A telephone call center where people place marketing calls to customers has a probability of success of . Find the number of calls

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1. The number of calls needed to ensure that there is a probability of at least 0.83 of obtaining eight or more successful calls is 210.

   In the question, we are given that a telephone call center where people place marketing calls to customers has a probability of success of 0.05. We are asked to find the number of calls needed to ensure that there is a probability of at least 0.83 of obtaining eight or more successful calls.

   We take the given case as a binomial probability distribution, with probability of success, p = 0.05, and the probability of failure, q = 1 – p = 1 – 0.05 = 0.95.

   The x-value to check is, x = 8.

   We know that the probability distribution of x successful trials in n number of trials with a probability of success of one trial as p is given using binomial distribution as:

   P(X = x) = nCx.pˣ.qⁿ⁻ˣ.

   Substituting the values, we get:

   P(X = 8) = nC8.(0.05)⁸(0.95)ⁿ⁻⁸.

   But we are given that the probability of 8 successful trials is 0.83.

   Thus, we can show this as:

   nC8.(0.05)⁸(0.95)ⁿ⁻⁸ = 0.83,

   or, n!/{ 8! (n – 8)! } (0.95)ⁿ = 0.83 * (0.95/0.05)⁸ = 14096357324.03,

   which on solving gives us n = 210.

   Thus, the number of calls needed to ensure that there is a probability of at least 0.83 of obtaining eight or more successful calls is 210.

   Learn more about binomial probability at

   https://brainly.com/question/24756209

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   The provided question is incomplete. The complete question is:

   “A telephone call center where people place marketing calls to customers has a probability of success of 0.05. Find the number of calls needed to ensure that there is a probability of at least 0.83 of obtaining eight or more successful calls.”

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