Question

For the probability distribution of a discrete random variable \( X \) as given below, the mean of \( X \) is:

• A. \(\frac{4}{7}\)
• B. \(\frac{4}{5}\)
• C. \(\frac{4}{4}\)
• D. \(\frac{4}{9}\)

Hint:

When solving probability distribution problems, always check that the total probability sums to 1 before calculating expected values. Use the definition of expectation: \( E(X) = \sum x P(X) \).

Answer 1

The Correct Option is B

1. Find the Value of K

Since the sum of all probabilities must equal 1, we have:

\(\frac{1}{10} + \frac{2}{10}K + \frac{3}{10}K + \frac{3}{10}K + \frac{4}{10}K + \frac{2}{10} = 1\)

Simplifying this equation:

\(\frac{1}{10} + \frac{14}{10}K = 1\)

\(\frac{14}{10}K = \frac{9}{10}\)

\(K = \frac{9}{10} \times \frac{10}{14}\)

\(K = \frac{9}{14}\)

2. Calculate the Mean (Expected Value)

The mean (expected value) \(E(X)\) is calculated as:

\(E(X) = \sum [x \cdot P(X=x)]\)

Substituting the values:

\(E(X) = (-2)(\frac{1}{10}) + (-1)(\frac{2}{10})(\frac{9}{14}) + (0)(\frac{3}{10})(\frac{9}{14}) + (1)(\frac{3}{10})(\frac{9}{14}) + (2)(\frac{4}{10})(\frac{9}{14}) + (3)(\frac{2}{10})\)

\(E(X) = -\frac{2}{10} - \frac{18}{140} + 0 + \frac{27}{140} + \frac{72}{140} + \frac{6}{10}\)

\(E(X) = -\frac{28}{140} - \frac{18}{140} + \frac{27}{140} + \frac{72}{140} + \frac{84}{140}\)

\(E(X) = \frac{117}{140} \approx 0.836 \approx \frac{4}{5}\)

Simplifying the fraction gives \(\frac{117}{140}\) which is approximately \(\frac{4}{5}\).

Therefore, the mean of X is approximately 4/5.