Class 9 Maths Chapter 12 Statistics – Exercise 12.2 NCERT Solutions

Introduction

Exercise 12.2 focuses on mean of grouped data using the direct method, assumed mean method, and step‑deviation method. Students learn how to calculate averages from frequency distributions and apply these techniques to real‑life data sets. This exercise builds analytical skills and prepares students for advanced statistics.

Key Formulas

1. Direct Method:

$$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}$$

where $f_i$ = frequency, $x_i$ = mid‑point of class interval.

2. Assumed Mean Method:

$$\bar{x}=A+\frac{\sum f_i{d}_i}{\sum f_i}$$

where $d_i=x_i-A$, $A$ = assumed mean.

3. Step‑Deviation Method:

$$\bar{x}=A+\frac{\sum f_i{u}_i}{\sum f_i}\cdot h$$

where $u_i=\frac{x_i-A}{h}$, $h$ = class size.

Common Mistakes

• Forgetting to calculate mid‑points correctly.

• Mixing up direct and assumed mean methods.

• Arithmetic errors in summation of frequencies.

• Not multiplying frequency with mid‑point.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Find mean of marks obtained by 10 students: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.

$$\bar{x}=\frac{10+20+30+40+50+60+70+80+90+100}{10}=\frac{550}{10}=55$$

Q2. Find mean of data: Class intervals 0–10, 10–20, 20–30, 30–40; frequencies 5, 7, 10, 8. Mid‑points: 5, 15, 25, 35.

$$\bar{x}=\frac{5\cdot 5+7\cdot 15+10\cdot 25+8\cdot 35}{30}=\frac{750}{30}=25$$

Q3. Find mean of data: Class intervals 0–20, 20–40, 40–60, 60–80; frequencies 10, 15, 20, 5. Mid‑points: 10, 30, 50, 70.

$$\bar{x}=\frac{10\cdot 10+15\cdot 30+20\cdot 50+5\cdot 70}{50}=\frac{1900}{50}=38$$

Q4. Find mean of data: Class intervals 0–10, 10–20, 20–30; frequencies 3, 4, 5. Mid‑points: 5, 15, 25.

$$\bar{x}=\frac{3\cdot 5+4\cdot 15+5\cdot 25}{12}=\frac{230}{12}=19.17$$

Q5. Find mean of data: Class intervals 10–20, 20–30, 30–40, 40–50; frequencies 6, 8, 10, 6. Mid‑points: 15, 25, 35, 45.

$$\bar{x}=\frac{6\cdot 15+8\cdot 25+10\cdot 35+6\cdot 45}{30}=\frac{870}{30}=29$$

Q6. Find mean of data: Class intervals 0–10, 10–20, 20–30, 30–40; frequencies 2, 6, 8, 4. Mid‑points: 5, 15, 25, 35.

$$\bar{x}=\frac{2\cdot 5+6\cdot 15+8\cdot 25+4\cdot 35}{20}=\frac{460}{20}=23$$

Q7. Find mean of data: Class intervals 0–20, 20–40, 40–60; frequencies 5, 10, 15. Mid‑points: 10, 30, 50.

$$\bar{x}=\frac{5\cdot 10+10\cdot 30+15\cdot 50}{30}=\frac{950}{30}=31.67$$

Q8. Find mean of data: Class intervals 10–30, 30–50, 50–70; frequencies 7, 9, 4. Mid‑points: 20, 40, 60.

$$\bar{x}=\frac{7\cdot 20+9\cdot 40+4\cdot 60}{20}=\frac{660}{20}=33$$

Q9. Find mean of data: Class intervals 0–25, 25–50, 50–75, 75–100; frequencies 5, 10, 15, 20. Mid‑points: 12.5, 37.5, 62.5, 87.5.

$$\bar{x}=\frac{5\cdot 12.5+10\cdot 37.5+15\cdot 62.5+20\cdot 87.5}{50}=\frac{3125}{50}=62.5$$

Q10. Find mean of data: Class intervals 0–10, 10–20, 20–30, 30–40, 40–50; frequencies 4, 6, 8, 10, 12. Mid‑points: 5, 15, 25, 35, 45.

$$\bar{x}=\frac{4\cdot 5+6\cdot 15+8\cdot 25+10\cdot 35+12\cdot 45}{40}=\frac{1100}{40}=27.5$$

FAQs (10)

FAQ1. What is mean in statistics? Average of data values.

FAQ2. What is grouped data? Data arranged in class intervals with frequencies.

FAQ3. What is direct method? Mean = $\frac{\sum f_i{x}_i}{\sum f_i}$.

FAQ4. What is assumed mean method? Uses assumed mean to simplify calculations.

FAQ5. What is step‑deviation method? Uses deviations divided by class size.

FAQ6. Why calculate mid‑points? They represent values of class intervals.

FAQ7. What is frequency? Number of times a value occurs.

FAQ8. Why is Exercise 12.2 important? It teaches calculation of mean for grouped data.

FAQ9. What is unit of mean? Same as unit of data values.

FAQ10. What is practical use of mean? Used in averages like marks, income, speed.

Conclusion

Exercise 12.2 covers mean of grouped data with solved examples and FAQs. Mastering these problems helps students analyze data effectively and prepares them for advanced statistics.

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