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1. You are given the following frequency distribution for the number of errors that
students made in a test Errors Frequency 0-2 11 3-5 18 6-8 13 9-11 7 12-14 19 Find the
mean, variance, and standard deviation.
Errors
0-2
3-5
6-8
9-11
12-14
Midvalue x Frequency f
1
11
4
18
7
13
10
7
13
19
68
x*f
11
72
91
70
247
491
x2f
11
288
637
700
3211
4847
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
2
Here N = 68,  xi fi = 491 and  xi f = 4847
Standard deviation =
S
Therefore,
Mean = 491/68 = 7.2206
Variance = 4847/68 –(7.2206)2
= 19.1425
Standard deviation = Sqrt(19.1425) = 4.3752
2. You are given the following data. x y 2 0 3 11 4 13 5 16 5 6 7 3 7 16 Find: - SS(x) SS(y) - SS(xy) - The linear correlation coefficient, r - The slope b1 - The y-intercept, b0 The equation of the line of best fit
Let y = b0 + b1 x be the equation of the line of best fit.
x
2
3
4
5
5
7
7
33
Thus from the given data
y
0
11
13
16
6
3
16
65
x*y
0
33
52
80
30
21
112
328
x2
4
9
16
25
25
49
49
177
y2
0
121
169
256
36
9
256
847
n
 xi =33,
n=7,
i 1
n
n
 yi =65,
 xi2 = 177,
i 1
i 1
n
 yi2 = 847 and
i 1
n
 x y = 328
i 1
i
i
2
 n 
2
  xi 
n
(33)
2
SS(x) =  xi   i 1  = 177 
= 21.4286
7
n
i 1
2
 n 
  yi 
n
(65) 2
2
SS(y) =  yi   i 1  = 847 
= 243.4286
7
n
i 1
n
n
n
 xi  yi
33*65
= 21.5714
7
n
i 1
We have the linear correlation coefficient,
SS( xy)
r
SS( x)SS( y)
21.5714
=
21.4286*243.4286
= 0.2987
SS(xy) =
x y 
i
The slope b1 =
i 1
i 1
i
= 328 
21.5714
SS( xy )
=
21.4286
SS( x)
= 1.0067
n
 yi
n
x
i
65
33
 (1.0067) *
= 4.54
7
7
n
n
The equation of the line of best fit is y = 4.54 +1.0067 x
The y-intercept, b0 =
i 1
 b1
i 1
=
3. - Use Chebyshev’s theorem to find what percent of the values will fall between 155
and 187 for a data set with mean of 171 and standard deviation of 4. - Use the Empirical
Rule to find what two values 95% of the data will fall between for a data set with mean
179 and standard deviation of 9.
a) The interval (155, 187) can be written as (171-4*4, 171+4*4) which is same as
(Mean -k*SD, Mean +k*SD), where k = 4.
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 - (1/4^2) = 1 – 0.0625= 0.9375
Thus 93.75 % of the values will fall between 155 and 187 for a data set with mean of 171
and standard deviation of 4.
b) According to Empirical rule, approximately 95% of the measurements (data) will fall
within two standard deviation of the mean.
There fore ( Mean -2*SD, Mean +2*SD) = (179-2*9, 179+2*9) = (161, 197) will
contain 95 % of the observations. Thus the two values are 161 and 197.
4. Nine households had the following number of children per household: 7, 9, 0, 0, 8, 6,
3, 2, 4 Find the mean, median, mode, range, and midrange for these data.
Here the number of observations, n = 9.
1 n
Mean x   xi = (7+9+0+0+8+6+3+2+4)/9 =31/9 = 3.4444
n i 1
Arranging the data in ascending order we get, 0, 0, 2, 3, 4, 6, 7, 8, 9
Medain = Size of (n+1)/2 th item = Size of (9+1)/2 th = 5 th item
=4
Mode = The most frequently occurred item = 0
Midrange = (highest value – lowest value)/2 = (9-0)/2 = 4.5
5. An aptitude test has a mean of 74 and standard deviation of 5. Find the corresponding z
scores for: - a test score of 80 - a test score of 11
Z score = (x score – mean) / standard deviation
Here mean = 74 and standard deviation = 5
Therefore,
a) z score for a test score of 80 = (80- 74)/5 = 1.2
b) z score for a test score of 11 = (11- 74)/5 = -12.6
6. Find P84 for the following data: 3 8 1 1 1 8 6 9 4 4 1 1
Arranging the data in ascending order we get, 1 1 1 1 1 3 4 4 6 8 8 9
 (n  1) 
P84 = 84 th Percentile = Size of 84
item
100 

th
 (12  1) 
= Size of 84
item = Size of 25*0.13 = 10.92 th item
100 

= 10th item + 0.92*(11th item -10th item)
= 8 + 0.92*(8-8) = 8
th
7. A class of 15 kindergarteners had the following weights in pounds: 54 52 38 49 35 46
47 44 47 42 36 52 49 47 44 Find the mean, variance, and standard deviation.
1 n
 xi = (54+52+38+49+35+46+47+44+47+42+36+52+49+ 47+44)/15
n i 1
=682/15 = 45.4667
1 n
Variance =  ( xi  x ) 2 =[(54-45.4667)²+(52-45.4667)²]+...+(44-45.4667)²/15
n i 1
= 30.7822
Standard deviation = Sqrt(Variance) = Sqrt(30.7822) = 5.5482
Mean x 
8. Starting with the data values 70 and 100, add three data values to the sample so that the
mean is 88, the median is 93, and the mode is 93.
Let x, y and z be the three data values added to the sample.
Since the mean is 88, we have
(70+x+y+z+100)/5= 88
That is, x+y+z = 88*5 – 170 = 270
Since mode is 93, at least two of them are 93 and all of them cannot be 93 because sum
of them is not equal to 270.
Therefore two of them are 93, the third one is 270 – 93 – 93 = 84
So three numbers are 84, 93, 93 it also has median 93 in data values 70, 84, 93, 93, 100.