Respuesta :
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.
