The aspirin content in the analgesic tablets is determined from 120 samples to have a ean of 250 mg with a standard deviation of 5 mg. from a production line, what percentage of the tablets are expected to contain between 243 and 262 mg of aspirin?What is the 95 % confidence interval with 120 samples? If in a random test of 5 samples from the production line, what is the 95% confidence interval?
To calculate the percentage of tablets expected to contain between 243 and 262 mg of aspirin, we can use the concept of the standard normal distribution or the z-score.
1. Standard Normal Distribution Approach:
First, we need to find the z-scores for both 243 mg and 262 mg using the formula:
z = (x - μ) / σ
Where:
x = value of interest (243 mg or 262 mg)
μ = mean (250 mg)
σ = standard deviation (5 mg)
For 243 mg:
z1 = (243 - 250) / 5 = -1.4
For 262 mg:
z2 = (262 - 250) / 5 = 2.4
Now, we can use a Standard Normal Distribution Table or an online calculator to find the probabilities associated with these z-scores.
From the standard normal distribution table, the corresponding probabilities are:
P(z < -1.4) = 0.08083
P(z < 2.4) = 0.9918
To find the probability between 243 and 262 mg, we subtract the probabilities:
P(243 ≤ x ≤ 262) = P(z < 2.4) - P(z < -1.4)
= 0.9918 - 0.08083
≈ 0.91097
So, approximately 91.097% of the tablets are expected to contain between 243 and 262 mg of aspirin.
2. Confidence Interval with 120 Samples:
To calculate the confidence interval, we can use the formula:
CI = x̄ ± (Z * (σ / √n))
Where:
x̄ = sample mean (250 mg)
Z = Z-value corresponding to the desired confidence level (95% confidence corresponds to Z ≈ 1.96)
σ = standard deviation (5 mg)
n = sample size (120)
CI = 250 ± (1.96 * (5 / √120))
≈ 250 ± (1.96 * 0.454)
≈ 250 ± 0.890
So, the 95% confidence interval with 120 samples is approximately (249.11, 250.89) mg.
3. Confidence Interval with 5 Samples:
To calculate the confidence interval with only 5 samples, we need to consider the smaller sample size and adjust the Z-value accordingly. The formula remains the same:
CI = x̄ ± (Z * (σ / √n))
Where:
x̄ = sample mean (250 mg)
Z = Z-value corresponding to the desired confidence level (95% confidence corresponds to Z ≈ 2.776)
σ = standard deviation (5 mg)
n = sample size (5)
CI = 250 ± (2.776 * (5 / √5))
≈ 250 ± (2.776 * 3.162)
≈ 250 ± 8.764
So, the 95% confidence interval with 5 samples is approximately (241.24, 258.76) mg.