Theoretical DistributionsMCQPYQ Sep 24Question 3557 of 230
All Questions

In a class of 100 students, the mean marks was 50 with Standard Deviation 14.9. Assuming the distribution of marks to be normal, find the number of students who obtained more than 70% marks (at Z=1.34\displaystyle Z = 1.34, area =0.4099\displaystyle = 0.4099).

Options

A9
B10
C8
D7
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 9

All Options:

  • A9
  • B10
  • C8
  • D7

Ad

Detailed Solution & Explanation

**Finding the Number of Students Scoring Above 70% Marks** Let X\displaystyle X represent the marks obtained by a student. Given: - Marks X\displaystyle X follow a normal distribution with mean μ=50\displaystyle \mu = 50 and standard deviation σ=14.9\displaystyle \sigma = 14.9 - Total number of students N=100\displaystyle N = 100 - 70%\displaystyle 70\% marks on a standard exam of 100 maximum marks corresponds to X=70\displaystyle X = 70. We want to find the number of students scoring more than 70\displaystyle 70: Number of Students=N×P(X>70)\text{Number of Students} = N \times P(X > 70) **Step 1: Convert X=70\displaystyle X = 70 to a standard normal Z-score** Z=X−μσZ = \frac{X - \mu}{\sigma} Z=70−5014.9=2014.9≈1.342≈1.34Z = \frac{70 - 50}{14.9} = \frac{20}{14.9} \approx 1.342 \approx 1.34 **Step 2: Find the probability P(Z>1.34)\displaystyle P(Z > 1.34)** Using the symmetry of the normal curve: P(Z>1.34)=0.5−P(0≤Z≤1.34)P(Z > 1.34) = 0.5 - P(0 \le Z \le 1.34) We are given the area under the curve from Z=0\displaystyle Z = 0 to Z=1.34\displaystyle Z = 1.34 is 0.4099\displaystyle 0.4099: P(0≤Z≤1.34)=0.4099P(0 \le Z \le 1.34) = 0.4099 Therefore: P(Z>1.34)=0.5−0.4099=0.0901P(Z > 1.34) = 0.5 - 0.4099 = 0.0901 **Step 3: Calculate the expected number of students** Expected Number=100×0.0901=9.01≈9\text{Expected Number} = 100 \times 0.0901 = 9.01 \approx 9 Thus, approximately 9 students scored more than 70%. 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

Standard Deviation

A measure of the dispersion or spread of values in a data set around the mean. It is the square root of Variance. A lower standard deviation indicates data points are clustered close to the mean; a higher value indicates greater spread.

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