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Q6. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of 35

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Q6. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of 35 and a standard deviation of 6. Assuming these raw scores form a normal distribution: a) What number represents the 65 percentile (what number separates the lower 65% of the distribution)? What number represents the 90" percentile? c) What is the probability of getting a raw score between 28 and 38? d) What is the probability of getting a raw score between 41 and 44? Q7. Scores on the SAT form a normal distribution with / = 500and o =100. (a) What is the minimum score necessary to be in the top 15% of the SAT distribution? (b) Find the range of values that defines the middle 80% of the distribution of SAT scores (372 and 628). Q8. For a normal distribution, find the z-score that separates the distribution as follows: (a) Separate the highest 30% from the rest of the distribution. (b) Separate the lowest 40% from the rest of the distribution. (c) Separate the highest 75% from the rest of the distribution. Q9. IQ scores have a mean of 100 and a standard deviation of 16. Albert Einstein reportedly had an IQ of 160. (a) What is the difference between Einsteins IQ and the mean? (b) How many standard deviations is that? (c) Convert Einstein's IQ score to a z score (d) If we consider "usual IQ scores to be those that convert z scores between -2 and 2, is Einstein's IQ usual or unusual? Q10. Students in MA238 spend on average 3.75 hours a week studying (with a population standard deviation o = 5.17). (1) Given this fact, describe the sampling distribution of the mean if all possible random samples of size 50 were chosen. (11) What can you say about the mean of all the possible sample means? (iii) If 250 such samples were chosen at random and for each you calculated a 95% C.I for the mean, how many such intervals would you expect to contain the true mean? (iv) If instead you calculated a 99% C.I. for the mean, how many intervals would you now expect to capture the true mean

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