Computer Science Statistics and Probability Assignment Solutions

This document provides a comprehensive solution to a statistics and probability assignment, covering a range of topics. The assignment includes problems on permutations and combinations, calculating the number of arrangements in a circle, and scenarios involving constraints. It delves into the binomial theorem, expanding and simplifying expressions. Furthermore, it addresses data analysis, including calculating mean, standard deviation, and variance for given datasets. The document also explores probability concepts, such as calculating the probability of winning a lottery with varying numbers of correct picks. Finally, the assignment includes a financial problem involving compound interest and calculating the necessary investment amount to reach a future goal. The solutions are detailed, providing step-by-step explanations for each problem.

COMPUTER SCIENCE
STUDENT ID:
[Pick the date]

Question 1
(a) (i) Formula for n objects arranged around a circle = (n-1)!
Hence, possible ways 12 people can be seated = (12-1)! = 11! = 39,916,800
ii) The first group of 6 men or 6 women can be seated in alternate seats around a circle in (6-
1)! = 5!
The remaining group can be now seated in the remaining six seats in 6! Ways
Hence, possible ways = 5!*6! = 86,400
iii) Consider Donald and Daffy Trump as one unit. Besides these, there are 10 more people.
Hence, in total there are 11 people who can be seated around a circle in (11-1)! Or 10!
However, Donald and Daffy Trump can switch their places and hence can be arranged in 2!
Ways
Hence, possible ways = 10!*2! = 7,257,600
iv) Now consider Bob, Donald and Daffy to be a single unit. Besides these, there are 9 more
people. Hence, in total there are 10 people who can be seated around a circle in (10-1)! Or 9!
However, Bob, Donald and Daffy Trump can be arranged in 3! Ways
Hence, possible ways = 9!*3! = 2,177,280
v) Possible ways = Total cases – Cases where Donald sits with Daffy or Bob
Hence, possible ways = 11! – 2*10!*2! = 7*10! = 25,401,600
b) i) Number of ways of arranging requisite books = 6C4*5C3*7! = 756,000
ii) Number of ways of arranging in which all the maths book are together = 6C4*5C3*4!*4!
= (6!/(4!*2!))*(5!/(3!*2!))*4!*4! = 86,400
iii) Number of ways in which the Maths book is the first on the shelf = 6C4*5C3*4*6! =
432,000

iv) Number of ways in which the Maths and Human Sciences book alternate = 6C4*5C3*4!
*3! = 21,600
v) Number of ways in which the Maths book is the first and Human Science is in the middle
on the shelf = 6C4*5C3*4*3*5! = 216,000
c) i) log6200 = log6(25*8) = log625 + log68 = log625 + log623 = log625 + 3log62 = a+3b
ii) log63.2 = log66 =1
iii) log675 = log6(25*3) = log625 + log63 = log625 + log6(6/2) = log625 + log66 - log62 = a+1-b
d) The given term can be extended using the binomial theorem.
(3 – 1/2x)5 = 5C0(3)5(– 1/2x)0 + 5C1(3)4(– 1/2x)1 + 5C2(3)3(– 1/2x)2 +5C3(3)2(– 1/2x)3 +
5C4(3)1(– 1/2x)4 +5C5(3)0(– 1/2x)5 = 243 – (405/2x) + (135/x2) – (45/4x3) + (15/16x4) –
(1/32x5)
Question 2
(a) Histogram

0 <10 10<20 20<25 25<30 30<35 35<40 40<80 80<100
0
5
10
15
20
25
30
35
Histogram
Number of smokers now
Number intended in the future
Cigrattes smoke per week
Frequency
(b) Mean for each of the distribution
 Number of smokers now
Mean= 1
∑ f ∑ f . x=3210
100 =32.10
 Number intended in the future

Mean= 1
∑ f ∑ f . x=3015
95 =31.74
(c) Standard deviation for each of the distribution
 Number of smokers now
Standard deviation= √∑ f .¿ ¿ ¿ ¿
 Number intended in the future
Standard deviation= √∑ f .¿ ¿ ¿ ¿
(d) Ogive for each of the distribution
 Number of smokers now

0 <10 10<20 20<25 25<30 30<35 35<40 40<80 80<100
0
20
40
60
80
100
120
Ogive: Number of smokers now
Number of smokers nowCigarettes smoke per week
Cumulative frequency
 Number intended in the future

0 <10 10<20 20<25 25<30 30<35 35<40 40<80 80<100
0
10
20
30
40
50
60
70
80
90
100
Ogive: Number intended in the future
Number intended in the future
Cigarettes smoke per week
Cumulative frequency
e) The median value for both the datasets based on the ogive would be around 27 cigarettes
considering the fact that the middle value tends to lie in the interval 25<30 for both the
datasets.
f) Considering the shape of both distributions, it can be concluded that both of these would be
quite similar with the only difference being that it can be expected that the number intended
in the future in almost all the cigarettes would be lower than the current smokers.
Question 3
Amount desired after 6 years = €50,000
Interest rate = 5% p.a. compounded quarterly or (5/4) i.e. 1.25% per quarter
Time period = 6 years or (6*4) = 24 quarters
Let the amount to be invested in the scheme at the present be € P
The following formula would be used
FV = PV (1+r)n
Here, FV = 50,000, r=1.25%, n=24
Hence, 50000 = P (1.0125)24
Solving the above, we get P = €37,109.85

Question 4
Set x
Range = Maximum value – Minimum value = 9-1 =8
Mean = sum of observation / number of observation = 58/10 =5.8
Standard deviation
Standard deviation = sqrt(57.6/(10-1)) = sqrt(6.4) = 2.53
Variance = 2.532 = 6.4
Coefficient of variation = standard deviation/ mean = 2.53/5.8 = 0.44
Set y

Range = Maximum value – Minimum value = 9-2 = 7
Mean = sum of observation / number of observation = 53/10 =5.3
Standard deviation
Standard deviation = sqrt(46.1/(10-1)) = sqrt(5.122) = 2.26
Variance = 2.262 = 5.122
Coefficient of variation = standard deviation/ mean = 2.26/5.3 = 0.43
Absolute dispersion is measured by standard deviation which is higher for Set x. The measure
of relative dispersion is coefficient of variation which is also higher for Set x.
Question 5
a) It is apparent that there are in total 47 numbers used in lottery out of which 6 are correct.
This implies that 41 numbers are wrong while 6 are correct.

i) Number of ways in which 3 out of 6 numbers are correct = 6C3*41C3
Total number of ways in which six numbers may be selected = 47C6
Hence, requisite probability = (6C3*41C3)/ (47C6)
ii) Number of ways in which 4 out of 6 numbers are correct = 6C4*41C2
Total number of ways in which six numbers may be selected = 47C6
Hence, requisite probability = (6C4*41C2)/ (47C6)
iii) Number of ways in which 5 out of 6 numbers are correct = 6C5*41C1
Total number of ways in which six numbers may be selected = 47C6
Hence, requisite probability = (6C5*41C1)/(47C6)
(iv) Number of ways in which 5 out of 6 numbers are correct = 6C6 =1
Total number of ways in which six numbers may be selected = 47C6
Hence, requisite probability = 1/(47C6) =6!*41!/47!