Question

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve fax messages, and consider a sample of 20 incoming calls.
(a) What is the expected number of calls among the 20 that involve a fax message? E(X)-5

(b) What is the standard deviation of the number among the 20 calls that involve a fax message? (Round your answer to three decimal places.) Ox= 1.936

(c) What is the probability that the number of calls among the 20 that involve a fax transmission exceeds the expected number by more than 2 standard deviations? (Round your answer to three decimal places.)

Answer 1

Answer:

Step-by-step explanation:

let us suppose that each call out of the 20 is indepent from each other. We have that the probability of having a fax message is 0.25. Let X the number of calls among the 20 that involve a fax message.

Then, the random variable X is distributed as a binomial random variable.

Recall the following for binomial random variables

$E[X] = np, Var[X] = np(1-p)$

and that the standar deviation is the square root of the variance. Then,

a) $E[X] = np = 20\cdot0.25 = 5$

b) $\sqrt[]{np(1-p)} = \sqrt[]{20\cdot 0.25 \cdot 0.75} = 1.936$

c) We want

$P(X-E[X]>1.936*2) = P(X>8.87) = P(X\geq 9) = 1-P(X$

We have that

$P(X\leq 8 ) = \sum_{k=0}^8 \binom{20}{k} 0.25^k 0.75^{20-k} =0.959$

Then, the desired probabilty is 0.041.