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?
Answers
GPT 3.5
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%.
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%.