Question

In an examination marks obtain by students in physics, maths and statistics denoted by x,y, and z are normally distributed with means 50,52 and 48 respectively and standard deviation 13,12,16, rape. the distribution of (x+y+x) is_____

• A. N(350,53)
• B. N(55,350)
• C. N(150,569)
• D. N(625,150)

Answer 1

The Correct Option is C

We need to determine the distribution of the sum \( X + Y + Z \), where \( X \), \( Y \), and \( Z \) represent the marks obtained by students in Physics, Maths, and Statistics, respectively. These are normally distributed with means 50, 52, and 48, and standard deviations 13, 12, and 16, respectively.

1. Properties of Normal Distributions:
If \( X \), \( Y \), and \( Z \) are independent normal random variables, their sum is also normally distributed. Specifically:
- \( X \sim N(50, 13^2) \)
- \( Y \sim N(52, 12^2) \)
- \( Z \sim N(48, 16^2) \)
We assume independence since the problem does not indicate any correlation between the subjects.

2. Mean of the Sum:
For normal random variables, the mean of the sum is the sum of the means:

\( \text{Mean of } (X + Y + Z) = \mu_X + \mu_Y + \mu_Z = 50 + 52 + 48 = 150 \)

3. Variance of the Sum:
Since \( X \), \( Y \), and \( Z \) are independent, the variance of the sum is the sum of the variances:

\( \text{Variance of } (X + Y + Z) = \sigma_X^2 + \sigma_Y^2 + \sigma_Z^2 \)
Calculate each variance:
- \( \sigma_X^2 = 13^2 = 169 \)
- \( \sigma_Y^2 = 12^2 = 144 \)
- \( \sigma_Z^2 = 16^2 = 256 \)
Sum the variances:

\( \text{Variance} = 169 + 144 + 256 = 569 \)
The standard deviation of the sum is:

\( \sqrt{569} \approx 23.85 \)

4. Distribution of the Sum:
Since \( X \), \( Y \), and \( Z \) are normally distributed and independent, their sum \( X + Y + Z \) is also normally distributed:

\( X + Y + Z \sim N(150, 569) \)

Final Answer:
The distribution of \( (X + Y + Z) \) is normal with mean 150 and variance 569, i.e., \( N(150, 569) \).