.
• Form the ratio: the weighted sum ∑ f_i u_i = -8 over the head count ∑ f_i = 50 gives the pure ratio -850=-0.16, a number with no units yet.
• Scale by the width: multiplying that ratio by h=10 converts it back to -1.6 kg, so the mean is x̄=125-1.6=123.4 kg.
• Why it matters: the busiest weight band 120–130 holds the most wrestlers, yet the mean still dips to 123.4 because the lighter tail outweighs the heavier one, and that gap between the mean and the modal class is exactly what central-tendency questions probe.

Mean weight =123.4 kg.

Mean first, judgement second. The computed number is what actually answers the manufacturer's claim.

• Set the anchors: with a=15 and h=2 the step deviations are -2,-1,0,1, giving a weighted sum ∑ f_i u_i=-13 over ∑ f_i=50 cars.
• Compute the mean: substituting yields x̄=15+2(-1350)=15-0.52=14.48 km per litre, the true average mileage of the sample.
• Deliver the verdict: since 14.48 falls below the advertised 16, the average mileage falls short, so we disagree with the manufacturer's claim.
• Why it matters: this is a model data-versus-advertising question, where the test data give a defensible average and any marketing figure above it must be challenged, so always anchor the verdict to the computed mean rather than the single best class.

x̄=14.48 km/litre; the 16 km/litre claim is not agreed.

Running-total discipline. Build the column from the top and never re-sum it from scratch at each row.

• Start the chain: take the first frequency, 4, as the opening cumulative total, since nothing lies below the lowest class.
• Add one at a time: fold each next frequency into the running total to produce 4,8,21,26,32,37,39,40 in a single downward pass.
• Check the end: the eighth total is 40, equal to the number of persons, which confirms the column is internally consistent.
• Why it matters: this less than column is exactly the set of plotting points for a less than ogive, with the upper boundary on the horizontal axis and the cumulative count on the vertical axis, which several board long-answer problems then ask you to graph.

Cumulative frequencies 4,8,21,26,32,37,39,40.

Differencing a cumulative column. A below column adds one class at a time as the marks rise, so reversing it means taking the gap between each pair of neighbouring totals.

• First row is special: no marks lie below 10, so the first cumulative total is also the first class frequency, making the frequency of 0–10 simply 10.
• Difference the rest: for every later class the frequency is the current below total minus the previous one, so 30–40 gives 270-130=140 and 40–50 gives 440-270=170.
• Read down the column: working steadily downward produces the full frequency list 10,40,80,140,170,130,100,70,40,20 in ascending class order.
• Verify the total: adding them gives 10+40+80+140+170+130+100+70+40+20=800, the number of students, so no row was mis-subtracted.
• Why it matters: the peak frequency 170 sits in the class 40–50, marking it as the modal class, and recovering this plain frequency column is the necessary first step before any mode, median, or histogram work on cumulative data.

Frequencies 10,40,80,140,170,130,100,70,40,20 (total 800).

Top-down differencing. A more than column shrinks as the lower limit rises, because fewer candidates clear a higher bar, so the natural reversal works from the top of the marks scale downward.

• Top class first: the smallest cumulative entry, 4 at more than or equal to 80, is the frequency of the top class 80–90, since no listed candidate scores above 90.
• Difference downward: for every lower class the frequency is its more-than total minus the next higher one, so 60–70 gives 11-6=5 and 40–50 gives 23-17=6.
• Collect in order: arranging the results from the lowest class upward gives the frequency list 2,2,3,4,6,6,5,2,4.
• Verify the total: their sum 2+2+3+4+6+6+5+2+4=34 matches the more than or equal to 0 entry exactly, so the differencing is consistent.
• Why it matters: a more than table is the data behind a more than ogive that falls from left to right, and recovering the plain frequencies lets you cross-check that curve against its less than partner, whose intersection pins the median.

Frequencies 2,2,3,4,6,6,5,2,4 (total 34).

Walk the running total. Each unknown is just one add or one subtract away from its neighbour in the table.

• Open the chain: the entry a is the first cumulative value, which equals the first frequency, so a=12 straight away.
• Fill the gaps: reading the cumulative gaps gives b=25-12=13, then c=25+10=35, then d=43-35=8, then e=48-43=5, one neighbour at a time.
• Close the chain: the last cumulative value f=48+2=50 equals the stated total, confirming every earlier unknown was found correctly.
• Why it matters: mixing the two columns is the common trap, since the frequency column is built from differences while the cumulative column is built from running sums, so keeping the two rules apart makes find-the-unknowns routine.

a=12, b=13, c=35, d=8, e=5, f=50.

Build both columns from the same totals. The less than column counts upward while the more than column counts back down from the grand total of 300 patients.

• Less than column: cumulate the frequencies from left to right, adding one class at a time, to reach the running totals 60,102,157,227,280,300 at the successive upper boundaries.
• More than column: begin at the full count 300 and peel off each frequency in turn as the lower limit rises, which gives the falling totals 300,240,198,143,73,20.
• Cross-check the pair: at any boundary the less than count plus the more than or equal count must equal 300, and at age 30 this reads 102+198=300 as expected.
• Why it matters: this pairing rule, where the two cumulative readings always add to the total at every boundary, is the fastest way to catch a slip in either column without recomputing the whole table from scratch.

Less than: 60,102,157,227,280,300; more than or equal: 300,240,198,143,73,20.

Difference the below column. The first entry stands on its own while every later entry is just a gap between neighbours.

• First class alone: since nothing lies below the lowest mark, the opening cumulative total is also the first class frequency, so the frequency of 0–20 is 17.
• Gaps for the rest: each later class frequency is its own below total minus the previous below total, which produces the successive values 5,7,8,13 down the table.
• Check the sum: adding everything gives 17+5+7+8+13=50, matching the class size, so no subtraction slipped.
• Why it matters: the wide classes of width 20 make the below jumps look small, yet the opening class carrying 17 of the 50 students shows most marks are low, a shape that the cumulative table hides until you difference it.

Frequencies 17,5,7,8,13 (total 50).

Locate, then interpolate. The whole computation hinges on placing the half-way mark n2 correctly in the cumulative column.

• Locate the class: with n2=300 the cumulative column 250,440, shows 300 sits in the second class, so the median class is 1000–2000 with l=1000, cf=250, f=190, and h=1000.
• Find the position: the fraction of the way into that class is 300-250190=50190≈ 0.263, a little over a quarter of the interval.
• Scale and add: the median is then 1000+0.263× 1000≈ 1263, that is about Rs 1263.16.
• Why it matters: this median sits far below the top brackets, reflecting that most families earn under Rs 2000, and because the median resists the pull of a few high incomes it is preferred to the mean for such skewed income data.

Median income ≈ Rs 1263.16.

Interpolate inside the second class. The cumulative column points straight at the median class with no guesswork.

• Locate the class: with n2=16.5 the cumulative run 11,20, shows the half-way point falls in 100–115, so l=100, cf=11, f=9, and h=15.
• Find the position: the fraction of the way into that class is 16.5-119=5.59≈ 0.611, just over halfway across the interval.
• Scale and add: the median is therefore 100+0.611× 15≈ 100+9.17=109.17 km/h.
• Why it matters: although the slowest band 85–100 holds the most players, the median speed still lands in the next band because more than half the players bowl above 100 km/h, a clear case of the modal class and median class differing.

Median bowling speed ≈ 109.17 km/h.

Plug the four numbers in once. The mode needs only the modal class together with its two immediate neighbours.

• Read the values: the modal class 10000–15000 supplies l=10000, f₁=41, f₀=26, f₂=16, and the width h=5000, which is everything the formula asks for.
• Form the pieces: the numerator f₁-f₀ comes to 15 and the denominator 2(41)-26-16 comes to 40, so the fraction is ready.
• Finish in one line: the mode is 10000+1540× 5000=10000+1875=11875 rupees.
• Why it matters: this mode tells us the single most common income band peaks just past Rs 10000, and for income policy the mode answers what most families actually earn, a figure the mean would overstate because rare high incomes drag it upward.

Modal income = Rs 11875.

Narrow classes, small correction. With a width of just one gram the mode sits barely inside the modal class.

• Read the values: the modal class 201–202 gives l=201, f₁=26, f₀=12, f₂=20, and the narrow width h=1, the figure that keeps the correction tiny.
• Form the pieces: the numerator 26-12 is 14 and the denominator 2(26)-12-20 is 20, so the fraction is simple.
• Finish in one line: the mode is 201+1420× 1=201+0.7=201.7 grams.
• Why it matters: the mode of 201.7 g shows most packets carry just over 201 g, a useful quality-control reading that the filling machine settles slightly above the 201 g mark rather than dead on it.

Modal weight =201.7 g.