Question

The probability distribution for the number of automobiles lined up at a Lakeside Olds dealer at opening time (7:30 a.m.) for service is: Number Probability 1 0.05 2 0.30 3 0.40 4 0.25 On a typical day, how many automobiles should Lakeside Olds expect to be lined up at opening time?

Answer 1

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 1*0.05 +2* 0.3 +3* 0.4 +4*0.25 = 2.85$

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:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 1*0.05 +2* 0.3 +3* 0.4 +4*0.25 = 2.85$

So we are going to expect about 2,85 automobiles for this case.