Question

For the following probability distribution:

X	3	4	5
P(X)	0.5	0.2	0.3

The mean, variance, and standard deviation respectively are:

• A. 4, 3.8, and 0.87
• B. 4, 3.8, and 0.76
• C. 3.8, 4, and 0.76
• D. 3.8, 0.76, and 0.87

Answer 1

Answer: (D)

3.8, 0.76, and 0.87

Answer 2

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