NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.2

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.