A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug is given to 40 people who have fevers. It is found that the mean time that it takes for the fever to get back to normal for this test group is 320 minutes with a standard deviation of 60 minutes. Find the 90% confidence interval for the mean time that the drug will take to reduce all fevers for all people.

The 90% confidence interval for the mean time that the drug will take to reduce all fevers for all people is between 304 minutes and 336 minutes

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 40 - 1 = 39

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 39 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.685

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 320 - 16 = 304 minutes

The upper end of the interval is the sample mean added to M. So it is 320 + 16 = 336 minutes

The 90% confidence interval for the mean time that the drug will take to reduce all fevers for all people is between 304 minutes and 336 minutes

0.8997 = 89.97% probability of a bulb lasting for at most 622 hours.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 590 hours, standard deviation of 25 hours.

This means that

Find the probability of a bulb lasting for at most 622 hours.

This is the p-value of Z when X = 622.

has a p-value of 0.8997.

0.8997 = 89.97% probability of a bulb lasting for at most 622 hours.