Solution Found!
Solved: (a) Determine the standard deviation of the
Chapter 10, Problem ACT(choose chapter or problem)
(a) Determine the standard deviation of the homework data from Problem 1.
(b) By hand, determine and interpret the interquartile range of the homework data from Problem 1.
(c) Which of these two measures of dispersion is resistant? Why?
Questions & Answers
QUESTION:
(a) Determine the standard deviation of the homework data from Problem 1.
(b) By hand, determine and interpret the interquartile range of the homework data from Problem 1.
(c) Which of these two measures of dispersion is resistant? Why?
ANSWER:Answer :
Step 1 of 2 :
Given, a random sample of eight students enrolled in Sullivan’s Intermediate Algebra course spent on the homework from Factoring Polynomials.
48 |
88 |
57 |
109 |
111 |
93 |
71 |
63 |
-
Mean =
= 640/8 = 80
x |
( |
|
48 |
-32 |
1024 |
88 |
8 |
64 |
57 |
-23 |
529 |
109 |
29 |
841 |
111 |
31 |
961 |
93 |
13 |
169 |
71 |
-9 |
81 |
63 |
-17 |
289 |
Total |
3958 |
Variance = =
=
= 565.42
Std dev = =
= 23.8
b) Quartiles
We have n = 8, so for lower quartile we have to add 2nd and 3rd value and divide it by 2
=
= 60
we have to add 4th and 5th value and divide it by 2
=
= 79.5
for upper quartile we have to add 6th and 7th value and divide it by 2
=
= 101
Inter quartile range (IQR) = -
= 101 - 60
= 41 min
the middle 50% of all study times has a range of 41 min.
c) The interquartile range is resistant, the standard deviation is not resistant.