Question

The probability distribution for the number of students being absent in a class on a Saturday is as follows: \[ \begin{array}{|c|c|} \hline X & P(X)
\hline 0 & p
2 & 2p
4 & 3p
5 & p
\hline \end{array} \] Where \( X \) is the number of students absent. (i) Calculate \( p \).
(ii) Calculate the mean of the number of absent students on Saturday.

Hint:

For any probability distribution, ensure that the sum of all probabilities is 1. To calculate the mean, multiply each value of \( X \) by its corresponding probability \( P(X) \) and sum them up.

Answer 1

(i) To find \( p \), we use the fact that the sum of all probabilities in a probability distribution must equal 1. Therefore: \[ p + 2p + 3p + p = 1 \] \[ 7p = 1 \implies p = \frac{1}{7} \] (ii) The mean of the number of absent students is given by the formula: \[ \text{Mean} = E(X) = \sum (X \cdot P(X)) \] Substitute the values from the table: \[ E(X) = 0 \cdot p + 2 \cdot 2p + 4 \cdot 3p + 5 \cdot p \] \[ E(X) = 0 + 4p + 12p + 5p = 21p \] Substitute \( p = \frac{1}{7} \): \[ E(X) = 21 \cdot \frac{1}{7} = 3 \] Thus, the mean number of absent students on Saturday is 3.