Download 1 - JustAnswer

1. You are given the following frequency distribution for the number of errors that
students made in a test Errors Frequency 0-2 12 3-5 14 6-8 18 9-11 12 12-14 5 Find the
mean, variance, and standard deviation.
Errors
0-2
3-5
6-8
9-11
12-14
Midvalue x Frequency f
1
12
4
14
7
18
10
12
13
5
61
x*f
12
56
126
120
65
379
x2f
12
224
882
1200
845
3163
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 = 61,  xi fi = 379 and  xi f = 3163
Standard deviation =
S
Therefore,
Mean = 379/61 = 6.2131
Variance = 3163/61 –(6.2131)2
= 13.2497
Standard deviation = Sqrt(13.2497) = 3.64
2. You are given the following data. x y 2 0 3 13 4 9 5 15 5 0 7 8 7 8 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
13
9
15
0
8
8
53
x*y
0
39
36
75
0
56
56
262
x2
4
9
16
25
25
49
49
177
y2
0
169
81
225
0
64
64
603
n
n
 xi =33,
n=7,
n
 yi =53,
i 1
 xi2 = 177,
i 1
i 1
n
 yi2 = 603 and
i 1
n
 x y = 262
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
(53) 2
2
SS(y) =  yi   i 1  = 603 
= 201.7143
7
n
i 1
n
n
n
 xi  yi
33*53
= 12.1429
7
n
i 1
We have the linear correlation coefficient,
SS( xy)
r
SS( x)SS( y)
12.1429
=
21.4286*201.7143
= 0.1847
SS(xy) =
x y 
i
The slope b1 =
i 1
i 1
i
= 262 
12.1429
SS( xy )
=
21.4286
SS( x)
= 0.5667
n
 yi
The y-intercept, b0 =
i 1
n
n
 b1
x
i 1
n
i
=
53
33
 (0.5667)*
7
7
= 4.9
The equation of the line of best fit is y = 4.9 +0.5667 x
3. Answer the following: - Use Chebyshev’s theorem to find what percent of the values
will fall between 161 and 229 for a data set with mean of 195 and standard deviation of
17. - Use the Empirical Rule to find what two values 95% of the data will fall between
for a data set with mean 106 and standard deviation of 19.
a) The interval (161, 229) can be written as (195-2*17, 195+2*17) 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 - (1/2^2) = 1 – 0.25= 0.75
Thus 75 percent of the values will fall between 161 and 229 for a data set with mean of
195 and standard deviation of 17.
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) = (106-2*19, 106+2*19) = (68, 144)
will contain 95 % of the observations. Thus the two values are 68 and 144.
4. Nine households had the following number of children per household: 1, 5, 2, 1, 0, 7,
4, 8, 3 Find the mean, median, mode, range, and midrange for these data.
Here the number of observations, n = 9.
1 n
Mean x   xi = (1+5+2+1+0+7+4+8+3)/9 =31/9 = 3.4444
n i 1
Arranging the data in ascending order we get, 0, 1, 1, 2, 3, 4, 5, 7, 8
Medain = Size of (n+1)/2 th item = Size of (9+1)/2 th = 5 th item
=3
Mode = The most frequently occurred item = 1
Midrange = (highest value – lowest value)/2 = (8-0)/2 = 4
5. An aptitude test has a mean of 184 and standard deviation of 17. Find the
corresponding z scores for: - a test score of 137 - a test score of 139
Z score = (x score – mean) / standard deviation
Here mean = 184 and standard deviation = 17
Therefore,
a) z score for a test score of 137 = (137- 184)/17 = -2.7647
b) z score for a test score of 139 = (139- 184)/17 = -2.6471
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) 
P25 = 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 92, the median is 91, and the mode is 91.
Let x, y and z be the three data values added to the sample.
Since the mean is 87, we have
(70+x+y+z+100)/5= 92
That is, x+y+z = 92*5 – 170 = 290
Since mode is 91, at least two of them are 91. And all of them can not be 91 because sum
of them is not equal to 290.
Therefore two of them are 91, the third one is 290 – 91 – 91 = 108
So three numbers are 91, 91, 108, it also has median 91 in data values 70, 91, 91, 100,
108.