Arandom sample of 25 tablets of buffered aspirin contains, on average, 325.05 mg of aspirin per tablet, with a standard deviation of 0.5 mg. find the 95% confidence interval for the mean content of this brand of buffered aspirin. assume that the aspirin content is normally distributed.

Step-by-step explanation:

We want to construct a 90% confidence interval for the mean content of this brand of buffered aspirin.

Number of sample, n = 25

Mean, u = 325.05 mg

Standard deviation, s = 0.5 mg

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

325.05 ± 1.96 × 0.5/√25

= 325.05 ± 1.96 × 0.1

= 325.05 ± 0.196

The lower end of the confidence interval is 325.05 - 0.196 =324.854

The upper end of the confidence interval is 325.05 + 0.196 =325.246

Therefore, with 95% confidence interval, the mean content of this brand of buffered aspirin is between 324.854 mg and 325.246 mg

the reason why your question keeps getting deleted because this is not a school related question : )

i will answer your question though.

i never tried church's

y=1/4x-4

step-by-step explanation: