Dr. Hernandez plans to measure FEV in a random sample of n young women from a certain popu- lation, and to use the sample mean y ̄ as an estimate of the population mean. Let A be the event that Hernandez’s sample mean will be within ±100 ml of the popula- tion mean. Assume that the population is normal with mean 3,000 ml and standard deviation 400 ml. Find Pr(A) if
(a) n = 15
(b) n = 60;
Please help, this isn't for homework points just so I can understand the concept for my midterm! I'm very confused.
To find Pr(A), we need to determine the probability that the sample mean y ̄ falls within ±100 ml of the population mean.
We know that the population mean (μ) is 3,000 ml and the standard deviation (σ) is 400 ml.
(a) For n = 15:
To calculate Pr(A), we need to find the standard error of the sample mean (SE), which is given by:
SE = σ / √n
In our case, σ = 400 ml and n = 15:
SE = 400 / √15 ≈ 103.28 ml
Now, we can find the probability using the standard normal distribution. However, since we need to consider both sides of the interval (within ±100 ml), we will use the z-distribution.
To find the z-score corresponding to ±100 ml, we use the formula:
z = (x - μ) / σ
For the lower bound:
z_lower = (3000 - 100 - 3000) / 400 = -0.25
For the upper bound:
z_upper = (3000 + 100 - 3000) / 400 = 0.25
Now, we can find the cumulative probabilities corresponding to these z-scores using a standard normal distribution table or a calculator.
Pr(A) = P(-0.25 < z < 0.25)
Using a table or calculator, we find that the probability is approximately 0.0968, or 9.68%.
(b) For n = 60:
Using the same formulas, we can calculate the standard error:
SE = 400 / √60 ≈ 51.64 ml
Calculate the z-scores:
For the lower bound:
z_lower = (3000 - 100 - 3000) / 400 = -0.25
For the upper bound:
z_upper = (3000 + 100 - 3000) / 400 = 0.25
Find Pr(A) = P(-0.25 < z < 0.25) using a table or calculator.
The probability is approximately 0.9612, or 96.12%.
Therefore, for sample size n = 15, Pr(A) is approximately 0.0968, or 9.68%, and for sample size n = 60, Pr(A) is approximately 0.9612, or 96.12%.
To find the probability Pr(A), we need to determine the probability that Dr. Hernandez's sample mean, y ̄, falls within ±100 ml of the population mean.
Note that when the sample size is large enough (typically n > 30), the sample mean is approximately normally distributed, regardless of the shape of the population distribution. In this case, since n > 30, we can assume that the sample mean follows a normal distribution.
Since we know the population mean (μ) is 3,000 ml and the standard deviation (σ) is 400 ml, we can use these values to calculate Pr(A).
(a) When n = 15:
First, we need to find the standard deviation of the sample mean, also known as the standard error (SE). The formula for the standard error is:
SE = σ / √n
where σ is the population standard deviation and n is the sample size.
SE = 400 / √15
≈ 103.28
Next, we need to find the z-scores for the boundaries of the interval ±100 ml from the population mean. The z-score formula is:
z = (x - μ) / σ
where x is the value and μ is the mean.
For the upper bound:
z_upper = (3,100 - 3,000) / 103.28
≈ 0.97
For the lower bound:
z_lower = (2,900 - 3,000) / 103.28
≈ -0.97
Using a z-table or a calculator, we can find the probability associated with each z-score. Let's assume the probability for z_upper is P_upper and the probability for z_lower is P_lower.
Pr(A) = P_lower + P_upper
= 2 * P(z ≤ 0.97) - 1
≈ 2 * 0.8365 - 1
≈ 0.673
(b) When n = 60:
SE = 400 / √60
≈ 51.96
z_upper = (3,100 - 3,000) / 51.96
≈ 1.93
z_lower = (2,900 - 3,000) / 51.96
≈ -1.93
Pr(A) = P_lower + P_upper
= 2 * P(z ≤ 1.93) - 1
Using a z-table or a calculator, we can find the probability associated with each z-score.
I hope this helps you understand how to calculate Pr(A) when the population is normal and the sample mean is used as an estimate of the population mean. Good luck with your midterm! Let me know if you have any other questions.