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Assume that the population of heights of female college studens is apporximately normally distributed with a mean of 66 inches and standard deviation of 4.0 inches; find the proportion of female college students whose height is greater than 65 inches. Then find the proportion of female college students whose height is no more than 65 inches.
Does this result support original claim, or is population percentage considerably less than 45%. Use alpha=.05 for level of significance.
A sample of n=9 scores is obtained from a normal population distribution with o-=12. The sample mean is M=60.
Calculate s(k) for each natural number k from 1 through 15. Are the numbers SQRT(5), PI, and -6 in the domain of the function s? What is the domain of the function s?
Solve inequality and graph the solution. Please explain how to how to solve this linear graph
If an overrun contract was bid at 12 million dollars, what does the least-squares line predict for the cost of overrun (as a percent of bid price)?
The maker of refrigerators buys bolts from two suppliers, and it is very important that the mean widths of the bolts received from both suppliers are equal since they must be used interchangeably.
Assume that stock is currently selling for $100. The change in the stock's price throughout the nest year follows a normal random variable with a mean of $10 and a standard deviation of $20.
martindate were tester which can compare two materials in sigle run, weights losses from seven runs. Examine these data by creating the 90% confidence interval for difference of means. Define your assumption.
Let Z be a standard normal random variable. Determine the value of c such that P (Z > c) = 0.1251 is there a way to do this step by step so i can understand it?
As it is a 2 × 2 table, try also two-tailed two-sample z test for π 1 = π 2 and verify that z 2 is same as your chi-square statistic. Which test do you prefer? Explain why?
Based on a scatter diagram, would you estimate the linear correlation coefficient to be:
Test whether the relationship between the plan at retirement and residential area using Chi square test.
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