Question:

For the following probability distribution:
X345
P(X)0.50.20.3
The mean, variance, and standard deviation respectively are:

Updated On: Mar 27, 2025
  • 4, 3.8, and 0.87
  • 4, 3.8, and 0.76
  • 3.8, 4, and 0.76
  • 3.8, 0.76, and 0.87
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

To calculate the mean, variance, and standard deviation for the given probability distribution, follow these steps:

Mean (\(\mu\))

The mean is given by:

\[ \mu = \sum X \cdot P(X). \]

Substituting the values:

\[ \mu = (3)(0.5) + (4)(0.2) + (5)(0.3) = 1.5 + 0.8 + 1.5 = 3.8. \]

Variance (\(\sigma^2\))

The variance is given by:

\[ \sigma^2 = \sum (X^2 \cdot P(X)) - \mu^2. \]

First, calculate \(\sum X^2 \cdot P(X)\):

\[ \sum X^2 \cdot P(X) = (3^2)(0.5) + (4^2)(0.2) + (5^2)(0.3) = (9)(0.5) + (16)(0.2) + (25)(0.3) = 4.5 + 3.2 + 7.5 = 15.2. \]

Now calculate the variance:

\[ \sigma^2 = 15.2 - (3.8)^2 = 15.2 - 14.44 = 0.76. \]

Standard Deviation (\(\sigma\))

The standard deviation is the square root of the variance:

\[ \sigma = \sqrt{0.76} \approx 0.87. \]

Final Results:

Mean: 3.8, Variance: 0.76, Standard Deviation: 0.87.

Final Answer: (4) 3.8, 0.76 and 0.87

Was this answer helpful?
0
0