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The average age of trees in a large local park is 60 years with a standard deviation of 2.2 years. A simple random sample of 400 trees is selected, and the sample mean age of these trees is computed.
We know the random variable has an approximately normal distribution because of
a. the law of large numbers.b. the central limit theorem.c. the 68-95-99.7 rule.d. the fact that probability is the long-run proportion of times an event occurs.
According to the survey, 37% of the adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that:
For the following statement, write the null hypothesis and the alternative hypothesis. Also label which one is the claim.
Given a p-cap of 0.16, n = 100, construct a Confidence Interval of 95%. What does the Range created mean? What happends if n changes to 10? What does the new Range mean?
Ten pairs of points yielded a correlation coefficient y of 0.79. If a=0.05, which of the following statements is correct if H 0:P=0? (Do not calculate a t-value.)
Student who scores in the top 5% of statewide scores. How high should a student score be to win this award? Give your answer to the nearest integer.
Suppose you obtain a 100 bag random sample by taking 25 subgroups of 4 bags each, the mean weight of the 25 subgroups was 16.00 ounces with a total of the subgroup ranges of 87.5 oz.
For each of the following populations, would a score of X = 50 be considered a central score (near the middle of the distribution) or an extreme score (far out in the tail of the distribution)?
There is concern about the percent of on-time flights. A recent survey of two major airlines yielded the following results. At α = 0.05 test the claim that there is a difference in proportions of on-time flights.
From past experience, a server at a restaurant knows that her tips follow a right-skewed distribution with mean $6 and standard deviation $4.
Find the probability that fewer than 5 are satisfied with the airlines. (c) Find the probability that 8 or more are satisfied with the airlines.
Compute a 90% confidence interval of the mean for a Weschler intellegence test with mean of 100.00 and standard deviation of 15.00 for a sample of 100 people. Use two decimal places.
Design a 95 percent confidence interval for the proportion of defectives in a large lot, if a sample of 200 showed 12 defectives.
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