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First show that $ has this sampling distribution for sample size n = 3: 13 %3 31 37 31 P($ 125 125 125 125 125...

Question

First show that $ has this sampling distribution for sample size n = 3: 13 %3 31 37 31 P($ 125 125 125 125 125

First show that $ has this sampling distribution for sample size n = 3: 13 %3 31 37 31 P($ 125 125 125 125 125



Answers

Random samples of size $n$ produced sample proportions $\hat{p}$ as shown. In each case decide whether or not the sample size is large enough to assume that the sample proportion $\hat{p}$ is normally distributed. a. $n=50, \hat{p}-0.45$ b. $n=50, \hat{p}-0.12$ c. $n=100, \hat{p}-0.12$

The formula for means of a binomial distribution is n times t. So if n equals three and we are not sure what he is just gonna say will PSP What is the mean? The mean would be three times t Because you're gonna multiply the the end, the number of trials, the number of people surveyed, whatever that end is described with whatever number gonna multiply it by P to find the me.

No. There are two formulas that you need to use to To say that a sample size is big enough to support a normal distribution. The first one is. You multiply the sample size by the p value. And if it's greater than equal to 10, your answer would be yes. But I also need these books so then your second one would be 10 times one mice. P. Also needs to be less than a quarter to 10. So you need both say yes. So for the first one you have then is 30 And the P value is .72. So in your first equation And through would be 21.6. So so far it's good. We have to the second to confirm That one would be 8.4. So it's what is not going to work. Yeah. And for B you have 10 is 30. The p value is pointing for what do the first one That gives you 25.2 And then your second one is 4.8. So that one's possibly going to be known. Then for the last one you have a separate size 75 P. Value .84. So for the love location is the first equation. You did 63 as the product. The second one You get 12. So this one is good. So baby would not be a normal distribution but see probably what

This problem works only have a binomial distribution foreigners, three and P is one half. And what we would like to do is find the cumulative distribution function for this. Now what that means is we first need to find the probability of zero probability of one probability of two And the probability of three, The probability of zero Will be 3-0 Times 1/2 to the zero Times 1/2 to the 3rd power. So evaluating this will give us 184.125, but I'll use 18. The probability of one Is three, choose 1 Times 1/2 to the one Times 1/2 squared, Which gives us 3/8. Yeah, Probability of two is 3, choose to Times 1/2 to the second, Times 1/2 to the first, Which is also 3/8 And probably three is 3, choose three Times 1/2 to the third, times one half to the zero, which is one Now for a cumulative distribution function. And Now if that's zero, The probability of that happening is 1/8. Now remember when X is one? We want anything that is less than or equal to one. And so we need to add those two values together and 1/8 plus three eight. As for AIDS. Yeah. Which is 1/2 now affects us too. That means we need to actually be less than or equal to two. So we had in that value of 3/8 And that gives us 78. And if X is three, We had in our final value here of 18 Which gives us one. And so there's a cumulative distribution function. Mhm.


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