Question

Based on historical data, your manager believes that 40% of the company's orders come from first-time customers. A random sample of 126 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.24 and 0.46?

Answer 1

To calculate the probability that the sample proportion is between 0.24 and 0.46, we need to find the z-scores for both ends of the interval and then calculate the area between these two z-scores.

The sample proportion, denoted as p̂, is assumed to follow a normal distribution with a mean equal to the population proportion (p) and a standard deviation calculated as the square root of (p*(1-p))/n, where n is the sample size.

In this case, p = 0.40 (as given in the problem) and n = 126 (the sample size).

To find the z-score for p̂ = 0.24, we can calculate it using the formula:
z1 = (p̂ - p) / sqrt((p*(1-p))/n)
= (0.24 - 0.40) / sqrt((0.40*(1-0.40))/126)
= -0.16 / sqrt((0.24*0.60)/126)
= -0.16 / sqrt(0.144/126)
= -0.16 / sqrt(0.00114286)
= -0.16 / 0.03383085
= -4.72992047

Similarly, for p̂ = 0.46, the z-score can be computed as follows:
z2 = (p̂ - p) / sqrt((p*(1-p))/n)
= (0.46 - 0.40) / sqrt((0.40*(1-0.40))/126)
= 0.06 / sqrt((0.40*0.60)/126)
= 0.06 / sqrt(0.144/126)
= 0.06 / sqrt(0.00114286)
= 0.06 / 0.03383085
= 1.77237602

Next, we need to find the area under the standard normal distribution curve between these two z-scores. We can use a standard normal distribution table or a calculator to find these probabilities.

Using the table or a calculator, we can find that the area to the left of z1 (-4.72992047) is almost 0 and the area to the left of z2 (1.77237602) is approximately 0.9625.

Therefore, the probability that the sample proportion is between 0.24 and 0.46 is given by:
P(0.24 < p̂ < 0.46) = P(-4.72992047 < z < 1.77237602)
= P(z < 1.77237602) - P(z < -4.72992047)
= 0.9625 - 0
= 0.9625

So, the probability is approximately 0.9625 or 96.25%.