Answer:
(a) The mean or expected value of <em>X </em>is 2.2.
(b) The standard deviation of <em>X</em> is 1.3.
Step-by-step explanation:
Let <em>X</em> = number of times the traffic light is red when a commuter passes through the traffic lights.
The probability distribution of <em>X</em> id provided.
The formula to compute the mean or expected value of <em>X </em>is:

The formula to compute the standard deviation of <em>X </em>is:

The formula of E (X²) is:

(a)
Compute the expected value of <em>X</em> as follows:

Thus, the mean or expected value of <em>X </em>is 2.2.
(b)
Compute the value of E (X²) as follows:

Compute the standard deviation of <em>X</em> as follows:

Thus, the standard deviation of <em>X</em> is 1.3.