Final answer:
To find the probability that more than 6% of 460 calls are wrong numbers, calculate the standard deviation of the sampling distribution, determine the z-score for 6%, and consult a standard normal distribution table. The probability is the area to the right of the z-score.
Step-by-step explanation:
To calculate the probability that the proportion of wrong numbers in a sample of 460 phone calls would be greater than 6%, we can assume that the distribution of the sample proportion is approximately normal if the sample size is large enough because of the Central Limit Theorem. Since the sample size is 460, which is a large sample, we can use the normal approximation.
First, we will determine the mean and standard deviation of the sampling distribution. The mean of the sampling distribution (p) is the assumed probability of getting a wrong number, which is 0.05. The standard deviation (σ_p) of the sampling distribution is calculated using the formula:
σ_p = √[p(1 - p) / n]
Where n is the sample size. Substituting the given values, we have:
σ_p = √[0.05 * (1 - 0.05) / 460]
σ_p = 0.0104
We can now find the z-score for the proportion of 0.06 by using the formula:
Z = (X - p) / σ_p
Where X is the sample proportion. We have:
Z = (0.06 - 0.05) / 0.0104
Z = 0.96
Finally, we find the probability that Z is greater than 0.96 using z-score tables or calculators. The corresponding probability is 1 - P(Z < 0.96), which can be looked up in standard normal distribution tables or calculated using statistical software.
Assuming we find this value to be 0.1685, then the probability that the proportion of wrong numbers in a sample of 460 phone calls would be greater than 6% is 0.1685.