mail-attachment.googleusercontent.com (E) the effect depends on the type of dist

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mail-attachment.googleusercontent.com (E) the effect depends on the type of distribution II. Short-Answers (1) A t of 1000 test scores has a symmetric, bell-shaped distribution. If the mean is 175 and the standard deviation is 10 give the approximate number of scores between 175 and 195 of 2 (2) If you were given a graph that was skewed right severely, which summary statistics would you recommend? Explain your reasoning. (3) For any distribution, determine the percentile value for each number in the five-number sum- (4) If the mean of 75 values is 52.6 pounds and the mean of 25 values is 48.4 pounds, find the mean (5) Which grade is better? A 78 on a test whose mean is 72 and standard deviation is 6.5, or am (6) Lee's score on his test was 90. The class average was 62.1 and the standard deviation was 6.76 (7) A symmetric and bell-shape distribution has a mean of 70 and a standard deviation of 10. Find mary of all 100 values. 83 on a test whose mean is 77 and standard deviation is 8.4? Justify your answer. Is his score unusual? Why or why not? the 16th and 95% percentile scores.

Explanation / Answer

1) the probability that any one score is between 175 and 195 is:

Pr(175 x 195) = (area between x = 175 and x = 195 under the bell curve)

= (total area under curve) - (area to the left of x = 175) - (area to the right of x = 195)

= 1 - Pr(x 175) - Pr(x 195)

= 1 - Pr(z (175 - 175)/10) - Pr(z (195 - 175)/10), via u = (x - u)/

= 1 - Pr(z 0) - Pr(z 2)

= 0.5 - Pr(z 2), since Pr(z 0) = 0.5 (half of the area under the bell curve).

we have:

Pr(z 2) = Pr(z -2) (via symmetry) = 0.0228.

So, Pr(175 x 195) = 0.5 - 0.0228 = 0.4772, and approximately 0.4772 * 1000 = 477 (rounded down) test scores will be between 175 and 195.

2) Since it is severely skewed, do not use the mean or standard deviation.

Use the "median" for central tendency and then use the "interquartile range" for the variation

4)we know that the mean is the sum of all the values divided by the sample size

for the first set we have X / 75 = 52.6

for the second set we have Y / 25 = 48.4

the average of all 100 points would be (X + Y) / 100

(52.6 * 75 + 48.4 * 25) / 100 = 51.55

5) 78 is better because it is farther away from the mean towards the high end.

(78-72)/6.5=0.923

(83-77)/8.4=0.714

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