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

The Correct Option is D

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

1. Mean (\(\mu\)):
   The mean of a probability distribution is calculated using \(\mu = \sum (X \times P(X))\).
   Calculate:
   \(\mu = (3 \times 0.5) + (4 \times 0.2) + (5 \times 0.3) = 1.5 + 0.8 + 1.5 = 3.8\).
2. Variance (\(\sigma^2\)):
   Variance is calculated using \(\sigma^2 = \sum ((X - \mu)^2 \times P(X))\).
   Calculate:
   \(\sigma^2 = ((3 - 3.8)^2 \times 0.5) + ((4 - 3.8)^2 \times 0.2) + ((5 - 3.8)^2 \times 0.3)\)
   \(\sigma^2 = (0.64 \times 0.5) + (0.04 \times 0.2) + (1.44 \times 0.3)\)
   \(\sigma^2 = 0.32 + 0.008 + 0.432 = 0.76\).
3. Standard Deviation (\(\sigma\)):
   Standard deviation is the square root of the variance:
   \(\sigma = \sqrt{0.76} \approx 0.87\).

Thus, the mean, variance, and standard deviation are 3.8, 0.76, and 0.87 respectively.

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