Plz explain and help me answer this question
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Answer:
And replacing we got:
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:
And replacing we got:
So we are going to expect about 2,85 automobiles for this case.