1.Find the mean and standard deviation of the following list of quiz scores:87, 88, 65, 90.

Round the standard deviation to two decimal places.

mean

standard deviation

2.It is found that the number of raisins in a box of a popular cereal is normally distributed, with a mean of 133 raisins per box and a standard deviation of 10 raisins. My cereal box has157raisins.What is thez-scorefor this box of cereal? Round your answer to one decimal place.

z=

3.It is known that under ordinary conditions, 15% of people will not contract a certain disease. Consider the situation where test groups of 500 were selected and the percentages of those who did not contract the disease were recorded.

According the Central Limit Theorem, percentages of groups of 500 who don't contract the disease are approximately normally distributed. Find the mean and standard deviation of this normal distribution. Round the standard deviation to one decimal place.

mean = 250%, standard deviation = 2.6%

mean = 15%, standard deviation = 1.6%

mean = 85%, standard deviation = 1.6%

mean = 15%, standard deviation = 2.6%

4.It is known that under ordinary conditions,17% of people will not contract a certain disease. Consider the situation where test groups of 500 were selected and the percentages of those who did not contract the disease were recorded.

For what percentage of such groups (of 500) will88or fewer not contract the disease? Round your answer to one decimal place as a percentage

5.In a drug trial, patients showed improvement with ap-value of0.03. Explain the meaning of thep-value in this trial.

There is only a% probability that the outcome would have occurred by chance alone if the drug had no effect.

Would such a result normally be considered to be statistically significant?

Yes, such a result would normally be considered to be statistically significant.

No, such a result would not normally be considered to be statistically significant.