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1. Calculate the mean and standard deviation of the sample. 2. Calculate the z-score for each sample. 3. Compare the z-scores to find which sample is more representative of the population.
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Find the z-score for the following means taken from a population with mean 10 and standard deviation 2: a. X̅ = 8, n = 12 b. X̅ = 8, n = 30 c. X̅ = 20, n = 4 d. X̅ = 20, n = 16
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A Standard Division of two is a population with a mean of thin. We have Meal to be close to 10 and we have our sigma to be close to two. We have to find the Z score. To get the Z score, we have to use the sample data that says Z is of course to X bar minus meal divided by sigma. The next four questions are going to be solved. The question is whether our expert is equal to eight or 12. The X bar is equal to eight and the N is equal to 2 12. Our Z is going to be divided by two by the crowd of 12. Let's use our calculator because this gives us -2 -2. So to be developed by square root of 12. We have minus two when we play the match. That means we have minus three. Please pardon me for a second question. It's equal to wait. Our n. is the same as 30 X. There's a B. There is a letter A. Our end is close to 30 and the cost is eight. Z is equal to eight minus stand divided by Zygmunt. There are 30 scrolls. We get -2 divided by 30. It needs to be 0.36 five. We will do the man's miners by 0.365. We have the answer to be -5 0.48. When our experts have close to 20 and our energy cost is four, our 3rd question is that. The cost to 20 and N is the cost to four. We have a Z. to be divided by two. Growth of four is what the vibe is. We have 10 divided by two and four. That makes us one. In this case, the car is equal to 10. The X bar is close to 20 for the fourth question. Our end is between 20 and 16. Our end is of course to 16. We have a Z2 because 220 -10 is divided by two by square root of 16. There are 10 divided by So and developed by square root. That gives us 10 and just 20. This girl is older than 20.
If a team's average score is 76, and the team got a score of 89, then it's easy to see that the team is performing better than average. But how much better than average is the team performing? This is where your standard deviation comes in.
In this problem, the standard deviation is 6. If you subtract 76 from 89, you get a difference of 13 points. You can then divide 13 by 6 to get a z-score of approximately 2.2. How good is a Z-score of +2.2? Please refer to Figure 6.4 on page 166. A Z-score of +2.2 only happens in about 2.28% of the time. What does that look like for a regular NCAA Division 1 team that plays 29 games per season? It means that the chance of a NCAA Division 1 Team having an ultra-performing night with Z-score of +2.2 is only 0.63, or less than one game. In other words, most teams in NCAA Division 1 do not experience this type of superb performance, not even for exhibition games.
So yes, a score of 89 for a team with Mean=76, SD=6 statistics is super, ultra, out-of-the world lucky good.
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