ersity of allu exceed 5 mean H levels that the p th obtain the ercentage fall ap

Question

ersity of allu exceed 5 mean H levels that the p th obtain the ercentage fall apply to of the sampled ut Find the interval the (c) microg r drinks of wells rule did you the t males What deviation for rs were mean and standard percentage of ss (Ani estimate? wells Queen (d) levels of the sampled the the rs were MTBE fall the uid loss. the levels that obtain is with you apply to interval. What rule did Explain. at Wit students D LOSS 1.94 Anxiety scale. the August 018 tenberg completed Reports, Scale questionnaire from 0 (no 020 Scores on the scale range was 11 017 to 20 (extreme anxiety). The mean score that 024 and the standard deviation was 3.5. Assume 020 of all scores on the Dental Anxiety 024 the distribution 11 and a is normal with 003 (a) Suppose you score a 16 on the Dental Anxiety 001 Find the z-value for this score 009 Scale. (b) Find the probability that someone scores Vol. between a 10 and a 15 on the Dental Anxiety k, and Scale, drink (c) ind the probability that someone scores above 5, with a 17 on the Dental Anxiety Scale .95. Improving SAT scores. Refer to the Chance (win piders ter 2001) study of students who paid a private tutor to help them improve their Standardized Admis sion Test (SAT) scores, table summarizing the changes in bot the sAT nectar These in the Mathematics and SAT-Verbal score for students is reproduced here. Assume that losses bothndis tributions of SAT score changes are a attern t into normal evap-

Explanation / Answer

= 11, = 3.5

(a) z = (x - )/ = (16 - 11)/3.5 = 1.43

(b) z1 = (10 - 11)/3.5 = -0.2857 and z2 = (15 - 11)/3.5 = 1.1429

P(10 < x < 15) = P(-0.2857 < z < 1.1429) = 0.486

(c) z = (17 - 11)/3.5 = 1.7143

P(x > 17) = P(z > 1.7143) = 0.0432

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