Class 12 Maths Chapter 13 Probability – Exercise 13.10 NCERT Solutions

Introduction

Exercise 13.10 focuses on the Normal Distribution. Students learn how to apply the probability density function, calculate probabilities using standard normal tables, and understand the properties of the bell‑shaped curve. This exercise is crucial for statistics, data analysis, and real‑world probability modeling.

Key Concepts

1. Normal Distribution Formula:

$$f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}$$

where $\mu$ = mean, $\sigma$ = standard deviation.

2. Standard Normal Variable (Z‑score):

$$Z=\frac{X-\mu }{\sigma }$$

3. Properties:

   • Symmetrical about mean.

   • Mean = Median = Mode.

   • Total area under curve = 1.

Students Frequently Make Mistakes

• Forgetting to convert to Z‑score.

• Misreading standard normal tables.

• Confusing variance with standard deviation.

• Skipping symmetry property.

• Errors in probability calculation using cumulative distribution.

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

Q1. If $X\sim N(10,4^2)$, find $P(X=10)$. For continuous distribution, probability at a point = 0.

$$P(X=10)=0$$

Q2. If $X\sim N(20,9)$, find mean and variance. $\mu =20,{\sigma }^2=9$.

Q3. If $X\sim N(30,25)$, find standard deviation. $\sigma =\sqrt{25}=5$.

Q4. If $X\sim N(50,16)$, find $P(46<X<54)$. Convert to Z:

$$Z=\frac{X-\mu }{\sigma },\mu =50,\sigma =4$$

For 46: $Z=\frac{-4}{4}=-1$. For 54: $Z=\frac{4}{4}=1$.

$$P(-1<Z<1)=0.6826$$

Q5. If $X\sim N(100,100)$, find $P(90<X<110)$. $\mu =100,\sigma =10$. For 90: $Z=-1$. For 110: $Z=1$.

$$P(-1<Z<1)=0.6826$$

Q6. If $X\sim N(50,25)$, find $P(45<X<55)$. $\mu =50,\sigma =5$. For 45: $Z=-1$. For 55: $Z=1$.

$$P(-1<Z<1)=0.6826$$

Q7. If $X\sim N(60,36)$, find $P(54<X<66)$. $\mu =60,\sigma =6$. For 54: $Z=-1$. For 66: $Z=1$.

$$P(-1<Z<1)=0.6826$$

Q8. If $X\sim N(40,16)$, find $P(36<X<44)$. $\mu =40,\sigma =4$. For 36: $Z=-1$. For 44: $Z=1$.

$$P(-1<Z<1)=0.6826$$

Q9. If $X\sim N(70,49)$, find $P(63<X<77)$. $\mu =70,\sigma =7$. For 63: $Z=-1$. For 77: $Z=1$.

$$P(-1<Z<1)=0.6826$$

Q10. If $X\sim N(80,64)$, find $P(72<X<88)$. $\mu =80,\sigma =8$. For 72: $Z=-1$. For 88: $Z=1$.

$$P(-1<Z<1)=0.6826$$

FAQs (10)

FAQ1. What is normal distribution formula? $f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}$.

FAQ2. What is Z‑score formula? $Z=\frac{X-\mu }{\sigma }$.

FAQ3. What are properties of normal distribution? Symmetrical, mean=median=mode, total area=1.

FAQ4. What is probability at a point in continuous distribution? Always 0.

FAQ5. What is probability between $\mu -\sigma$ and $\mu +\sigma$? 0.6826.

FAQ6. What is probability between $\mu -2\sigma$ and $\mu +2\sigma$? 0.9544.

FAQ7. What is probability between $\mu -3\sigma$ and $\mu +3\sigma$? 0.9973.

FAQ8. What is mean of $N(20,9)$?

FAQ9. What is variance of $N(20,9)$?

FAQ10. Why is Exercise 13.10 important? It builds mastery of normal distribution and Z‑score applications.

Conclusion

Exercise 13.10 has 10 solved questions and 10 FAQs that strengthen your understanding of normal distribution. This completes the Probability chapter in Class 12 Maths.

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