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1. For a particular sample of 50 scores on a psychology exam, the following results were
obtained. First quartile = 42 Third quartile = 77 Standard deviation = 10 Range = 68
Mean = 65 Median = 68 Mode = 71 Midrange = 65 Answer each of the following; show
all work. - What score was earned by more students than any other score? Why? - What
was the highest score earned on the exam? - What was the lowest score earned on the
exam? - According to Chebyshev's Theorem, how many students scored between 45 and
85? - Assume that the distribution is normal. Based on the Empirical Rule, how many
students scored between 35 and 95?
The score that was earned by more students than any other score is 71 because mode is
the value of the item that repeats more number of times.
We have, Range = Max Value – Min Value = 68
(1)
Also, Midrange = (Min Value + Max Value)/2 = 65
That is, Min Value + Max Value = 130
(2)
Adding (1) and (2) we get, 2*Max Value = 198
Therefore, Max Value = 198/2 = 99.
Substituting this value in (1) we get,
Min Value = 99 – 68 = 31
So, the highest score earned on the exam is 99 and the lowest score earned on the exam is
31.
The interval (45, 85) can be written as (65-2*10, 65+2*10) which is same as
(Mean -k*SD, Mean +k*SD), where k = 2.
According to Chebyshev’s theorem, at least 1 - (1/k-squared) of the measurements
will fall within (Mean -k*SD, Mean +k*SD)
But 1 - (1/k-squared) = 1 - (1/4) = 1 – 0.25= 0.75
Therefore, 75% of the students scored between 45 and 85 for a data set with mean of 65
and standard deviation of 10.
Thus, the number of students scored between 45 and 85 is 37.5 = 38 (Approximately)
The interval (35, 95) can be written as (65-3*10, 65+3*10) which is same as
(Mean -3*SD, Mean +3*SD).
According to Empirical rule, approximately 99.7 % of the measurements (data)
will fall within three standard deviation of the mean.
Therefore, 99.7% of the students scored between 35 and 95 for a data set with mean of 65
and standard deviation of 10.
Thus, the number of students scored between 35 and 95 is 49.85 = 50 (Approximately)
2. Answer the following: - If the correlation coefficient is 0.91, what is the sign of the
slope of the regression line? - As the correlation coefficient decreases from -0.21 to -0.27,
do the points of the scatter plot move toward the regression line, or away from it?
The sign of the slope of the regression line is same as the sign of the correlation
coefficient and hence it is positive.
As the correlation coefficient decreases from -0.21 to -0.27, the points of the
scatter plot move toward the regression line
3. A math test was given; a sample of the scores yielded the following results: 19, 81, 48,
78, 42, 52, 7, 3, 73, 95, 94, 78, 25, 93, 66, 96 Find the range, standard deviation, and
variance for the sample scores
Range = Max Value – Min Value = 96 – 3 = 93
1 n
( xi  x )2 = 32.19705

n  1 i 1
Variance = (Standard deviation)^2 = 1036.65
Standard deviation =
4. In terms of the mean and standard deviation: - What does it mean to say that a
particular value of x has a standard score of +6.0?
It means that the score "x" was 6.0 standard deviations above the mean.
- What does it mean to say that a particular value of x has a z-score of -4.1
It means that the score was 4.1 standard deviations below the mean.
5. A student scored 81 percent on a test, and was in the 67th percentile. Explain these two
numbers
It means that 67% of students scored lower than 81% on the test.
6. An animal trainer obtained the following sample data (Table A) in a study of reaction
times of dogs (in seconds) to a specific stimulus. He then selected another group of dogs
that were much older than the first group and measure their reaction times to the same
stimulus. The sample data is shown in Table B.
Table A
Table B
Classes Frequency
Classes Frequency
2.3-2.9 32
2.3-2.9 13
3.0-3.6 17
3.0-3.6 12
3.7-4.3 25
3.7-4.3 20
4.4-5.0 14
4.4-5.0 13
5.1-5.7 4
5.1-5.7 3
5.8-6.4 20
5.8-6.4 18
Find the variance and standard deviation for the two distributions above. Compare the
variation of the data sets. Decide if one data set is more variable than the other.
We have,
Mean = x 
1
N
x f
Variance = S 2 
i i
1
N
x
f  x2
2
i i
1
 xi2 fi  x 2 , where N   fi
N
For Table A, N = 112,  xi fi = 448.7 and  xi2 f = 1971.55
Therefore,
Mean = 448.7 /112 = 4.0063
Standard deviation = S 
Variance = 1971.55/112 –(4.0063)2 = 1.5531
Standard deviation = Sqrt(1.5531) = 1.2462
For Table B, N = 79,  xi fi = 340.5 and  xi2 f = 1582.99
Therefore,
Mean = 340.5 /79 = 4.3101
Variance = 1582.99/79 –(4.3101)2 = 1.4607
Standard deviation = Sqrt(1.4607) = 1.2086
Here the variation is minimum for data given in Table B
7. You are given the following data. Number of Absences Final Grade 0 97 1 96 2 75 2
72 3 78 3 69 4 67 5 51 6 41 • Make a scatter plot for the data; • Find the correlation
coefficient for the data. • Find the equation for the regression line for the data, and predict
the final grade of a student who misses 3.5 days.
Scatter Plot
y = -9.2154x + 98.4
120
Final Grade
100
80
60
40
20
0
0
1
2
3
4
5
6
7
Number of Absences
The correlation coefficient for the data, r = -0.9565
The fitted regression equation is
Final Grade = 98.4 – 9.2154 Number of Absences
The final grade of a student who misses 3.5 days is
Final Grade = 98.4 – 9.2154*3.5 = 66.1462
8. Indicate which would be the independent variable and which would be the dependent
variable. Speed = dependent, Travel time required = independent Speed = independent,
Travel time required = dependent
Speed = independent, Travel time required = dependent
9. Indicate which would be the independent variable and which would be the dependent
variable: Crop yield = dependent, Rainfall = independent Crop yield = independent,
Rainfall = dependent
Crop yield = dependent, Rainfall = independent