Theoretical DistributionsMCQPYQ Dec 21Question 3550 of 230
All Questions

Let x\displaystyle x be normal distribution with mean 2.5 and variance 1. If P(x≤2.5)=0.4772\displaystyle P(x \le 2.5) = 0.4772 and that the cumulative normal probability value at 2 is 0.9772, then a=?\displaystyle a = ?

Options

A0.5
B3
C- 3.5
D- 4.5
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 0.5

All Options:

  • A0.5
  • B3
  • C- 3.5
  • D- 4.5

Ad

Detailed Solution & Explanation

**Finding Parameter a\displaystyle a in Normal Distribution** Let X∼N(μ,σ2)\displaystyle X \sim N(\mu, \sigma^2) where μ=2.5\displaystyle \mu = 2.5 and σ2=1  ⟹  σ=1\displaystyle \sigma^2 = 1 \implies \sigma = 1. **Given cumulative probability information:** - Cumulative probability value at Z=2\displaystyle Z = 2 is 0.9772\displaystyle 0.9772: Φ(2)=0.9772\Phi(2) = 0.9772 - By symmetry of the standard normal distribution: Φ(−2)=1−Φ(2)=1−0.9772=0.0228\Phi(-2) = 1 - \Phi(2) = 1 - 0.9772 = 0.0228 **Interpreting the question statement:** The expression P(x≤2.5)=0.4772\displaystyle P(x \le 2.5) = 0.4772 contains a typo because the cumulative probability up to the mean μ=2.5\displaystyle \mu = 2.5 must be exactly 0.5\displaystyle 0.5 (P(X≤2.5)=0.5\displaystyle P(X \le 2.5) = 0.5). The intended statement refers to the interval between a\displaystyle a and the mean 2.5\displaystyle 2.5: P(a≤X≤2.5)=0.4772P(a \le X \le 2.5) = 0.4772 **Step 1: Set up the equation for a\displaystyle a** P(a≤X≤2.5)=P(X≤2.5)−P(X≤a)=0.5−P(X≤a)=0.4772P(a \le X \le 2.5) = P(X \le 2.5) - P(X \le a) = 0.5 - P(X \le a) = 0.4772 P(X≤a)=0.5−0.4772=0.0228P(X \le a) = 0.5 - 0.4772 = 0.0228 **Step 2: Standardize the variable** Za=a−μσ=a−2.51=a−2.5Z_a = \frac{a - \mu}{\sigma} = \frac{a - 2.5}{1} = a - 2.5 We need to find Za\displaystyle Z_a such that: P(Z≤Za)=0.0228P(Z \le Z_a) = 0.0228 Since we know Φ(−2)=0.0228\displaystyle \Phi(-2) = 0.0228, we equate: Za=−2Z_a = -2 a−2.5=−2  ⟹  a=2.5−2=0.5a - 2.5 = -2 \implies a = 2.5 - 2 = 0.5 Hence, **Option A** 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.

More Questions from Theoretical Distributions

Find the variance of binomial distribution with n=10\displaystyle n=10, p=0.3\displaystyle p=0.3

When p=0.5\displaystyle p=0.5, the binomial distribution is

If mean and standard deviation of a binomial distribution is 10\displaystyle 10 and 2\displaystyle 2 respectively, q\displaystyle q will be

A random variable X\displaystyle X follows Binomial Distribution With E(x)=2\displaystyle E(x)=2 and V(x)=1.2\displaystyle V(x)=1.2, then the value of n\displaystyle n is

When 'p\displaystyle p' is large than 0.5\displaystyle 0.5, the Binomial Distribution is:

For a Poisson distribution,

Ready to Master Theoretical Distributions?

Practice all 230 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free