Question

Let X be a random variable, and let P(X = x) denote the probability that X takes the value x. Suppose that the points (x, P(X = x)), x = 0,1,2,3,4, lie on a fixed straight line in the xy -plane, and P(X = x) = 0 for all x ∈ R - {0,1,2,3,4}. If the mean of X is \(\frac{5}{2}\) , and the variance of X is α, then the value of 24α is ______.

Answer 1

Correct Answer: 42

Probability Distribution and Variance Calculation 

1. Given Probabilities

Let:

• \( P(X = 0) = a \)
• \( P(X = 1) = b \)
• \( P(X = 2) = c \)
• \( P(X = 3) = d \)
• \( P(X = 4) = e \)

Since the points \( (0, a), (1, b), (2, c), (3, d), (4, e) \) lie on a straight line, the probabilities form an Arithmetic Progression (A.P.).

2. Total Probability Condition

By the probability sum rule:

\[ a + b + c + d + e = 1 \]

3. Mean Condition

The mean of \( X \) is:

\[ \mu = \sum X P(X) = 0 \cdot a + 1 \cdot b + 2 \cdot c + 3 \cdot d + 4 \cdot e = b + 2c + 3d + 4e \]

Given that \( \mu = \frac{5}{2} \), we get:

\[ b + 2c + 3d + 4e = \frac{5}{2} \]

4. Arithmetic Progression Condition

Since \( a, b, c, d, e \) are in A.P., they satisfy:

\[ b - a = c - b = d - c = e - d = k \]

Rewriting in terms of \( a \) and \( k \):

• \( b = a + k \)
• \( c = a + 2k \)
• \( d = a + 3k \)
• \( e = a + 4k \)

5. Solving for \( a \) and \( k \)

Using total probability condition:

\[ a + (a + k) + (a + 2k) + (a + 3k) + (a + 4k) = 1 \]

\[ 5a + 10k = 1 \Rightarrow a + 2k = \frac{1}{5} \]

Using mean condition:

\[ (a + k) + 2(a + 2k) + 3(a + 3k) + 4(a + 4k) = \frac{5}{2} \]

\[ 10a + 30k = \frac{5}{2} \Rightarrow 2a + 6k = \frac{1}{4} \]

Solving equations:

\[ 2k = \frac{5}{20} - \frac{8}{20} = -\frac{3}{20} \Rightarrow k = -\frac{3}{40} \]

\[ a + 2 \times -\frac{3}{40} = \frac{1}{5} \Rightarrow a = \frac{7}{20} \]

6. Find \( b, c, d, e \)

\[ b = a + k = \frac{7}{20} - \frac{3}{40} = \frac{11}{40} \]

\[ c = a + 2k = \frac{7}{20} = \frac{14}{40} \]

\[ d = a + 3k = \frac{7}{20} - \frac{9}{40} = \frac{1}{8} \]

\[ e = a + 4k = \frac{7}{20} - \frac{12}{40} = \frac{1}{10} \]

7. Variance of \( X \)

Variance formula:

\[ \sigma^2 = \sum X^2 P(X) - (\text{mean})^2 \]

\[ \sum X^2 P(X) = 0 + \frac{11}{40} + 4 \times \frac{7}{20} + 9 \times \frac{1}{8} + 16 \times \frac{1}{10} \]

\[ = \frac{11}{40} + \frac{28}{40} + \frac{45}{40} + \frac{64}{40} = \frac{148}{40} \]

Mean squared:

\[ \mu^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]

\[ \sigma^2 = \frac{148}{40} - \frac{25}{10} = \frac{37}{10} - \frac{25}{10} = \frac{12}{10} \]

8. Compute \( 24 \sigma^2 \)

\[ 24 \times \frac{12}{10} = 42 \]

Final Answer: 42