Question

#3 – Suppose a random sample of text-message lengths (in characters) resulted in the following observations: 59,60,52,66,73. What was the sample’s standard deviation?

ANSWER:

#5 – In a marketing study participant were measured as to: 1) the number of mobile devices registered to their wireless plan; and 2) the amount of data (in average GB per month) they use. The first measurement would result in _____ data and the second measurement would result in _____ data.

ANSWER:

#6 – Pat was among 170 college students participating in a quantitative skills assessment. It has been determined that there were 34 scores that were lower than Pat’s. Pat’s score therefore was the _____ percentile.

ANSWER:

#8 – For a given population, the mean is equal to 65 and the standard deviation equal to 10. An observation from this population with a standardized value of

Z = -13 would have an actual value of _______.

ANSWER:

Answer 1

Result:

#3 – Suppose a random sample of text-message lengths (in characters) resulted in the following observations: 59,60,52,66,73. What was the sample’s standard deviation?

ANSWER: 7.9057

n	5
sum	310.00
sum of squares	19,470.00

= 7.9057

#5 – In a marketing study participant were measured as to: 1) the number of mobile devices registered to their wireless plan; and 2) the amount of data (in average GB per month) they use. The first measurement would result in discrete data and the second measurement would result in continuous data.

ANSWER: discrete, continuous

#6 – Pat was among 170 college students participating in a quantitative skills assessment. It has been determined that there were 34 scores that were lower than Pat’s. Pat’s score therefore was the _____ percentile.

ANSWER:19.54

(34/174)*100 =19.54023

#8 – For a given population, the mean is equal to 65 and the standard deviation equal to 10. An observation from this population with a standardized value of

Z = -13 would have an actual value of _______.

ANSWER: 52

Let given z value -1.3

X= mean+z*sd = 65-1.3*10

= 52