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Probability & Statistics Sec-3 with Dr. Kishore Kumar P. K. for Sem II 2020
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EN MATH4130-3 KK 2020-2
OFA - ONLINE FINAL ASSIGNMENT
Final Online Assignment Questions
Question 1
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A1.1. The following are the points to draw a less than ogive curve from a frequency distribution.
(7, 4), (14, 12), (21, 24), (28, 39), (35, 49), (42, 56), (49, 61), (56, 64), (63, 65)

a)      Construct the corresponding frequency distribution from the given points.

b)      Calculate the points to draw a more than ogive curve from the same frequency distribution.

c)      Draw the more than ogive curve in the graph paper attached.
[6 Marks]

A1.2. There are 5 red, 4 blue, 6 green and 3 black balls in a box. Ahmed draws 5 balls from the box at random. What is the probability that at least three of them are blue?
[4 Marks]

Answer in the Word document (ANSWER SHEET) and Upload as attachment below

Question 2
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B3.1. The median and mode of the following wage distribution of 320 workers are given as 48.15 and 53.25 respectively. Find the missing frequencies from the corresponding relative frequency distribution given below.

Wages
(in RO) 0 - 9.5 10 - 19.5 20 - 29.5 30 - 39.5 40 - 49.5 50 - 59.5 60 - 69.5 70 - 79.5 80 - 89.5 90 - 99.5
Relative
frequency 0.0375 0.0625 0.09375 0.175 x y z 0.06875 0.05 0.0375
[8 Marks]

B3.2. How many ways a team of five salespersons can be formed from a group of 10 salesmen and 8 saleswomen, if the team contains at most 2 saleswomen?
[2 Marks]

Answer in the Word document (ANSWER SHEET) and Upload as attachment below

Question 3
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C2.1. The table below shows the number of absences, x, in a Calculus course and the final exam marks, y for 8 students.

X

2

1

2

6

4

3

5

3

Y

93

90

94

60

72

82

70

84

a) Calculate the regression coefficients and hence find the correlation coefficient.

b) Find the regression equation of y on x.

c) Predict the mark of a student whose number of absence is 0.

Answer 1

Solution B 3.2

For these type of questions, we need to make use of combinations

So, we need to find ways a team of five salespersons can be formed from a group of 10 salesmen and 8 saleswomen if the team contains at most 2 saleswomen

There are three ways in which a team of five salespersons can be made having at most two saleswomen

1st - Team having 0 saleswomen and 5 salesmen

2nd - Team having 1 saleswoman and 4 salesmen

3rd - Team having 2 saleswomen and .3 salesmen

so adding all these possible ways we get,

= 8C0 X 10C5 + 8C1 X 10C4+ 8C2 X 10C3

= 1 X 252 + 8 X 210 + 28 X 120

= 252 + 1680 + 3360

= 5292

So, there are 5292 ways in which a team of five salespersons can be formed from a group of 10 salesmen and 8 saleswomen if the team contains at most 2 saleswomen.