NCERT Solutions Class 10 Maths Chapter 13 Statistics

How an examiner reads this answer. Full marks here depend on a clean three-column table and a stated reason for the method.

• Quick check first: the seven classes are equal in width and the frequencies add to exactly 20 houses, so confirm that total before any arithmetic begins.
• Class-mark idea: taking each class mark as the centre of its interval treats every house in the 8–10 class as if it had exactly 9 plants, which is the standard grouped-data assumption.
• Why direct: the products f_i x_i run 1,6,5,35,54,22,39 and total 162; with the largest class mark only 13 and the largest frequency only 6, these multiplications are trivial, so the direct method is fastest and least error-prone.
• Presentation: write the Total row in bold so the examiner can verify both column sums at a glance, then quote the single division that produces the mean.
• Sanity check: the answer 8.1 sits between the smallest class mark 1 and the largest 13, which earns confidence.

x̄=16220=8.1 plants per house, by the direct method.

Pick the assumed mean from the middle row. The single decision that makes this problem easy is choosing a=550, the central class mark, so the u_i values are the small balanced integers -2,-1,0,1,2.

• Why step deviation: with wages in the hundreds, direct products like 12× 510 = 6120 invite slips; the one-digit u_i values keep the heaviest multiplication at just 12×(-2)=-24.
• Scaling back: the column f_i u_i sums to -12, and the factor h=20 outside the bracket converts the small average back to real rupees, giving a correction of -4.8 added to 550.
• Read the sign: the correction is negative because more workers, 26 of the 50, sit in the two lowest wage classes and pull the mean below the central value.
• Common slip: write a, h and the sign of ∑ f_i u_i down explicitly, since a dropped minus sign is the usual way to lose the final mark here.

x̄=550+20(-1250)=545.2 rupees.

Treat the unknown as an ordinary symbol and the algebra is short. The whole question reduces to one linear equation, so the marks come from setting it up correctly rather than from heavy arithmetic.

• Fixed plus unknown: the fixed part of ∑ f_i x_i is 84+84+144+234+110+96=752, and the class containing the unknown contributes 20f because its class mark is 20; likewise the fixed part of ∑ f_i is 7+6+9+13+5+4=44, and the missing class adds f to that total.
• One equation: putting the ratio equal to the stated mean of 18 gives 18(44+f)=752+20f; expanding to 792+18f=752+20f and moving like terms to opposite sides yields 40=2f, so f=20.
• Verify: with f=20 the total frequency is 64 children and the product sum is 1152, and dividing gives exactly 18, matching the given mean, so the answer is confirmed not just produced.
• Mark-earning habit: writing that one-line check at the end signals to the examiner that you have tested the value, which is worth a presentation mark on this type of missing-frequency question.

18(44+f)=752+20f⇒ f=20; verified since 1152÷ 64=18.

Equal class width is the green light for step deviation. Every class here spans exactly three beats, so dividing the deviations by 3 turns them into the clean integers -3,-2,-1,0,1,2,3.

• Balanced choice: choosing a=75.5 at the centre balances the positive and negative contributions, since the lower three classes give -17 and the upper three give 21, leaving a small net deviation sum of 4.
• Scaling: multiplying by the width h=3 and dividing by the 30 women scales this back to a correction of 0.4 beats just above the assumed mean, so the mean works out to 75.9.
• Why it is sensible: the result being close to 75.5 is exactly what we expect, because the frequencies are nearly symmetric about the middle class.
• Presentation: quote the assumed mean and class size on their own line before the table, because a mislabelled u_i column is the usual reason this style of answer loses a mark.

x̄=75.5+3(430)=75.9 beats per minute.

Large frequencies, small class marks, so reduce the heavy column. The frequencies run into the hundreds, so anything that keeps the multiplications one-digit is worth using, and that is exactly what step deviation does here.

• Inclusive but evenly spaced: even though the printed classes are inclusive, the class marks are evenly spaced three apart, so u_i takes the tidy values -2,-1,0,1,2 and the heaviest product drops from 110× 54 = 5940 to just 110×(-1)=-110.
• Scaling: the column f_i u_i totals 25, and scaling by h=3 over the 400 boxes gives a correction of only 0.1875, so the mean lands just above the central class mark at about 57.19.
• Why close to 57: that closeness is sensible because the two middle classes alone hold 250 of the 400 boxes, concentrating the data near the centre.
• Presentation: note the class width once, state the assumed mean, then show the single scaling step, which keeps the work compact even though the data set is large.

x̄=57+3(25400)≈ 57.19 mangoes per box.

Cancel the class width against the total before you multiply. The neatest route to 211 avoids decimals entirely by simplifying the factor 5025 to 2 at the very start.

• Centre the assumed mean: half the households sit in the 200–250 class, so choosing a=225 puts the biggest frequency on u=0 and forces the deviation sum to be small.
• Net deviation: the lower classes contribute -13 and the upper classes only 6, leaving a deviation sum of -7 to scale.
• Cancel then multiply: the correction becomes simply 2×(-7)=-14, so the mean drops to 211, below the central mark, which matches the data being skewed toward lower expenditure.
• Name the method: the question asks for a suitable method, so state step deviation and the reason of large class marks, because naming it shows the examiner the choice was deliberate and earns a presentation mark.

x̄=225+2(-7)= 211 per household per day.

Small decimals are still easier as integers. Even though the class marks are below one, the step deviation method is worth using because it replaces decimal multiplications with the integers -2,-1,0,1,2,3.

• Balanced column: choosing a=0.10 and h=0.04 gives a u_i column whose products nearly cancel, since the lower classes give -17 and the upper classes give 16, leaving a deviation sum of -1.
• Tiny correction: the correction is therefore very small, about -0.0013, so the mean lands just below the assumed mean at about 0.099 ppm.
• What it tells you: the near cancellation is a clue that the data is fairly balanced around 0.10 ppm.
• Rounding care: keep the final division as a fraction until the last step, then round once, and report the answer to three decimal places, since the data is given to two decimals and the correction sits in the third place.

x̄=0.10+0.04(-130)≈ 0.099 ppm.

Unequal class widths force the direct method. The trap here is the tempting but wrong step deviation table; because the widths jump between 6, 4, 6, 8, 10 and 2, there is no common h.

• Class marks still work: each class has a well-defined class mark, the midpoint of its own limits, even when the widths differ, so the direct method handles them all.
• The arithmetic: the products f_i x_i sum to 499, and dividing by the 40 students gives 12.475, which rounds to 12.48 days.
• Why it is low: the mean sits well below the upper classes because most students, 21 of the 40, fall in the two lowest absence bands.
• Checkpoint and reasoning: the mean must lie between the smallest class mark 3 and the largest 39, and 12.48 does, so state explicitly that the unequal classes are why the direct method is used, because an examiner rewards the reasoning not just the number.

x̄=49940=12.475≈ 12.48 days.

A small deviation sum means the answer hugs the assumed mean. With a deviation sum of -2 over 35 cities, the correction is tiny, so the mean sits a fraction below 70%.

• Near-symmetric data: the central class 65–75 holds the most cities at 11, and the two wings almost balance each other.
• Exploit the symmetry: choosing a=70 puts the largest frequency on u=0 and leaves only -2 in the deviation column, which keeps the work short.
• The result: scaling by h=10 over 35 gives a correction of about -0.57, so the mean is about 69.43%, almost exactly the central mark, which doubles as a sanity check for balanced data.
• Rounding care: keep the division as a fraction until the final line and round once, because rounding the intermediate division early can shift the last digit.

x̄=70+10(-235)≈ 69.43%.

Two measures, two different stories about the same patients. The mode comes from one class only, while the mean uses all six, which is exactly why they differ.

• Mode reasoning: spotting the modal class is the whole battle, and 35–45 carries the top frequency 23 with neighbours 21 and 14; the small numerator over a moderate denominator pushes the mode only about 1.8 above the lower limit, giving 36.8 years.
• Mean reasoning: the products f_i x_i total 2830 over the 80 patients, giving an average age of about 35.37 years.
• What the gap shows: the two numbers being close, with the mode just above the mean, tells you the age distribution is roughly symmetric with a slight bunching near 40 years.
• Interpretation mark: write one comparison sentence saying the most frequent age is about 36.8 while the average is about 35.4; quoting the numbers without that sentence usually loses the interpretation mark.

Mode ≈ 36.8 years, mean ≈ 35.37 years; the most common age is a little above the average age.

Lock the three frequencies to their positions, then it is one substitution. The only place students lose marks here is swapping f₀ and f₂, so fix them by position before touching the formula.

• Fix by position: the modal class 60–80 has f₁=61; the class immediately before gives f₀=52 and the class immediately after gives f₂=38, never the other way round.
• One substitution: the numerator 9 and the denominator 32 are both straightforward, so the mode comes out as 65.625 hours in a single step.
• Why it sits there: the answer landing a quarter of the way up from 60 is exactly what the formula should give, since the modal frequency is much closer to the class after than the class before.
• Avoid the swap: label the three frequencies above their numbers in the substitution, which is the surest way to prevent the common swap error and to show the examiner your reasoning.

Mode=60+932× 20=65.625 hours.

Mode reads one class, mean reads all eight. The contrast here is striking: the mode at about 1848 rupees is far below the mean at 2662.50 rupees, and that gap is the real lesson.

• Mode: the modal class 1500–2000 has the top frequency 40 with neighbours 24 and 33, which places the mode at about 1847.83 rupees.
• Mean: choosing a=2750 keeps the u_i values in the range -3 to 4; the deviation column sums to -35, and scaling gives a correction of -87.5, so the mean is 2662.50 rupees.
• Read the skew: the mode being far below the mean signals a right-skewed distribution, where most families cluster at lower expenditure but a tail of higher-spending families pulls the average up.
• Full marks: recognising that skew and saying so in words is what distinguishes a complete answer from one that just lists the two numbers.

Mode ≈  1847.83; mean =  2662.50 (data is right-skewed).

Keep the empty classes in place. The trap here is dropping the two zero-frequency rows, which would corrupt the u_i ladder and the mode's neighbours, so leave them in the table.

• Mode: the modal class 30–35 has f₁=10 with neighbours 9 and 3; the small numerator over denominator 8 gives a mode just 0.625 above 30, namely 30.625.
• Mean: taking a=32.5 produces a deviation column summing to -23, and the scaling gives a correction of about -3.29, so the mean is about 29.2 students per teacher.
• Read the shape: the mode being a little above the mean, with a long thin tail at 50–55, again points to a mild right skew in the data.
• Interpretation: write that a typical state runs at roughly 30 students per teacher while the all-India average is a touch lower at about 29, the difference coming from the handful of states with very high ratios.

Mode ≈ 30.6, mean ≈ 29.2 students per teacher.

A tall modal class with a low left neighbour pushes the mode well up. The jump from f₀=4 to f₁=18 is large, so the mode sits high inside the modal class.

• Big numerator: the numerator 14 is large relative to the denominator 23, so the fraction is substantial; multiplied by the big class size h=1000 it adds about 609 runs, giving 4608.7.
• Why it sits high: the mode landing more than halfway up the modal class is the formula's way of saying the data rises steeply into this class from the sparse band below and falls more gently afterwards.
• Mode is local: only the three frequencies around the modal class matter, so the long thin tail out to 11000 runs is irrelevant to the mode, a useful reminder of what the mode measures.
• Common slip: write h=1000 explicitly in the substitution, which avoids the frequent mistake of multiplying the fraction by 1 or by 100 instead of the true class size.

Mode=4000+1423× 1000≈ 4608.7 runs.

One global maximum, despite the bumpy data. The frequencies rise and fall several times, so the key skill is to ignore the minor peaks and lock onto the single tallest class.

• Pick the global peak: that class is 40–50 with f₁=20, and its neighbours 12 and 11 are close to each other, so the mode sits near the middle of the modal class.
• The value: the numerator 8 over denominator 17 times the class size 10 gives about 4.7, so the mode is about 44.7 cars.
• Why near the midpoint: because the two neighbours are almost equal, the formula places the mode only a little below the class midpoint 45, the sensible outcome for a near-symmetric peak.
• Exam habit: on multi-peak data, circle the maximum frequency first, then read its two neighbours straight from the table, so the substitution cannot pick the wrong class.

Mode=40+817× 10≈ 44.7 cars.

All three measures agree, which is the headline. The same class 125–145 is both the median class and the modal class, and the mean lands right beside them, so the data is close to symmetric.

• Median first: the cumulative column is the workhorse, since it pins the median class as the first one whose cumulative frequency passes the halfway value 34, which here is the class 125–145 with cumulative frequency 42; plugging l=125, cf=22, f=20, h=20 gives a clean median of 137 units.
• Mode and mean: the mode from the same tall central class is about 135.77, and the step-deviation mean works out to about 137.06, both sitting right beside the median.
• Why they converge: their closeness is not a coincidence, because a single dominant central class forces all three measures to cluster together rather than spread apart.
• Empirical check: the relation that three times the median equals the mode plus twice the mean is a nice safety check, since here 411 matches 409.9 to within rounding, confirming no slip crept in.
• Interpretation: state that this symmetry lets any one of the three measures stand in for the others as a fair typical consumption, which is the comment an examiner is looking for.

Median =137, mode ≈ 135.77, mean ≈ 137.06 units; nearly symmetric data.

Two unknowns need two independent equations. One comes from the total frequency and one from the median formula, and the order in which you use them matters.

• Total alone is not enough: the total gives the relation x+y=15 straight away, but it cannot find either unknown on its own, so the median formula has to supply the second equation.
• Median equation: the median 28.5 falls in the class 20–30, whose cumulative frequency from the class before is 5+x; substituting l=20, f=20, h=10 and the halfway value 30 leads to 25-x=17, so x=8.
• Then the total: only after finding x does the total relation deliver y=15-8=7, which is why the order of use matters here.
• Verify: with x=8 the cumulative frequencies reach 13 before the median class and 33 after, so the halfway count 30 does fall inside 20–30, consistent with the stated median.
• Reliable order: set up the cumulative column symbolically and solve the median equation before the total, which is the route least likely to misplace the cumulative frequency.

x=8, y=7 (since x+y=15 and the median equation gives x=8).

The hidden step is rebuilding ordinary frequencies. The table hands you cumulative totals, so the first job is to difference them; only then can the median formula be used.

• Difference the totals: subtracting successive below values gives the class frequencies 2, 4, 18, 21, 33, 11, 3, 6, 2, and the first class is 18–20 because the question restricts policies to age 18 onwards.
• Find the median class: with the halfway value 50, the cumulative frequency first crosses 50 in the 35–40 class, which has cumulative frequency 78, so that is the median class.
• Substitute: putting l=35, cf=45, f=33 and h=5 into the formula gives a median of about 35.76 years.
• Why it sits early: the median falling early inside its class is sensible, since the cumulative frequency only just passes 50 there, with 45 before and 78 after.
• Reliable habit: write the cumulative column explicitly and underline the first value that reaches the halfway count, so the median class is never misread.

Median=35+2533≈ 35.76 years.

Continuity first, formula second. The single decision that makes this answer correct is converting the inclusive classes to continuous ones before anything else.

• Shift the limits: after subtracting and adding 0.5, each class spans 9 mm and the lower limit of the fourth class becomes 144.5, which is the value the formula needs.
• Locate the median class: the cumulative column reaches 17 before the median class and 29 at its end, so with the halfway value 20 the median class is 144.5–153.5.
• Substitute: putting l=144.5, cf=17, f=12 and h=9 into the formula gives a median length of 146.75 mm.
• Why a quarter in: the median landing a quarter of the way into the class fits the fact that only 3 of the 12 leaves there are needed to pass the halfway mark.
• Presentation: show the converted class column alongside the original, so the examiner sees the continuity correction was applied deliberately rather than by accident.

Median=144.5+2712=146.75 mm.

Continuous data means no conversion, just a clean cumulative column. The lamps data is already in continuous classes, so the whole answer hinges on reading the median class correctly from the cumulative column.

• Locate the class: with the halfway count 200, the cumulative frequency jumps from 130 to 216 across the 3000–3500 class, so that class straddles the 200 mark and is the median class.
• Substitute: putting l=3000, cf=130, f=86 and h=500 into the formula gives a median lifetime of about 3406.98 hours.
• Rounding care: keep the scaling as one fraction until the last division, which avoids rounding drift when the numbers run into the thousands.
• Why a little before centre: the median sitting just before the middle of the class reflects that all 86 lamps are spread across it while only 70 of them are needed to cross the halfway point.
• Reliable habit: underline the first cumulative value that reaches the halfway count, which keeps the median class unambiguous.

Median=3000+3500086≈ 3406.98 hours.

Same class serves median and mode; the mean confirms them. The 7–10 class is both the median class and the modal class, so the three measures are bound to come out close.

• Median: the cumulative column makes the median class obvious, since it jumps from 36 to 76 across 7–10 and straddles the halfway count 50; with l=7, cf=36, f=40 and h=3 this gives a median of 8.05 letters.
• Mode: the same class is the tallest with frequency 40 and neighbours 30 and 16, so the mode comes out as about 7.88 letters.
• Mean: the step-deviation mean, with a=8.5 and a deviation sum of -6 over the 100 surnames, is 8.32 letters.
• What it shows: all three falling between 7.9 and 8.3 means the distribution is nearly symmetric with a single central peak, so a typical surname has about eight letters.
• Empirical check: the relation that three times the median roughly equals the mode plus twice the mean holds well here, 24.15 against 24.52, a quick way to confirm no arithmetic slip crept into any of the three calculations.

Median =8.05, mode ≈ 7.88, mean =8.32 letters per surname.

Median class is about position, not height. The frequent mistake here is to pick the tallest class 50–55; the median actually lands in the next class, 55–60.

• Position decides it: the cumulative frequency reaches 13 at the end of 50–55, just short of the halfway count 15, and then 19 at the end of 55–60, so the median class is 55–60.
• Substitute: with l=55, cf=13, f=6 and h=5, the median weight comes out as about 56.67 kg.
• Why a third in: the median sitting only a third of the way into its class reflects that just 2 of the 6 students there are needed to cross the halfway mark.
• Shape check: the frequencies mirror around the centre, so the data is symmetric and the median near 56.7 kg is a fair central weight.
• Dependable routine: mark the halfway count first, then scan the cumulative column top-down for the first value that meets or exceeds it, because that class, not the tallest one, is always the median class.

Median=55+106≈ 56.67 kg.