According to given,
p = 0.12 (success)
q = 1 - 0.12
= 0.88 (failure)
x = 7
By applying binomial probability distribution we get -
P(x) = nCx.p^x.q^(n-x)
⇒ P(x=7) = nC7.(0.12)^7.(0.88)^(n-7)
Also given,
Probability of 7 successful trials = 0.77
Thus,
nC7.(0.12)^7.(0.88)^(n-7) = 0.77
⇒ n!/(7! (n-7)!).(0.12)^7.(0.88)^(n-7) = 0.77
⇒ n!/(7! (n-7)!).(0.88)^n = 0.77 * (0.88/0.12)^7
⇒ n!/(7! (n-7)!).(0.88)^n = 878214.711
⇒ n = 130
Thus, the number of calls needed to ensure that there is a probability of at least 0.77 of obtaining 7 or more successful calls is 130.
Learn more about binomial probability distribution : brainly.com/question/24756209
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Disclaimer : This question provided is incomplete, the complete question is given below
Question : A telephone call centre where people place marketing calls to customers has a probability of success of 0.12. Find the number of calls needed to ensure that there is a probability of at least 0.77 of obtaining 7 or more successful calls.
