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CBSE · NCERT

Class 10 Maths – Chapter 14: Statistics

Statistics helps organize, analyze, and interpret data. Learn three methods of finding the mean (Direct, Assumed Mean, Step Deviation), plus how to find the mode and median of grouped data, and draw ogives (cumulative frequency curves).

Exercises: 14.1–14.4·Total Questions: 25

Exercise 14.1 9 Questions – Mean by Direct Method

ðŸ" Direct Method: xÌ„ = Σfáµ¢xáµ¢ / Σfáµ¢. Find class mark xáµ¢ = (lower+upper)/2, multiply by frequency fáµ¢, sum, divide by total frequency.
Q 1Daily Wages

A survey was conducted to find daily wages of workers. Find mean daily wages by direct method:

Wages: 100-120(12), 120-140(14), 140-160(8), 160-180(6), 180-200(10).

xáµ¢: 110,130,150,170,190. Σfáµ¢xáµ¢ = 1320+1820+1200+1020+1900 = 7260. Σfáµ¢ = 50. xÌ„ = 7260/50 = ₹145.20

Q 2–9Applications

Q3: Mean age of 100 students. Use direct method with class marks. â†' Calculate Σfáµ¢xáµ¢/100 for given distribution.

Q5: Find mean number of mangoes per box. (Use step-deviation or direct.)

Q9: The following table gives literacy rate (in %) of 35 cities. Find mean literacy rate. â†' Class mark = (45+55)/2=50 for first class etc. Σfáµ¢xáµ¢/35 ≈ 69.43%

Exercise 14.2 6 Questions – Assumed Mean & Step Deviation Method

ðŸ" Assumed Mean: xÌ„ = a + Σfáµ¢dáµ¢/Σfáµ¢ where dáµ¢=xáµ¢âˆ'a. Step Deviation: xÌ„ = a + (Σfáµ¢uáµ¢/Σfáµ¢)×h where uáµ¢=(xáµ¢âˆ'a)/h.
Q 1Step Deviation

Find mean number of days a student was absent using step deviation method. Class: 0-6(11), 6-10(10), 10-14(7), 14-20(4), 20-28(4), 28-38(3), 38-40(1).

Let a=17, h (varies), but using step: Take a=17, calculate uáµ¢=(xáµ¢âˆ'17)/h. Σfáµ¢uáµ¢ = ... Final mean ≈ 12.48 days

Q 2–6All Three Methods

Q2: Find mean concentration of SOâ‚‚ in air (30 localities). â†' Use all three methods, verify answer matches.

Q5: Find mean number of days a student has attended school out of 220. â†' Use assumed mean or direct method.

Exercise 14.3 7 Questions – Mode & Median

ðŸ" Mode: Mode = l + ((f₁âˆ'fâ‚€)/(2f₁âˆ'fâ‚€âˆ'fâ‚‚))×h. Median: Median = l + ((n/2âˆ'cf)/f)×h. l=lower limit of modal/median class, f₁=freq of modal class, fâ‚€=freq before, fâ‚‚=freq after, cf=cumulative freq before median class.
Q 1Mode

Find mode of ages of 100 patients: 0-5(6), 5-10(11), 10-15(21), 15-20(23), 20-25(14), 25-30(5). Modal class = 15-20 (f₁=23). f₀=21, f₂=14, l=15, h=5.

Mode = 15 + ((23âˆ'21)/(46âˆ'21âˆ'14))×5 = 15 + (2/11)×5 = 15 + 0.91 = 15.91 years

Q 2Median

If median of distribution 60, find x and y: Given total frequency=50 and median class determined.

Build cf table. Median class is where cf ≥ n/2=25. Solve for x and y using total frequency constraint.

Q 3–7Applications

Q3: Find mean, median, mode of shopping hours data. Compare them. â†' x̄≈75.8, median≈76.7, mode≈78.3. Mean < Median < Mode (negatively skewed).

Q5: Find median of student percentages.

Q7: Find mean, median, mode of distribution and compare.

Exercise 14.4 3 Questions – Ogives (Cumulative Frequency Curves)

ðŸ" Ogive: Less than ogive – plot upper limit vs cumulative frequency. More than ogive – plot lower limit vs descending cumulative frequency. Intersection gives median.
Q 1Draw Ogive

The following distribution gives daily income of 50 workers. Convert to less than type and draw its ogive. Find median from graph.

Income: 100-120(12), 120-140(14), 140-160(8), 160-180(6), 180-200(10). Less than cf: 12, 26, 34, 40, 50. Plot points (120,12),(140,26)... Draw smooth curve. n/2=25, from graph median ≈ ₹138.50

Q 2–3Both Types

Q2: During medical check-up of 35 students, weights recorded. Draw less than and more than type ogives. Find median from graph.

Q3: Give less than ogive data, find median for 60 observations. â†' x-coordinate at y=30 on less than curve.

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