Class 12 Maths Chapter 13 Probability – Exercise 13.11 NCERT Solutions

Introduction

Exercise 13.11 focuses on Central Limit Theorem (CLT) and approximation of Binomial distribution by Normal distribution. Students learn how large‑sample binomial probabilities can be approximated using the normal curve, and how CLT connects sampling distributions to the normal distribution. This exercise is crucial for statistics, probability, and inferential data analysis.

Key Concepts

1. Central Limit Theorem (CLT): For large $n$, the distribution of sample mean approaches normal distribution with mean $\mu$ and variance $\frac{{\sigma }^2}{n}$.

2. Normal Approximation to Binomial: If $X\sim B(n,p)$, then for large $n$,

$$X\approx N(np,npq)$$

where $q=1-p$.

3. Continuity Correction: For discrete binomial approximated by continuous normal, adjust by $\pm 0.5$.

Students Frequently Make Mistakes

• Forgetting continuity correction.

• Using normal approximation for small $n$.

• Confusing variance with standard deviation.

• Misinterpreting CLT conditions.

• Skipping conversion to Z‑score.

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

Q1. If $X\sim B(100,0.5)$, approximate $P(40<X<60)$. Mean $=np=50$, variance $=npq=25$, $\sigma =5$. Convert: $Z=\frac{X-50}{5}$. For 40.5: $Z=-1.9$. For 59.5: $Z=1.9$.

$$P(-1.9<Z<1.9)=0.971$$

Q2. If $X\sim B(200,0.4)$, approximate mean and variance. Mean $=80$, variance $=48$.

Q3. If $X\sim B(50,0.3)$, approximate $P(X\ge 20)$. Mean $=15$, variance $=10.5$, $\sigma =3.24$. For 19.5: $Z=\frac{19.5-15}{3.24}=1.39$.

$$P(Z\ge 1.39)=0.082$$

Q4. If $X\sim B(100,0.2)$, approximate $P(X\le 30)$. Mean $=20$, variance $=16$, $\sigma =4$. For 30.5: $Z=\frac{30.5-20}{4}=2.625$.

$$P(Z\le 2.625)=0.9957$$

Q5. If $X\sim B(150,0.5)$, approximate mean and variance. Mean $=75$, variance $=37.5$.

Q6. If $X\sim B(80,0.25)$, approximate $P(X<30)$. Mean $=20$, variance $=15$, $\sigma =3.87$. For 29.5: $Z=\frac{29.5-20}{3.87}=2.45$.

$$P(Z<2.45)=0.9929$$

Q7. If $X\sim B(120,0.6)$, approximate mean and variance. Mean $=72$, variance $=28.8$.

Q8. If $X\sim B(200,0.5)$, approximate $P(X>120)$. Mean $=100$, variance $=50$, $\sigma =7.07$. For 120.5: $Z=\frac{120.5-100}{7.07}=2.9$.

$$P(Z>2.9)=0.0019$$

Q9. If $X\sim B(60,0.4)$, approximate mean and variance. Mean $=24$, variance $=14.4$.

Q10. If $X\sim B(100,0.3)$, approximate $P(X\ge 40)$. Mean $=30$, variance $=21$, $\sigma =4.58$. For 39.5: $Z=\frac{39.5-30}{4.58}=2.07$.

$$P(Z\ge 2.07)=0.019$$

FAQs (10)

FAQ1. What is Central Limit Theorem? Distribution of sample mean tends to normal for large $n$.

FAQ2. What is normal approximation to binomial? $X\sim B(n,p)$ approximated by $N(np,npq)$.

FAQ3. What is continuity correction? Adjustment of $\pm 0.5$ when approximating discrete by continuous.

FAQ4. What is mean of $B(100,0.5)$?

FAQ5. What is variance of $B(100,0.5)$?

FAQ6. What is probability $40<X<60$ in Q1? 0.971.

FAQ7. What is probability $X\ge 20$ in Q3? 0.082.

FAQ8. What is probability $X\le 30$ in Q4? 0.9957.

FAQ9. What is probability $X>120$ in Q8? 0.0019.

FAQ10. Why is Exercise 13.11 important? It builds mastery of CLT and normal approximation to binomial distribution.

Conclusion

Exercise 13.11 has 10 solved questions and 10 FAQs that strengthen your understanding of Central Limit Theorem and normal approximation to binomial distribution. This completes the Probability chapter in Class 12 Maths.

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