Assignment SBST1303

8/10/2019 Assignment SBST1303

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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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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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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:-


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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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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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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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


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/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


/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


/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

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Assignment SBST1303