Question

It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. What is the probability that of today's 1,300 callers at least 5 received a busy signal?

Answer 1

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• 0.54% of the calls receive a busy signal, hence  p = 0.0054.

• A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

$P(X \geq 5) = 1 - P(X < 5)$

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009$

$P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062$

$P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218$

$P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513$

$P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903$

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295$

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

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