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 determine the percentage of tablets expected to contain between 243 and 262 mg of aspirin, we need to use the normal distribution. Since we know the mean (250 mg) and the standard deviation (5 mg), we can calculate the z-scores for the lower (243 mg) and upper (262 mg) limits.
The z-score for the lower limit (243 mg) can be calculated using the formula:
Z = (X - μ) / σ
where X is the lower limit, μ is the mean, and σ is the standard deviation. Plugging in the values:
Z = (243 - 250) / 5 = -1.4
Similarly, the z-score for the upper limit (262 mg) is:
Z = (262 - 250) / 5 = 2.4
Now, we can use a z-table or a statistical calculator to find the area under the normal curve between these two z-scores. This area represents the percentage of tablets expected to contain between 243 and 262 mg of aspirin.
Next, let's calculate the 95% confidence interval with 120 samples. The formula to calculate the confidence interval is:
CI = X̄ ± Z * (σ / √n)
where X̄ is the sample mean, Z is the critical z-value (from the standard normal distribution table corresponding to the desired confidence level), σ is the population standard deviation, and n is the sample size.
Since we don't have the population standard deviation, we can estimate it using the sample standard deviation (s) and the formula:
s = σ / √n
Therefore, the 95% confidence interval with 120 samples is:
CI = X̄ ± Z * (s / √n)
Now, let's determine the 95% confidence interval with a random test of 5 samples from the production line. The process is similar to the previous calculation, but with a different sample size (n).
Hope this helps!