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SBST1303 (Elementary Statistic)
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QUESTION 1
a. Construct a relative frequency distribution for the data in Table 1
Num. Of Children Frequency (f) Relative Frequency (%)
0 7 7/60*100 = 11.666 1 7 7/60*100 = 11.666 2 18 18/60*100 = 30 3 20 20/60*100 = 33.333 4 7 7/60*100 = 11.666 5 1 1/60*100 = 1.666
SUM 60 99.997
b. Describe the distribution of the data
The table above is Frequency Distribution Table of qualitative variable married couples. Thefirst row is the category of the variable; the second row is the frequency of each categorical
value with the total of 60 of couples married. The third row shows the relative frequency of
the class is the ratio of its frequency to the total frequency. And the fourth row shows the
relative frequency expressed in percentage by multiplying 100% to each relative frequency.
c. Use the frequency distribution table to construct a bar chart
0
5
10
15
20
25
0 1 2 3 4 5
F r e q u e n c y
Num Of Children
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d. Calculate the percentage of married couples having
I. Two children
(18) X 100= 30%
60
II. At least two children:-
No of children Couples Married
2 18
3 20
4 7
5 1
_____
46
_____
(46) X 100= 76.666% 60
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QUESTION 2
a. Using a class width of 7 and a first class lower limit of 30 , construct a frequency
distribution table using the data in Table 2.
Cholesterol Level Frequency
30-36 437-43 344-50 1051-57 1158-64 565-71 472-78 3SUM 40
Working:-
The number of class:
K = 6.29= 6
Class width : 7
72- 32 = 6.83 7
6 b. Determine the following
I. upper and lower boundaries and the class mid point for the second class
Lower Boundaries of second class = 37+362
= 36.5
Upper Boundaries of second class = 43+44
2
= 43.5
Mid point for the second class = 36.5 + 43.52
= 40
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SBST1303 (Elementary Statistic)
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II. relative frequency of the fifth class
Cholesterol Level Frequency Relative Frequency Relative Frequency (%)58-64 5 0.13 13SUM 40
Working:-
Relative frequency = 5/ 40
= 0.125
Relative frequency (%) = 0.13 x 100
= 13
III. Range of the data = 11- 3
= 8
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c. With reference to the data in Table 2, construct a cumulative frequency polygon on
a graph paper .
Working:-
Cholesterol Level Frequency UpperBoundary
CumulatingProcess
CumulatingFrequency
23-29 0
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SBST1303 (Elementary Statistic)
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Cholesterol Level Frequency
30-36 437-43 3
44-50 1051-57 1158-64 565-71 472-78 3SUM 40
The class mode is 51-57;
Lower Boundary is(50+51)/2 = 50.5
Class width C = 7 and= 11-10 = 1; = 11-5 = 6
III. Median
n + 1 = 40 + 1 = 20.5
2 2
= 50.5 + 7 (20.5 17)
11
= 50.5 + 24.5
11
= 52.727
The highest frequency = 11
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Working:-
Cholesterol Level Frequency
30-36 437-43 344-50 1051-57 1158-64 565-71 472-78 3SUM 40
= = 4+3+10= 17
(50 + 51)/2 = 50.5
Class Width = 37-30= 7
= 11
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SBST1303 (Elementary Statistic)
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QUESTION 3
a. The probability that a patient is allergic to penicillin is 0.20. Suppose this drug is administered to
three patients.
i. Illustrate the probabilities of this event on a tree diagram .
Z Allergic Penicillin = 0.2
Not Allergic Penicillin = 0.8
Z
Z
Z
0.2
0.8
0.20.2
0.8
0.8
0.2
0.8
0.2
0.8
0.2
0.8
0.2
0.8
Z
Z
Z
Z
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SBST1303 (Elementary Statistic)
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ii. Obtain the probability that all three patients are allergic to penicillin .
P (all three patients are allergic ) = 0.2x 0.2 x 0.2
=0.008
Z
Z
Z
0.2
0.8
0.20.2
0.8
0.8
0.2
0.8
0.2
0.8
0.2
0.8
0.2
0.8
Z
Z
Z
Z
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iii. Obtain the probability that at least one of them is not allergic to penicillin
P ( ) = 1 P (
= 1 (0.2 x 0.2 x 0.2)
= 0.992
Z
Z
Z
0.2
0.8
0.20.2
0.8
0.8
0.2
0.8
0.2
0.8
0.2
0.8
0.2
0.8
Z
Z
Z
Z
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b. An office building has two fire detectors. The probability is 0.02 that any fire detector ofthis type will fail to go off during a fire. Obtain the probability that both of these fire
detectors will fail to go off in case of a fire .
P ( ) = 0.02 x 0.02
= 0.0004
F
F
F
0.98
0.0.2
0.98
0.02
0.98
0.02
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SBST1303 (Elementary Statistic)
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QUESTION 4
a. Present the data in a cumulative frequency distribution table .
Salary/Gaji Number of employees
/BilanganPekerja Upper
Boundary
Cumulating
Process
Cumulative
Frequency
1300 - 1499 0 0 0
1500 1699 16 0+16 16
1700- 1899 31 16+31 47
1900- 2099 22 47+22 69
2100-2299 14 69+14 83
2300-2499 10 83+10 932500-2699 7 93+7 100
b. Calculate the mean and the median
Mean
Working:-
Salary/Gaji Lower
Boundary
Upper
Boundary
Class
Mid-point
(x)
Frequency
(f)
(f*x)
F Multiply X
1500-1699 1499.5 1599.5 16 25592
1700- 1899 1699.5 1799.5 31 55784.5
1900- 2099 1899.5 1999.5 22 43989
2100-2299 2099.5 2199.5 14 30793
2300-2499 2299.5 2399.5 10 23995
2500-2699 2499.5 2599.5 7 18196.5SUM 100 198350
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= 198350 = 1983.5 100
Median
Working:-
n + 1 = 100 + 1 = 50.5
2 2
SUM = f1+f2+f3= 16+31+22= 69> = 50.5
The third frequency makes sum greater than . So the third frequency will be the medianclass.
Median class = 1900-2099 ; = (1899+1900)/2= 1899.5 ; C = 1900-1700= 200 ; =16+31= 47 ; = 22
= 1899.5 + 200(50.5 47)
22
= 1899.5 + 700
22
= 1931.318
8/10/2019 Assignment SBST1303
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SBST1303 (Elementary Statistic)
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Salary/Gaji Number of employees
/BilanganPekerja
1500 1699 16
1700- 1899 31
1900- 2099 22
2100-2299 14
=2300-2499 10
2500-2699 7
c. Calculate the inter-quartile range
Working:-
1(n+1) = 100 + 1= 25.25 4 4
f1 +f2 = 47 (> 25.25 )
The second class is greater than . So the second class will be the class of the firstquartile.
Class = 1700-1899 ; = (1699+1700)/2= 1699.5 ; C = 1500-1700= 200 ; = 16 ; = 31
Salary/Gaji Number of employees
/BilanganPekerja
1500 1699 16
1700- 1899 311900- 2099 22
2100-2299 14
=2300-2499 10
2500-2699 7
f1+f2= 47
16
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1699.5 + 200 (25.25 16)31
1699.5 + 185031
= 1759.177
3(n+1) = (3*100) + (3*1) = 300+3 = 75.75 4 4 4
Working:-
f1 +f2 + f3 + f4 = 83 (> 75.75 )
The forth class is greater than . So the forth class will be the class of the third quartile.
Class = 2100-2299 ; = (2099+2100)/2= 2099.5 ; C = 1900-2100= 200 ;` = 69 ; = 14
Salary/Gaji Number of employees
/BilanganPekerja
1500 1699 16
1700- 1899 31
1900- 2099 22
2100-2299 14
=2300-2499 10
2500-2699 7
2099.5 + 200 (75.75 69)14
2099.5 + 135014
= 2195.928
IQR
= Q3 Q1
= 2195.928 1759.177 = 436.752
f1+f2+f3= 69