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Fourth root of 80y cubed
See your most recent post.
Yes, you are correct. Read about the Central Limit Theorem. It might help to clarify the concepts for you.
How about D?
Normal distribution: 99.7% is approximately +/- 3 standard deviations around the mean. 0.3% is outside those limits.
2.82843 from an online calculator. If you do this by hand, find the mean, then the variance. Standard deviation is the square root of the variance. Check your work.
I'm assuming your problem is this: (-6+i)/(-5+i) If so, then you can multiply both the numerator and denominator by an equivalent of 1: (-6+i)/(-5+i) * (-5 - i)/(-5 - i) = (30 - 5i + 6i - i^2)/(25 - i^2) = (30 + i + 1)/(25 + 1) = (31 + i)/26 Check the
Since n = 200, .6 = 120, .3 = 60, and .1 = 20 Therefore, the expected vale of X is: E(X) = sum x*p = 120(.6) + 60(.3) + 20(.1) = 92 To get the variance, subtract the mean (expected value) from each X. Next, square each of those values. Finally, add the
For 1). Your values are the following: p = .1, q = 1 - p = .9, and n = 25 To find mean and standard deviation: mean = np = (25)(.1) = ? standard deviation = √npq = √(25)(.1)(.9)= ? I'll let you finish the calculations to answer the question. For 2).
You'll need to calculate the mean and standard deviation. Once you have those values, try a t-test formula since your sample size is small. Formula: t = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
The Tukey HSD test looks at the probability of making one or more Type I errors (rejecting a true null hypothesis) in comparing population means and is sensitive to treatment errors. Could we use 2-sample t-tests? No. Multiple t-tests increase the
