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1.
The mean, median, and mode are the most common measures of
dispersion (spread). F (T/F)
2.
Each set of data has four quartiles; they divide the ranked data into
four equal quarters.
T (T/F)
3.
The standard score (or z-score) identifies the position of a particular
value of x has relative to the mean, measured in standard deviations. T
(T/F)
4.
For any distribution, the sum of the deviations from the mean
equals zero. T (T/F)
5.
In a data set, the mode will always be unique. F (T/F)
6.
Correlation coefficients range between 0 and +1. F (T/F)
7.
The line of best fit is used to predict the average value of y that can
be expected to occur at a given value of x. T (T/F)
8.
For which of the following situations is it appropriate to use a scatter
diagram or scatter plot?
A.
Presenting two qualitative variables
B.
Presenting one qualitative and one quantitative variable
C.
Presenting two quantitative variables
D.
All of the above
9.
The only measure of central tendency that can be found for nominal
data is the
A. mean
B. median
C. mode
D. midrange
10.
The measure of central tendency that is most affected by a few
large or small numbers is
A. mean
B. median
C. mode
D. range
11.
When a distribution is bell-shaped, approximately what percentage
of data values will fall within one standard deviation of the mean?
A. 50%
B
95%
C. 68%
D. 99.7%
12. What is the relationship between the variance and the standard
deviation? What are the symbols used to represent the sample variance
and standard deviation? Why is the unbiased estimator of variance used?
Standard Deviation is the square root of the variance. Sample
Variance is presented as s2, and sample standard deviation is
presented as s. An Unbiased estimator of variance is used
because it regards the sample as a population, hence it provides
almost the exact sample variance.
13. Nine households had the following number of children per
household:
2, 0, 2, 2, 1, 2, 4, 3, 2
Find the mean, median, mode, and midrange for these data.
Mean = 2
Median = 2
Mode = 2
Midrange = 2
Please see the attached excel sheet for calculations
14. a) Use Chebyshev's theorem to find what percent of the values
will fall between 120 and 150 for a data set with mean of 135 and
standard deviation of 7.5.
150 – 135 = 15 = 2
135 – 120 = 15 = 2
1
1
1
1
 1  2  1   1  0.25
2
k
(2)
4
= 0.75 = 75%
b) Use the Empirical Rule to find what two values 95% of the data will fall
between for a data set with mean 234 and standard deviation of 12.
According to the Empirical Theory, 95% of the data values fall
within -2 and 2 Standard deviations.
2 standard deviations = 2* 12 = 24
Therefore the values are:
234 – 24 = 210
234 + 24 = 258
15. Find P25 for the following data: 2
3
4
2
1
2
0
1 3
L p  (n  1)
6
6
3
p
100
Where Lp refers to the location of the percentile, n refers to the
number of observations, and P is the Desired percentile
L 25  (11  1)
25
25
 12 
100
100
= 3rd position
Therefore, the 25th Percentile lies in the 3rd position of the ordered
data values. Re-arranging the data – as per the excel sheet – the
3rd position is 2
The 25th Percentile = 2
16. You are given the following frequency distribution for the number of
errors 10 students made in a test
Errors
Frequency
0-2
14
3-5
13
6-8
11
9-11
8
12-14
9
Find: a. mean
b. variance
a)
Mean = x 
= 6.18
 FM 340

n
55
c. standard deviation.
b)
s2 
 f (M  x ) 2
n 1
Where:
M is the midpoint of the class
F is the Class Frequency
n is the number of observations in a sample
s2 
980.1818
55  1
= 18.15
c)
s=
s 2  18.15
= 4.26
17. An aptitude test has a mean of 220 and standard deviation of 10.
find the corresponding z score for: a) a test score of 232 b) a test score
of 212
z 
x 

a)
=
232  220
10
= 1.2
b)
=
212  220
10
= -0.8
18. You are given the following data.
x
2
3
4
5
5
7
7
y
14
13
11
8
9
4
3
Find
a.
SS(x)
SS x   x 2 
= 177 
( x ) 2
n
(33) 2
7
= 21.4286
b.
SS(y)
( y ) 2
SS y   y 
n
2
= 656 
(62) 2
7
= 106.857
c.
SS(xy)
SS xy   xy 
= 245 
( x )( y )
n
(33)(62)
7
= -47.2857
d.
The linear correlation coefficient, r
r
SS xy
SS x SS y
47.2857
(21.4286)(106.857)
=
= -0.020651
-0.98817
e.
b1 
The slope b1
SS xy
SS x

47.2857
21.4286
= -2.21
f.
The y-intercept, b0
b0  y  bx
= 8.86  ( 2.21  4.71)
= 19.25
g.
The equation of the line of best fit.
Y’ = -2.21 + 19.25x