Consider the following: A telephone exchange operator assumes that 8% of the phone calls are wrong numbers.
-If the operator is accurate, what is the probability that the proportion of wrong numbers in a sample of 726 would be less than 7%?
To find the probability that the proportion of wrong numbers in a sample of 726 would be less than 7%, we can use the normal distribution and z-scores.
Step 1: Set up the hypothesis:
Let's assume that the operator is accurate. This means that the true proportion of wrong numbers is indeed 8%.
Step 2: Calculate the standard deviation:
Since we are dealing with proportions, we need to calculate the standard deviation using the formula:
σ = √(p(1-p)/n)
where p is the true proportion (8%) and n is the sample size (726).
σ = √(0.08(1-0.08)/726)
= √(0.07264/726)
= √0.0001001
≈ 0.0100
Step 3: Convert the desired proportion (7%) to a z-score:
To calculate the z-score, we use the formula:
z = (x - μ) / σ
where x is the desired proportion (7%), μ is the mean proportion (8%), and σ is the standard deviation (0.0100).
z = (0.07 - 0.08) / 0.0100
= -0.01 / 0.0100
= -1.0
Step 4: Find the probability using standard normal distribution table:
We want to find the probability that the proportion of wrong numbers in the sample is less than 7%. This corresponds to finding the area to the left of the z-score (-1.0) on the standard normal distribution table.
Using the standard normal distribution table or a calculator, we can find that the probability associated with a z-score of -1.0 is approximately 0.1587.
Therefore, the probability that the proportion of wrong numbers in a sample of 726 would be less than 7%, assuming the operator is accurate, is approximately 0.1587 or 15.87%.