Theoretical DistributionsMCQMTP Nov 18Question 3564 of 230
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If X is normal variate with mean 6 and variance 16 then the value of the probability. P(2≤X≤10)\displaystyle P(2 \le X \le 10) is equal to.

Options

A2P(0≤X≤10)\displaystyle 2P(0 \le X \le 10)
B2P(6≤X≤10)\displaystyle 2P(6 \le X \le 10)
CP(0≤X≤6)\displaystyle P(0 \le X \le 6)
DP(0≤X≤10)\displaystyle P(0 \le X \le 10)
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Correct Answer

✅ Option b — 2P(6≤X≤10)\displaystyle 2P(6 \le X \le 10)

All Options:

  • A2P(0≤X≤10)\displaystyle 2P(0 \le X \le 10)
  • B2P(6≤X≤10)\displaystyle 2P(6 \le X \le 10)
  • CP(0≤X≤6)\displaystyle P(0 \le X \le 6)
  • DP(0≤X≤10)\displaystyle P(0 \le X \le 10)

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Detailed Solution & Explanation

**Probability for a Normal Variate** Let X\displaystyle X be a normal random variable with: - Mean μ=6\displaystyle \mu = 6 - Variance σ2=16  ⟹  Standard Deviation σ=4\displaystyle \sigma^2 = 16 \implies \text{Standard Deviation } \sigma = 4 We want to find the probability P(2≤X≤10)\displaystyle P(2 \le X \le 10). We can standardize the variable X\displaystyle X to the standard normal variable Z\displaystyle Z: Z=X−μσ=X−64Z = \frac{X - \mu}{\sigma} = \frac{X - 6}{4} **Step 1: Standardize the limits** - For X=2\displaystyle X = 2: Z1=2−64=−1Z_1 = \frac{2 - 6}{4} = -1 - For X=10\displaystyle X = 10: Z2=10−64=1Z_2 = \frac{10 - 6}{4} = 1 **Step 2: Express the probability in terms of Z\displaystyle Z** P(2≤X≤10)=P(−1≤Z≤1)P(2 \le X \le 10) = P(-1 \le Z \le 1) **Step 3: Apply the symmetry of the normal distribution** Since the standard normal curve is symmetric about Z=0\displaystyle Z = 0: P(−1≤Z≤1)=2P(0≤Z≤1)P(-1 \le Z \le 1) = 2 P(0 \le Z \le 1) **Step 4: Convert back to the original variable X\displaystyle X** - Z=0  ⟹  X=6\displaystyle Z = 0 \implies X = 6 - Z=1  ⟹  X=10\displaystyle Z = 1 \implies X = 10 Therefore, we have: P(0≤Z≤1)=P(6≤X≤10)P(0 \le Z \le 1) = P(6 \le X \le 10) Substituting this back: P(2≤X≤10)=2P(6≤X≤10)P(2 \le X \le 10) = 2 P(6 \le X \le 10) This corresponds to Option B. *Note:* The textbook key lists Option A, which is a common typo where the mean is mistakenly treated as 0 instead of 6. Hence, **Option B** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

Variance

A measure of dispersion equal to the average of the squared deviations of each data point from the mean. Formula: σ² = Σ(xᵢ − x̄)² / n. Standard Deviation = √Variance.

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