NCERT Exemplar Class 10 Maths Chapter 13 Statistics Exercise 13.1 Solutions

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).