Answer:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) [/tex]
And replacing we got:
[tex] E(X) = 1*0.05 +2* 0.3 +3* 0.4 +4*0.25 = 2.85[/tex]
So we are going to expect about 2,85 automobiles for this case.
Step-by-step explanation:
For this case we define the random variable X as "number of automobiles lined up at a Lakeside Olds dealer at opening time (7:30 a.m.)" and we know the distribution for X is given by:
X 1 2 3 4
P(X) 0.05 0.30 0.40 0.25
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete
For this case we can calculate the epected value with this formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) [/tex]
And replacing we got:
[tex] E(X) = 1*0.05 +2* 0.3 +3* 0.4 +4*0.25 = 2.85[/tex]
So we are going to expect about 2,85 automobiles for this case.
