Theoretical DistributionsMCQMTP Oct 21Question 3587 of 230
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For a normal distribution with mean as 500 and SD as 120, what is the value of k so that the interval [500, k] covers 40.32 per cent area of the normal curve? (Given ϕ(1.30)=0.9032\displaystyle \phi(1.30) = 0.9032)

Options

A740
B750
C656
D800
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Correct Answer

✅ Option c — 656

All Options:

  • A740
  • B750
  • C656
  • D800

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Detailed Solution & Explanation

**Finding Interval Boundary for a Normal Curve** We are given: - Mean μ=500\displaystyle \mu = 500 - Standard deviation σ=120\displaystyle \sigma = 120 - Area of interval [500,k]\displaystyle [500, k] is 40.32%=0.4032\displaystyle 40.32\% = 0.4032 **Step 1: Standardize the interval** Let Z=X−μσ\displaystyle Z = \frac{X - \mu}{\sigma}: - For X=500\displaystyle X = 500: Z=0\displaystyle Z = 0 - For X=k\displaystyle X = k: Zk=k−500120\displaystyle Z_k = \frac{k - 500}{120} The probability is given by: P(500≤X≤k)=P(0≤Z≤Zk)=0.4032P(500 \le X \le k) = P(0 \le Z \le Z_k) = 0.4032 **Step 2: Relate to the cumulative distribution function Φ(Z)\displaystyle \Phi(Z)** P(0≤Z≤Zk)=Φ(Zk)−Φ(0)=0.4032P(0 \le Z \le Z_k) = \Phi(Z_k) - \Phi(0) = 0.4032 Since Φ(0)=0.5\displaystyle \Phi(0) = 0.5: Φ(Zk)−0.5=0.4032  ⟹  Φ(Zk)=0.9032\Phi(Z_k) - 0.5 = 0.4032 \implies \Phi(Z_k) = 0.9032 **Step 3: Solve for k\displaystyle k** We are given that Φ(1.30)=0.9032\displaystyle \Phi(1.30) = 0.9032. Therefore: Zk=1.30Z_k = 1.30 k−500120=1.30  ⟹  k−500=1.30×120=156\frac{k - 500}{120} = 1.30 \implies k - 500 = 1.30 \times 120 = 156 k=500+156=656k = 500 + 156 = 656 This matches Option C. Hence, **Option C** 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.

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