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Question: FinAid is a new, free Web site that helps people obtain information on 180,000 college tuition aid awards. A random sample of 500 such awards revealed that 368 were granted for reasons other than financial need. They were based on the applicant's qualifications, interests, and other variables. Construct a 95% confidence interval for the proportion of all awards on this service made for reasons other than financial need.
A research journal states: "Rejection rate for submitted manuscripts: 86%." A prospective author believes that the editor's statement reflects the probability.
the lifetime of certain light bulbs has a mean of 2050 hours and a standard deviation of 405 hours. if you randomly
assume that womens heights are normally distributed with a mean of 63.5in and a standard deviation of 2.4in.a if a
you plan to conduct a survey to find what proportion of the workforce has two or more jobs. you decide on the 99
Determine the binomial distribution for any random data. Also determine the probability greater than 4.
If the data are normally distributed, what percentage of values will be mild outliers? - What percentage will be extreme outliers?
a recent survey asked 4381 individuals whats most important to you when choosing where to live? the responses are shown
Delivery Format The General Social Survey asks, "About how often did you have sex in the past 12 months?" About 47% of respondents indicated they had sex at least once a week.
The tolerance of the automatic filling machine, leads to weights of bags being approximately Normally distributed with standard deviation 5g.
find the probability that a card drawn from a standard deck of playing cards will be either a four or a
1.i read an article about a new drug which stated that the incidence of side effects was similar to placebo p-value gt
An accepted standard deviation for LSAT scores is 125. A random sample of 25 scores had an average of 1,375. Assume a significance level of 0.05. State the appropriate null and alternative hypotheses. Calculate the p-value to test the hypotheses.
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