Answer :
Answer:
0.8759 is the probability that no less than 4 of the entry forms will include an order.
Step-by-step explanation:
We are given the following information:
We treat subscription order with an entry form as a success.
P(subscription order with an entry form) = 0.4
Then the number subscription order with an entry form follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 14
We have to evaluate:
[tex]P(x \geq 4) =1 - P(x = 0) - P(x = 1) - P(x = 2) - P(x = 3) \\= 1 - (\binom{14}{0}(0.4)^0(1-0.4)^{14} + \binom{14}{1}(0.4)^1(1-0.4)^{13} + \binom{14}{2}(0.4)^2(1-0.4)^{12} + \binom{14}{3}(0.4)^3(1-0.4)^{11})\\=1-(0.0007+0.0074+0.0317+0.0846)\\= 0.8756[/tex]
0.8759 is the probability that no less than 4 of the entry forms will include an order.