A telephone exchange operator assumes that 8% of the phone calls are wrong numbers. If the operator is right, what is the probability that the proportion of wrong numbers in a sample of 421 phone calls would differ from the population proportion by greater than 3%? Round your answer to four decimal places.

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24-03-2020
Mathematics

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A telephone exchange operator assumes that 8% of the phone calls are wrong numbers. If the operator is right, what is the probability that the proportion of wrong numbers in a sample of 421 phone calls would differ from the population proportion by greater than 3%? Round your answer to four decimal places.

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warylucknow warylucknow · Answer 1 · 2020-03-26T07:20:05+01:00

Answer:

The probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

Step-by-step explanation:

Let X = number of phone calls that are wrong numbers.

The proportion of phone calls that are wrong numbers is, p = 0.08.

A sample of n = 421 phone calls is selected to determine the proportion of wrong numbers in this sample.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of a Binomial distribution is:

[tex]P(X=x)={421\choose x}0.08^{x}(1-0.08)^{421-x}[/tex]

Now, for the sample proportion to differ from the population proportion by 3% the value of the sample proportion should be:

[tex]\hat p-p=0.03\\\hat p-0.08=0.03\\\hat p=0.11[/tex] [tex]\hat p-p=-0.03\\\hat p-0.08=-0.03\\\hat p=0.05[/tex]

So when the sample proportion is less than 5% or greater than 11% the difference between the sample proportion and population proportion will be greater than 3%.

If sample proportion is 5% then the value of X is,

[tex]X=np=421\times 0.05=21.05\approx21[/tex]

Compute the value of P (X ≤ 21) as follows:

[tex]P(X\leq 21)=\sum\limits^{21}_{x=0}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0106[/tex]

If the sample proportion is 11% then the value of X is,

[tex]X=np=421\times 0.11=46.31\approx47[/tex]

Compute the value of P (X ≥ 47) as follows:

[tex]P(X\geq 47)=\sum\limits^{471}_{x=47}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0135[/tex]

Then the probability that the sample proportion differ from the population proportion by greater than 3% is:

[tex]P(\hat p-p>0.03)=P(X\leq 21)+P(X\geq 47)=0.0106+0.0135=0.0241[/tex]

Thus, the probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.