Questions: The random variable (X) has the probability distribution table shown below. (X) 3 5 7 9 11 --- --- --- --- --- --- (P(X=X)) 0.2 0.2 (a) (a) 0.2 (a) Assuming (P(X=7)=P(X=9)), find each of the missing values. (a=) (b) Calculate (P(X geq 7)) and (P(3<x<9)). (P(x geq 7) =) (P(3<x<9) =)

Solution Steps

Given the probability distribution table for the random variable \( X \):

\[ \begin{array}{|c|c|c|c|c|c|} \hline X & 3 & 5 & 7 & 9 & 11 \\ \hline P(X=X) & 0.2 & 0.2 & a & a & 0.2 \\ \hline \end{array} \]

Assuming \( P(X=7) = P(X=9) = a \), we can set up the equation for the total probability:

\[ 0.2 + 0.2 + a + a + 0.2 = 1 \]

This simplifies to:

\[ 0.6 + 2a = 1 \]

Solving for \( a \):

\[ 2a = 0.4 \implies a = 0.2 \]

Thus, the missing probability value is:

\[ \boxed{a = 0.2} \]

To find \( P(X \geq 7) \):

\[ P(X \geq 7) = P(X=7) + P(X=9) + P(X=11) = a + a + 0.2 = 0.2 + 0.2 + 0.2 = 0.6 \]

Thus,

\[ \boxed{P(X \geq 7) = 0.6} \]

To find \( P(3 < X < 9) \):

\[ P(3 < X < 9) = P(X=5) + P(X=7) = 0.2 + 0.2 = 0.4 \]

Thus,

\[ \boxed{P(3 < X < 9) = 0.4} \]

The mean \( \mu \) of the distribution is calculated as follows:

\[ \mu = 3 \times 0.2 + 5 \times 0.2 + 7 \times 0.2 + 9 \times 0.2 + 11 \times 0.2 = 7.0 \]

The variance \( \sigma^2 \) is calculated as:

\[ \sigma^2 = (3 - 7.0)^2 \times 0.2 + (5 - 7.0)^2 \times 0.2 + (7 - 7.0)^2 \times 0.2 + (9 - 7.0)^2 \times 0.2 + (11 - 7.0)^2 \times 0.2 = 8.0 \]

The standard deviation \( \sigma \) is:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{8.0} \approx 2.828 \]

Final Answer