Question

Please ignore the circled choices. I am double checking my work.

ame rangs, then what must be a) their high and low data valoes b) the entire Eistribations () their standard deviacions 0) the difference betweem their high and lew data values e) their means 2. nkewed distribution, the median is peefenred over the mean as a measure of the center becaune he median is less than the mean d) the men menures the spread in the data modian measures the arithmetie aveage of the data excluding outliers . If a distribution has a mean of 9 and five of the six data values are 6, 8, 11, 1S, und 18, them data valoo is 4 Gail received a grade of 75 on her chemistry exam. Rich received a grade of 80 on his lawexam. The mean Soe the chemistry exam is 70 with a standard deviation of 5 and the mean for the law exam is 7 with a standard deviation of 4. Whose grade is more outstanding? a) Rich's, since his grade of 80 is higher than Gail's 75 b) Gail's, ince the class mean in Chemistry is lower than the class mean in law GRich's, since he has a higher z score d) Gail's, since she has a higher z soore e) sone of these

Answer 1

Here' the answer to the question. Please don't hesitate to give a "thumbs up" in case you're satisfied with the answer

1.

If ranges are equal , it means that difference of their high and low data values is same.

range = high - low

So, d is correct

2. Due to skewness, mean may b influenced by outlier. To correct this, we prefer median over mean in case of skewed distributions.
Hence, b. is correct

3. Let the missing value be x. So:

(6+8+11+15+18+x)/6= 9
58+x =54
x = -4
So, c is correct

4. Lets check who has a higher Z score, his grade will be more outstanding

Gails' Z score = 75 - 70/5 = 1
Rich' Z score = 80-75/4 = 5/4 = 1.25
Since 1.25>1, we conclude that c. is correct.