NCERT Exemplar Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

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.