A commuter must pass through five traffic lights on her way to work, and she will have to stop at each one that is red. After years of commuting she has developed the following probability distribution for the number of red lights she stops at each day on her way to work: No. of red lights x 0 1 2 3 4 5 Probability .05 .25 .30 .20 .15 .05 Note that the standard deviation of the above probability distribution is SD(X) = 1.27.

Step-by-step explanation:

Hello!

You have the information about the number of red lights a commuter pass on her way to work and the probability of them stoping her:

Be x: number of red lights

X: 0,         1,        2,       3,      4,      5

hi: 0.05, 0.25, 0.30, 0.20, 0.15, 0.05

a. What is the expected number of red lights at which she will stop on her way to work?

The expected number of red lights is the sample mean, you can calculate it using the following formula:

She's expected to be stopped by 2.3 red lights on the way to work.

b. Suppose each red light delays the commuter 1.8min. What is the standard deviation od the number of minutes that she is delayed by red lights?

If each light delays the commuter 1.8 min then you can determine a new variable of interest:

Be Y: the time a commuter is delayed by red lights on the way work, then Y= X*1.8min

Meaning if X= 0, then Y=0 (the commuter will be delayed 0 min), if X=1, then Y= 1.8min, if X=2, then Y= 3.6min and to on....

The properties of variance state that if

Y= X*k (Where K= constant)

Then the sample variance of Y will be

V(Y)= V(X*k)= k²*V(X)

Then the standard deviation of Y will be the constant k by the standard deviation of X:

Sy= k*Sx= 1.8 * 1.27= 2.286

I hope it helps!