Answer:
0.215 ,10, 4
Step-by-step explanation:
Let X be the random variable of number of calls (out of total of 10 calls) during which the phone lines were occupied. X has the binomial distribution with parameters n = 10, p = 0.4.
P(X = 3) = (10 3)*0.4^3*0.67^7 = 0.215
P(X [tex]\leq[/tex] 9) = 1 — P(X > 9) = 1 — P(X = 10) =
= 1 — (10 10)*0.41^10*0.6^0
= 10
E(X) = n x p = 10 x 0.4
= 4