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γλώσσες
Σελίδες
Νομικός
STAMFORD UNIVERSITY
Department Of Business Administration
Master of Business Administration Program
744, Satmosjid Road
Dhanmondi, Dhaka
◊Statistics For Business (MAT 421)
Topic:
Submitted by:
Engr. Mohd. Abdus Sattar
ID. MBA 044 12301
Trimester: Spring-2011
Assignment on Statistics
1. Standard Deviation
Example-1: Find (i) Variance, σ² and (ii) Standard Deviation, σ from the following data: 51, 52, 53, 54, 55
Solution:X X- (X- )²
(i) Standard Deviation, σ =
=
= = 1.414
(ii) Variance, σ² = 2
5152535455
-2-1012
41014
= 265 10
= = = 53
Example-2: Sales on a number of shops are given in thousand taka as follows:Sales f
Requird: (i) Variance, σ² (ii) Standard Deviation, σ (iii) Coefficient of Variation, CV
0-5 1
5-10 2
10-15 4
15-20 2
20-25 1
Solution:Sales f Midpoint (x) fx X- f(X- )²
0-5 1 2.5 2.5 -10 100
5-10 2 7.5 15 -5 50
10-15 4 12.5 50 0 0
15-20 2 17.5 35 5 50
20-25 1 22.5 22.5 10 100∑ 10 125 300
= = = 12.5
(i) Standard Deviation, σ = (ii) Variance,
σ² = 30
= = (iii) CV =
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Assignment on Statistics
= 5.477 =
=43.82%
Example-3: Sales of company in January and February/2011 are recorded in thousand taka
and presented as follows:
Sales Frequency Requird:
(i) Variance, σ² for each month
(ii) Standard Deviation, σ for each month
(iii) Coefficient of Variation, CV for each
month
(iv) Which month’s average sale is more?
v) Which month’s sale is more variable?
January February
30-35 1 2
35-40 6 8
40-45 9 14
45-50 12 3
50-55 5 2
55-60 1 1
Solution:
Sales Midpoint
(X)
January February
(f1) (f1X) X- 1 f1(X- 1)² (f2) (f2X) X- 2 f2(X- 2)²
30-35 32.5 1 32.5 -12.5 156.25 2 65.0 -9.67 187.02
35-40 37.5 6 225 -7.5 337.5 8 300 -4.67 174.47
40-45 42.5 9 382.5 -2.5 56.25 14 595 0.33 1.52
45-50 47.5 12 570 2.5 75 3 142.5 5.33 85.23
50-55 52.5 5 262.5 7.5 281.25 2 105 10.33 213.42
55-60 57.5 1 57.5 12.5 156.25 1 57.5 15.33 235.11
∑ 34 1,530 1,062.50 30 1,265.00 896.77
(i) 1 = = = = 45
(ii) Standard Deviation, σ = = =5.59
(iii) Variance, σ² = 31.25
(i) 2 = = = = 42.17
(ii) S.D. σ = =5.45
(iii) Variance, σ² = 29.89
(iii) CV = =
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Assignment on Statistics
(iii) CV = = =12.42%=12.92%
iv) Average Sale of January is more. (v) Feb. month’s sale is more variable because CV of Feb > CV of Jan
2. Quartile, Decile & Percentile Deviation
Example 4: Sales of a number of shops in thousand taka are given as follows:3,11, 13, 12, 14, 9, 8, 18, 17 & 24.
Required: (i) Ordered Set, XL , XH, R (ii) Five number summary (XL , Q1, Q2, Q3, XH)(iii) Inter-Quartile Range (IQR), (iv) Quartile Deviation or Semi Inter-QuartileRange (SIQR) (v) D3 , D7 (vi) P29, P61 (vi) Inter-fractile Range(IFR) between P29
and P61
Solution:
(i) Ordered Set = 3, 8, 9, 11, 12, 13, 14, 17, 18, 24
XL = 3, XH =24, R = 24-3 =21
(ii) Q1 = P25 = X3 = 9 [ i = .25n= 0.25 x 10 = 2.5, Say 3]
Q2 = P50 = ½ (X5 + X6)= 12.5 [ i = .50n= 0.5 x 10 = 5]
Q3 = P75 = X8 = 17 [ i = .75n= 0.75 x 10 = 7.5, Say 8]
So Five number summary is 3, 9, 12.5, 17, 24
(iii) Inter-Quartile Range (IQR) = Q3 - Q1 = 17-9 =8
(iv) Quartile Deviation or Semi Inter-Quartile Range (SIQR) = ½ ( Q3- Q1) = 4
(v) D3 = P30 = ½ (X3 + X4)= ½ (9 +11)= 10 [ i = .30n= 0.3 x 10 = 3]
D7 = P70 = ½ (X7 + X8)= ½ (14 +17)= 15.5 [ i = .70n= 0.7 x 10 = 7]
(vi) P29 = X3 = 9 [ i = .29n= 0.29 x 10 = 2.9 Say 3]
P61 = X7 = 14 [ i = .61n= 0.61 x 10 = 6.1 Say 7]
(vii) Inter-fractile Range (IFR) between P29 and P61 = P61 – P29 = 14-9 = 5
Example 5: Sales of a number of shops in thousand taka are given in the following Frequency Distribution:
Sales f Required: (i) XL , XH, R (ii) Five number summary (XL , Q1, Q2, Q3, XH) (iii) Inter-Quartile Range (IQR), (iv) Quartile Deviation or Semi Inter-Quartile Range (SIQR) (v) D3 , D7 (vi) P29, P61 (vi) Inter-fractile Range(IFR) between P29 and P61
40-45 1
45-50 4
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Assignment on Statistics
50-55 8
55-60 9
60-65 565-70 3
Solution:
Sales f fc
40-45 1 1
45-50 4 5
50-55 8 13
55-60 9 22
60-65 5 2765-70 3 30
(i) XL=40, XH=70, R = 70-40 = 30
(ii) Q2 = Median, Me= L1 +
0.5 n = 0.5 x 30 = 15; So, Q2 lies on series 55-60 i.e. Median group is 55-60.
So, Q2 = L1 +
= 55 +
= 56.11
In the same way, 0.25 n = 0.25 x 30 = 7.5; So, Q1 lies on series 50-55 i.e. Quartile group is 50-55.
So, Q1 = L1 +
= 50 +
= 51.56
0.75 n = 0.75 x 30 = 22.5; So, Q3 lies on series 60-65 i.e. 3rd Quartile group is 60-65.
So, Q3 = L1 +
= 60 +
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Assignment on Statistics
= 60.5Five Number Summary = XL , Q1, Q2, Q3, XH = 40, 51.56, 56.11, 60.5, 70
(iii) Inter-Quartile Range (IQR) = Q3- Q1 = 60.5-51.56 = 8.94
(iv) Quartile Deviation or Semi Inter-Quartile Range (SIQR) = ½ ( Q3- Q1) = 4.47
(v) D3 = P30
0.30 n = 0.30 x 30 = 9; So, D3 lies on series 50-55
So, D3 = L1 +
= 50 +
= 52.5D7 = P70
0.70 n = 0.70 x 30 = 21; So, D7 lies on series 55-60
So, D7 = L1 +
= 55 +
= 59.44(vi) 0.29n = 0.29 x 30 = 8.7; So, P29 lies on series 50-55
So, P29 = L1 +
= 50 +
= 52.31
0.61n = 0.61 x 30 =18.3; So, P29 lies on series 55-60
So, P61 = L1 +
= 55 +
= 57.94
(vii) Inter-fractile Range (IFR) between P29 and P61 = P61 – P29 = 57.94-52.31 = 5.63
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Assignment on Statistics
Levin & Rubin Pg 111: SC-3-11
First we arrange the data in ascending order:
59 65 67 68 71 72 75 79 81 83
85 87 88 91 92 93 94 95 100 100
P80 = ½ (X16 + X17)= ½ (93 +94)= 93.5 [ i = .80n= 0.80 x 20 = 16 ]
Levin & Rubin Pg 111: SC-3-12
XL = 3,600, XH =20,300, R = 20,300-3,600 =16,700miles
Q1 = P25 = ½ (X10 + X11)= ½ (8,100 +8,300)= 8,200 miles [ i = .25n= 0.25 x 40 = 10]
Q3 = P75 = ½ (X30 + X31)= ½ (12,700 +12,900)= 12,800miles [ i = .75n= 0.75 x 40 = 30]
Inter-Quartile Range (IQR) = Q3- Q1 = 12,800-8,200 = 4,600miles
Levin & Rubin Pg 111: 3-52
First we arrange the data in ascending order:
33 45 52 54 55 61 66 68 69 7274 75 76 77 84 91 91 93 97 99
Q1 = P25 = ½ (X5 + X6)= ½ (55 +61)= 58 [ i = .25n= 0.25 x 20 = 5]
Q3 = P75 = ½ (X15 + X16)= ½ (84 +91)= 87.5 [ i = .75n= 0.75 x 20 = 15]
Inter-Quartile Range (IQR) = Q3- Q1 = 87.5-58 = 29.5
Levin & Rubin Pg 111: 3-53
First we arrange the data in ascending order:
2145 2200 2228 2268 2549 2598 2653 2668 2697 28413249 3268 3362 3469 3661 3692 3812 3842 3891 3897
a) XL = 2,145, XH =3,897, R = 3,897-2,145 =1752
b) P20 = ½ (X4 + X5)= ½ (2,268 +2,549)= 2,408.5 [ i = .20n= 0.20 x 20 = 4]
P80 = ½ (X16 + X17)= ½ (3,692 +3,812)= 3752 [ i = .80n= 0.80 x 20 = 16]
Inter-fractile Range (IFR) between P20 and P80 = P80 – P20 = 3752-2408.5 = 1,343.5
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Assignment on Statistics
c) Q1 = P25 = ½ (X5 + X6)= ½ (2,549 +2,598)=2,573.5 [ i = .25n= 0.25 x 20 = 5]
Q3 = P75 = ½ (X15 + X16)= ½ (3,661 +3,692)= 3676.5 [ i = .75n= 0.75 x 20 = 15]
Inter-Quartile Range (IQR) = Q3- Q1 = 3676.5-2573.5 = 1103
Levin & Rubin Pg 111: 3-54
First we arrange the data in ascending order:
69 78 82 84 84 86 87 87 88 8888 88 89 89 89 92 92 94 94 9494 95 96 97 98 99 99 102 102 105
P70 = ½ (X21 + X22)= ½ (94 +95)= 94.5 Degrees [ i = .70n= 0.70 x 30 = 21]
Levin & Rubin Pg 111: 3-55
First we arrange the data in ascending order:
51 83 92 93 93 95 101 115 123 125126 127 129 132 133 135 143 147 157 185
XL = 51, XH =185, R = 185-51 =134
Comment: It is one of the measures of dispersion. But there are more useful measures are
available.
Levin & Rubin Pg 111: 3-56
First we arrange the data in ascending order:
0.10 0.12 0.23 0.32 0.45 0.48 0.50 0.51 0.53 0.580.59 0.66 0.67 0.69 0.77 0.83 0.89 0.95 1.10 1.20
XL = 0.10, XH =1.20, R = 1.20-0.10 =1.10 minutes
Q1 = P25 = ½ (X5 + X6)= ½ (0.45 +0.48)=0.465 minutes [ i = .25n= 0.25 x 20 = 5]
Q3 = P75 = ½ (X15 + X16)= ½ (0.77 +0.83)= 0.800 minutes [ i = .75n= 0.75 x 20 = 15]
Inter-Quartile Range (IQR) = Q3- Q1 = 0.800-0.465 = 0.335 minutes
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Assignment on Statistics
1. What is Dispersion? What are the different measures of dispersion?
Dispersion:
The extent to which the observations in a sample or in a population vary about their mean is known as
Dispersion. A quantity that measures the dispersion in a sample or in a population is known as the
measure of dispersion.
The main measures of dispersions are the range, the semi-inter quartile range or the quartile deviation, the
mean deviation or the average deviation, the variance and the standard deviation.
2. Distinguish between Absolute and Relative measures of dispersion.
Absolute Dispersion is one that measures the dispersion in terms of the same units or the square
units as the units of data. For example: If the units of the data are in taka, meters, kilogram etc, the units
of measures of dispersion will also be in taka, meters, kilogram etc.
Cannot be used to compare the variation of two or more series. For example: the SD of
the heights of students in inches cannot be compared with the SD of the weights of the
students in pound. Even if the units are identical, for e.g. the comparison of ht. in cm and
length of their noses in cm.
On the contrary, Relative dispersion, sometimes called the coefficient of variation, is one that is
expressed in the form of ratio, percentage and is independent of the measure of units.
3. Explain the following with example:
a) Mean Deviation
The mean deviation, also known as mean absolute deviation or average deviation, of a statistical data is
defined as the arithmetic mean of the numerical values of the deviations of observations from its mean.
M.D. = for ungrouped data
= for grouped data
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Assignment on Statistics
b) Standard Deviation
Standard deviation is the positive square root of the mean square deviation of the observation from their
arithmetic mean.
SD, For ungrouped data
SD, For grouped data
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Assignment on Statistics
c) Quartiles
d) Percentiles
e) Inter-Quartile Range (IQR)
f) Semi Inter-Quartile Range (SIQR)
g) Inter-fractile Range (IFR)
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Assignment on Statistics
Skewness, Moments & KutosisSkewness:
Skewness refers to the lack of symmetry. Measures of skewness can be both absolute as well as
relative. Since in a symmetrical distribution mean, median and mode are identical, the more the
mean moves away from the mode, the larger the asymmetry or skewness.
Skewness is measured by the co-efficient of skewness.
1. Karl Pearson’s Co-efficient of Skewness:
Skp = or
Skp =
2. Bowley’s Co-efficient of Skewness (based on quartiles) :
SkB = =
3. Kelly’s Co-efficient of Skewness (based on Percentiles or Deciles) :
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Assignment on Statistics
SkK = (Based on Percentiles)
SkK = (Based on Deciles)
If Co-efficient of Skewness,
Sk = 0 , it means the distribution is symmetrical.
Sk > +1 , it means the distribution is Significant positively skewed or skewed to the right.
Sk < -1 , it means the distribution is Significant negatively skewed or skewed to the left..
Sk is more than zero to less than 1 , it means the distribution is moderate positively skewed.
Sk is less than zero to less than -1 , it means the distribution is moderate negatively skewed.
Moments:
Moments are certain mathematical constants used to ascertain the nature and form of a
distribution. The Greek letter μ(mu) is generally used to denote moments.
The rth moment of a variable x about the arithmetic mean is given by:
μr = for un-grouped data
= for grouped data
Sk =
=
=
=
Kurtosis:
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Assignment on Statistics
Kurtosis refers to the degree of peaked ness or flatness of the frequency curve of a frequency
distribution and is denoted by the co-efficient of kurtosis, .
=
= 3, means the distribution is normal and the curve is mesokurtic.
> 3, means the distribution is leftokartic.
< 3, means the distribution is Platykartic.
Example-1: Sales on a number of shops are given in thousand taka as follows:Sales f
Requird: (i) Find , , and (ii) Compute , and Sk (iii) Comments on the nature and form of the distribution.
0-5 1
5-10 2
10-15 4
15-20 2
20-25 1
Solution:
Sales f Mid-point
x
x-
0-5 1 2.50 2.50 -10.00 -10.00 100.00 -1,000.00 10,000.005-10 2 7.50 15.00 -5.00 -10.00 50.00 -250.00 1,250.0010-15 4 12.50 50.00 0.00 0.00 0.00 0.00 0.0015-20 2 17.50 35.00 5.00 10.00 50.00 250.00 1,250.0020-25 1 22.50 22.50 10.00 10.00 100.00 1,000.00 10,000.00
10 125.00
0.00 300.00 0.00 22,500.00
12.50
(i) = 0
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Assignment on Statistics
(ii)
; So, the distribution is flaty-kartic.
Sk = means the distribution is symmetrical.
(iii) The distribution is flaty-kartic and the curve is symmetrical.
Example-2: Sales on a number of shops are given in thousand taka as follows:Sales f
Requird: (i) Find , , and (ii) Compute , and Sk (iii) Comments on the nature and form of the distribution.
0-5 1
5-10 1
10-15 1
15-20 6
20-25 1
Solution:
Sales f Mid-point
X
x-
0-5 1 42.5 42.50 -12.50 -12.50 156.25 -1,953.13 24,414.065-10 1 47.5 47.50 -7.50 -7.50 56.25 -421.88 3,164.0610-15 1 52.5 52.50 -2.50 -2.50 6.25 -15.63 39.0615-20 6 57.5 345.00 2.50 15.00 37.50 93.75 234.3820-25 1 62.5 62.50 7.50 7.50 56.25 421.88 3,164.06
10 550.00 0.00 312.50 -1,875.00 31,015.63
55.00
(i) = 0
(ii)
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Assignment on Statistics
; So, the distribution is lefto-kartic.
Sk = means the curve is significant negatively skewed to the left.
(iii) The distribution is lefto-kartic and the curve is significant negatively skewed to the left.
Example-3: Sales on a number of shops are given in thousand taka as follows:Sales f
Requird: (i) Find , , and (ii) Compute , and Sk (iii) Comments on the nature and form of the distribution.
0-5 1
5-10 6
10-15 1
15-20 1
20-25 1
Solution:
Sales f Mid-point
X
x-
0-5 1 42.5 42.50 -7.50 -7.50 56.25 -421.88 3,164.065-10 6 47.5 285.00 -2.50 -15.00 37.50 -93.75 234.3810-15 1 52.5 52.50 2.50 2.50 6.25 15.63 39.0615-20 1 57.5 57.50 7.50 7.50 56.25 421.88 3,164.0620-25 1 62.5 62.50 12.50 12.50 156.25 1,953.13 24,414.06
10 500.00 12.50 0.00 312.50 1,875.00 31,015.63
50.00
(i) = 0
(ii)
; So, the distribution is lefto-kartic.
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Assignment on Statistics
Sk = means the curve is significant positively skewed to the right.
(iii) The distribution is flaty-kartic and the curve is significant positively skewed to the right.
Example-4: The Capital investment of a number of firms recorded in lac taka are presented as follows:
Capital f
Requird: (i) Find , , and (ii) Compute , and Sk (iii) Comments on the nature and form of the distribution.
40-45 2
45-50 5
50-55 9
55-60 12
60-65 8
65-70 2
Solution:
Capital f Mid-point, x x-
40-45 2 42.5 85.00 -13.29 -26.58 353.25 -4,694.67 62,392.1545-50 5 47.5 237.50 -8.29 -41.45 343.62 -2,848.61 23,615.0150-55 9 52.5 472.50 -3.29 -29.61 97.42 -320.50 1,054.4555-60 12 57.5 690.00 1.71 20.52 35.09 60.00 102.6060-65 8 62.5 500.00 6.71 53.68 360.19 2,416.89 16,217.3665-70 2 67.5 135.00 11.71 23.42 274.25 3,211.45 37,606.04
38 2,120.00 -0.02 1,463.82 -2,175.44 140,987.60
55.79
(i)
(ii)
; So, the distribution is Platy-kartic.
Sk = means the curve is moderate positively skewed to the right.(iii) The distribution is Platy-kartic and the curve is moderate positively skewed to the right.
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Assignment on Statistics
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Assignment on Statistics
Correlation Analysis
Correlation:
Study of relationship between variables which are conceptually related.
There are three types of correlation.
1. Simple Correlation-Between two variables
2. Multiple Correlation-Among more than two variables.
3. Spurious Correlation-
Relationship between variables that cannot be explained with conceptually related
variables.
Not expandable.
Positive Correlation:
If the relationship between x and y are such that y is directly proportional to x or y increases with
the increase of x, then the relationship is called Positive Relationship.
Negative Correlation
If the relationship between x and y are such that y is inversely proportional to x or y decreases
with the increase of x, then the relationship is called Negative Relationship.
Co-efficient:
Correlation coefficient is the measure of the degree of relationship between the variables.
Correlation is measured by co-efficient of correlation suggested by Karl Pearson’s Product
Moment Correlation defined by:
r=-1 means Perfect negative correlation
-1<r<0 means weak negative correlation
r=0 means no correlation
0<r<1 means strong positive correlation
r=1 means perfect positive correlation.
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Assignment on Statistics
Co-efficient of determination,
refers to the proportion of the dependant variable that can be explained by the independent
variable.
Example:
Shops Capital
(Lac tk.)
Profit
(Lac Tk)
Reqd:
(i) Compute the Karl Pearson’s Product moment
correlation coefficient,r
(ii) Evaluate the strength correlation
(iii) Compute the coefficient of determination
(iv) Compute the coefficient of non-determination
(v) Interpret the result.
1 5 2
2 10 3
3 15 5
4 20 8
5 25 12
Solution:
Shops Capital(Lac tk.)
(x)
Profit (Lac Tk)
(y)
xy
1 5 2 25.00 4.00 10.002 10 3 100.00 9.00 30.003 15 5 225.00 25.00 75.004 20 8 400.00 64.00 160.005 25 12 625.00 144.00 300.00
75 30 1,375.00 246.00 575.00(i) Karl Pearson’s Product moment correlation coefficient
= 0.973
(ii) as 0<r<1, strong positive correlation.
(iii) Coefficient of determination, = 0.947
(iv) Coefficient of non-determination, 1- =1- 0.947=0.053
(v) 94.7% of Profit can be explained by capital investment and 5.3% of profit cannot be
explained by capital investment.
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Assignment on Statistics
Regression Analysis
Regression:
Regression is the statistical tool with the help of which we are in a position to estimate (or
predict) the unknown values of one variable from known values of another variable. It is used for
estimation and development of model or to study the nature of relationship between the
variables.
y = a + bx of Y on X
Where, b =
a =
Standard error, Sp =
Scatter Diagram:
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Assignment on Statistics
Example-1:
Shop Capital (x) Profit (y) x2 xy yE =-1.5 + 0.5x (y-yE)2
1 5 2 25 10 1 1
2 10 3 100 30 3.5 0.25
3 15 5 225 75 6 1
4 20 8 400 160 8.5 0.25
5 25 12 625 300 11 1
75 30 1375 575 3.5
b = = = 0.5
= = =15; = = =6
a = = 6-0.5 x 15 = -1.5
y = -1.5 + 0.5x
yx=30 = -1.5 +0.5 x 30 = 13.5
yx=35 = -1.5 +0.5 x 35 = 16
yx=40 = -1.5 +0.5 x 40 = 18.5
Standard error, Sp =
=
= 1.08
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Assignment on Statistics
Example-2 (Ref. Levin Rubin Pg 671 SC 12-2:
Sl x y x2 xy1 13 6.2 169 80.62 16 8.6 256 137.63 14 7.2 196 100.84 11 4.5 121 49.55 17 9 289 1536 9 3.5 81 31.57 13 6.5 169 84.58 17 9.3 289 158.19 18 9.5 324 17110 12 5.7 144 68.4
140 70 2038 1035
b = = = 0.705
= = =14 = = =7
a = = 7-0.705 x 14 = -2.87
y = -2.87 + 0.705x
yx=10 = -2.87 +0.705 x 10 = 4.18
yx=13 = -2.87 +0.705 x 13 = 6.295
yx=20 = -2.87 +0.705 x 20 = 11.23
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Assignment on Statistics
Probability
Event :
An event is a possible outcomes of an experiment or a result of a trial or an observation.
Experiment:
An experiment is any process or study which results in the collection of data, the outcome of
which is unknown. In statistics, the term is usually restricted to situations in which the researcher
has control over some of the conditions under which the experiment takes place.
Relative Frequency Definition of Probability:
The ratio of the number of occurances of an event to the number of possible occurances in an
experiment is referred to as the relative frequency. Two definitions in terms of relative frequency
can be given as follows:
a) If an experiment is performed n times under the same conditions and there are ‘a’
outcomes, a≤n, favoring an event, then an estimate of the probability of that event is the
ratio a/n.
b) The estimate of probability of event, a/n approaches a limit, the true probability of the
event when n approaches to infinity is given by,
P(E) = Limit
n→
Random Variables
When the numerical value of a variable is determined by a chance event, that variable is called a random variable.
Random Variables are of two types:
1. Discrete (Integer)
Continuous (Integer or Non-integer)
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Assignment on Statistics
The Classical Definition of Probability:
This definition is for equally likely outcomes. If an experiment can produced mutually
exclusive and equally likely outcomes out of which outcomes are favorable to the occurrence
of event , then the probability of is denoted by and is defined as the ratio . Thus
the probability of is given by
0 P(X) 1
Condition for Classical Probability:
mutually exclusive
and equally likely outcomes
Each outcome of an experiment has the same chance of appearing as any other and,
therefore, can be assigned same weight (probability) for its occurrence as any other.
Subjective Probability:
A probability derived from an individual’s personal judgment about whether a specific outcome
is likely to occur. Subjective probabilities contain no formal calculations and only reflects the
subjects opinions and past experience.
Sample Space:
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Assignment on Statistics
Example 1: Toss a coin. What is the probability of getting head or tail?
Solution: P(H or T) = ½ =0.5
Example-2:
Economic
Condition
f
S 3
G 4
D 2
B 1
n= 10
Reqd.
(i) P(D) = =0.2 = 20%
(ii) P(G) = =0.4 = 40%
(iii) P(S) = =0.3 = 30%
(iv) P(B) = =0.1 = 10%
(v) P(W) = =0.0 = 0%
Example-3:
P(R) = =0.625 = 62.5%
P(R) = =0.375 = 37.5%
P (B) = =0.0 = 0.0%
Example-4:
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5R 3W
Assignment on Statistics
Opinion M F
Strongly Support 9 10
Mildly Support 11 3
Undecided 2 2
Mildly Opposed 4 8
Strongly Opposed 4 7
n= 30 30
(i) P( M) Strongly Support =
=0.30 = 30%
(ii) P( M) Mildly Support =
=0.367 = 36.7%
(iii) P( M) Undecided =
=0.0667 = 6.67%
(iv) P( M) Mildly Opposed =
=0.133 = 13.3%
(v) P( M) Strongly Opposed =
=0.133 = 13.3%
(vi) P( F) Strongly Support =
=0.333 = 33.3%
(vii) P( F) Mildly Support =
=0.1 = 10%
(viii) P( F) Undecided = =0.0667
= 6.67%
(ix) P( F) Mildly Opposed =
=0.267 = 26. 7%
(x) P( F) Strongly Opposed =
=0.233 = 2.33%
(xi) P( W) Strongly Support =
=0.317 = 31.7%
(xii) P( W) Mildly Support =
=0.233 = 23.3%
(xiii) P( W) Undecided =
=0.0667 = 6.67%
(xiv) P( W) Mildly Opposed = =
=0.20 = 20%
(xv) P( W) Strongly Opposed =
=0.183= 18.3%
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Assignment on Statistics
Ref. Levin & Ruvin. Pg 168
Example-5
(i) P(A Seven) = =0.0769 = 7.69%
(ii) P (A Black Card) = =0.50 = 50%
(iii) P ( An ace or a king) = =0.1538 = 15.38%
(iv) P ( A Black Two or a Black three)=
=0.0769 = 7.69%
(v) P ( A red face card-King, Queen, or
Jack) = =0.1154 = 11.54%
Example-6
Worksheet for Calculation:
Deposit (Lac Tk.)
Days P(X) = fc F(X) = X. P(X) X2 X2. P(X)
(X) (f)
1 2 3 4 5 6 7 8
5 10 0.10 10 0.10 0.50 25 2.50
10 20 0.20 30 0.30 2.00 100 20.00
15 40 0.40 70 0.70 6.00 225 90.00
20 20 0.20 90 0.90 4.00 400 80.00
25 10 0.10 100 1.00 2.50 625 62.50
n= 100 E(X)=∑XP(X) 15.00 ∑ X2. P(X) 255.00
(i) P (X=10) = =0.20 = 20%
(ii) P (X =15) = = 0.40 = 40%
(iii) P (X 15) = = 0.70 = 70%
(iv)P ( X 20) = = 0.90 = 90%
(v) P (10 X 20) = = 0.80 =80%
(vi) P(X) = [Done in
Column (3) ]
(vii) F(X) = [Done in Column
(5) ]
(viii) E(X) = ∑X.P(X) [ Done in
Column (6) ]
(ix) Var (X) = ∑X2.P(X) –
[E(X)]2
= 255 - 152
= 30
(x) S.D. (X) =
=
= 5.47
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Assignment on Statistics
Example-7
Sonali Bank Head Office has kept a record of the number of grievances of the employees filed
per week for last 50 weeks. The results are shown as below:
Worksheet for Calculation:
No of grievancesNo of weeks
P(X) = fc F(X) = X. P(X) X2 X2. P(X)
(X) (f)
1 2 3 4 5 6 7 8
0 2 0.04 2 0.04 0.00 0 0.00
1 18 0.36 20 0.40 0.36 1 0.36
2 25 0.50 45 0.90 1.00 4 2.00
3 4 0.08 49 0.98 0.24 9 0.72
4 1 0.02 50 1.00 0.08 16 0.32
n= 50 E(X)=∑XP(X) 1.68 ∑ X2. P(X) 3.40
(i) P (X=3) = =0.08 = 8%
(ii) P(X) = [Done in Column (3) ]
(iii) F(X) = [Done in Column (5) ]
(iv) E(X) = ∑X.P(X) [ Done in Column (6) ]
(v) Var (X) = ∑X2.P(X) – [E(X)]2
= 3.40 – 1.682
= 0.5776
(vi) S.D. (X) =
=
= 0.76
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Assignment on Statistics
2. Rules of Probability:
1. Total Rule of Probability
P (A B ) = P (A) + P(B) – P (A B) For not mutually exclusive event
= P (A) + P(B) For mutually exclusive event
Example 1: Determine the probabilities of the following events in drawing a card from a
standard deck of 52 cards.
(i) P (S K ) = P (S) + P(K) – P (S K) = + - =
(ii) P (S C ) = P (S) + P(K) – P (S K) = + - 0=
(iii) P (D Q ) = P (S) + P(K) – P (S K) = + - =
(iv) P (Q J ) = P (S) + P(K) – P (S K) = + - 0=
2. Multiplication Theorem of Probability:P (A B) = P (A) . P (B/A)
= P (B). P (A/B)
Or, P (B/A) = P (A B)/P (A)
Or, P (A/B) = P (A B)/P (B)
Example-2:Gender Efficient (E) Inefficient (I) Total
Male (M) 28 12 40
Female (F) 55 5 60
Total 83 17 100
(i) P(M) = 0.40 = 40%(ii) P(F) = 0.60 = 60%(iii) P(I) = 0.17 = 17%(iv) P(E) = 0.83 = 83%
(v) P (M E) = 0.28=28%
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Assignment on Statistics
(vi) P (M I) = 0.12 = 12%(vii) P (F E) = 0.55 = 55%(viii) P (F I) = 0.05 =5%(ix) P (M E ) = 40% + 83%-28% = 95%(x) P (M I ) = 40% + 17%-12% = 45%(xi) P (F E ) = 60% + 83%-55% = 88%(xii) P (F I ) = 60% + 17%-5% = 72%(xiii) P (M/E ) = P (M E)/P(E) = 0.28/0.83 = 33.73% (xiv) P (M/I ) = P (M I)/P(I) = 0.12/0.17 = 70.59%(xv) P (F/E ) = P (F E)/P(E) = 0.55/0.83 = 66.27%(xvi) P (F/I ) = P (F I)/P(I) = 0.5/0.17 = 29.41%
Sl. 2 Ref: Supplied SheetAn Experiment consists of counting the number of dishonored cheques received by a large bank branch at Motijheel area of Dhaka city. The management of the Bank branch has kept a record of such cheques received per day a period of 200days. The data is shown below:
Number of Dishonored Cheques Number of Days0 81 122 203 604 405 306 207 10
Total 200
(i) Develop a probability distribution for the above data.(ii) Determine the cumulative probability distribution F(X)(iii) What is the probability that in given day the branch receives less than or equal to 4
dishonored cheques?(iv) What is probability that in a day the bank receives more than3 bad cheques?(v) What is the expected number of dishonored cheques per day?(vi) Determine the standard deviation.
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Assignment on Statistics
Sl. 3 Ref: Supplied SheetThe Record of Janata Bank kept over last 300 days shows the following number frauds occurred in different branches over the country:
Number of Frauds Number of Days0 451 752 1203 454 15
Total 300 (i) Develop a probability distribution for the daily frauds occurred in bank branches.(ii) Determine the cumulative probability distribution F(X)(iii) What is the probability that in given day there will be no fraud occurred?(iv) What is the probability that in given day there will be atleast 1 fraud occurred?(v) What is probability that in a day the bank receives more than3 bad cheques?(vi) Determine the expected number of frauds and its variances.
Sl. 4 Ref: Supplied SheetIDCL is lease financing firm. The number of new clients that they have obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution which is shown below:
Number of New Clients Probability 0 0.051 0.102 0.153 0.354 0.205 0.106 0.05
Total 1.00
(i) What is the expected number of new clients per month?(ii) Determine the variance and standard variation.
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Assignment on Statistics
Sl. 5 Ref: Supplied SheetA company sells its products to wholesaler in batched of 1000 units only. The daily demand for their respective probabilities are given below:
Demands Probability0 0.1
1000 0.22000 0.33000 0.24000 0.1Total 0.90
(i) Determine the expected daily demand.(ii) Assume that the sells their product at Tk. 3.75 per unit. What is the expected daily
revenue?(iii) Determine the standard deviation of daily demand.(iv) It is known that the company has a Tk. 3000 per day fixed cost and the variable cost
per unit is Tk. 0.90. If the company produces a number equal to the expected daily demand, what is the expected total cost?
(v) What is the expected daily profit or loss?
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