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Accelerated Math III

Accelerated Math III. Statistics Summary. One Minute Question. Of f(x) = x 2 – 2x – 15 and g(x) = x 2 – x – 12, write the coordinates of the hole in. Answer:. So the x-coordinate of the hole is -3 and the reduced y = and y(-3) =. Sample Quiz Problem.

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Accelerated Math III

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  1. Accelerated Math III Statistics Summary

  2. One Minute Question • Of f(x) = x2 – 2x – 15 and g(x) = x2 – x – 12, write the coordinates of the hole in .

  3. Answer: • So the x-coordinate of the hole is -3 and the reduced • y = and y(-3) =

  4. Sample Quiz Problem At the Alpharetta Math Tournament, 380 varsity students took the written test. Their mean score was 108 with a standard deviation of 22. We believe these scores are normally distributed. • Are these numbers statistics or parameters? How do you know? • What is the z-score for a test grade of 100 and what does that mean? • How many people should have scored above 120? • How many people should have scored between 84 and 128? • If 90% of the mathletes scored within an evenly spaced interval about the mean, what are the endpoints of that interval?

  5. Solutions: • A) These are parameters because we have the entire population of the mathletes at the tournament. • B) A z-score = which means that score is 0.3636 standard deviations below the mean. Furthermore, by finding normalcdf(-9999, -0.3636), we also know that 35.81% of these mathletes scored lower than 100. • C) This is a normal distribution and , so we can find normalcdf(6/11, 9999) = 0.2927. So 0.2927 ∙380 ≈ 111 students scored above 120. • D) These values are 1 standard deviation from the mean. Using the Empirical rule, the answer should be 68%. Using normalcdf(-1, 1), we get 68.27%, so the answer is 258 or 259 mathletes. • E) The upper z-score will be invNorm(.95) = 1.645, and 1.645 standard deviations from the mean give you an interval from 72 to 144.

  6. Problem #2 • I looked up Math II EOCT scores for 18 students in one of the Math II classes at Lassiter: • A) Are these numbers statistics or parameters? • Statistics • B) What is the mean and standard deviation of this data? • The mean is 457 with standard deviation of 37.19 • C) What type of sample is this? (Last year) • Convenience or cluster sample – probably not random

  7. Problem #2 • I looked up Math II EOCT scores for 18 students in one of the Math II classes at Lassiter: • D) What is our best estimate for the mean and standard deviation of all Math II EOCT scores? • The mean is 457 with standard deviation of 37.19 • E) What is our best estimate for the mean of all such samples and the standard deviation of the means? • The mean is 457 with standard deviation of 8.766

  8. Problem #2 • I looked up Math II EOCT scores for 18 students in one of the Math II classes at Lassiter: • F) What would the Central Limit Theorem say about this situation? • The distribution of the means is approx. normal with mean approx. equal to the population mean and standard deviation approx. equal to the population standard deviation divided by the square root of 18.

  9. Problem #2 • I looked up Math II EOCT scores for 18 students in one of the Math II classes at Lassiter: • G) What is the probability that one of these 18 scores is greater than 500? • 2/18 = 11.11% • H) If the population mean is 457 and the population standard deviation is 37.19, what is the probability a randomly chosen Math II EOCT score will be greater than 500? • normalcdf((500-457)/37.19, 9999)= 12.38%

  10. Problem #2 • I looked up Math II EOCT scores for 18 students in one of the Math II classes at Lassiter: • G) What is the probability that one of these 18 scores is greater than 500? • 2/18 = 11.11% • I) What is the probability that a random sample of 18 students will have a mean that is greater than 500? • normalcdf((500-457)/(37.19/, 9999)= .0000004667

  11. Problem #2 • I looked up Math II EOCT scores for 18 students in one of the Math II classes at Lassiter: • J) Write a 99% confidence interval for the mean of the Math II EOCT scores. • InvT(.995, 17) = 2.898, so the margin of error is 2.898 ∙ 37.19/ = 25.4, so the 99% confidence interval is [432, 482].

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