B.Sc. II Statistics Practical Paper -II
In this article we see all the practical problem Of B.Sc. II in Practical Paper -II
Practical Number 1.
Padmbhushan Vasantraodada Patil Mahavidyalaya, Kavathe Mahankal
Department of Statistics
Title: Model sampling from Discrete Uniform distribution
Questions:
1. Draw a model sample of size 10 from the following Discrete Uniform Distribution
P(X) = 1/8; x=1,2,3,...,8
= 0 otherwise.
Calculate A.M. and H. M. of your sample.
2. Draw a model sample of size 15 from the following Discrete Uniform Distribution Taking values 10,15,20,25,30,35,40,45,50,55. Find mean deviation from mode of your sample.
3. Draw a model sample of size 8 from the following Discrete Uniform Distribution
P(X) = 1/13; x=1,2,3,...,13
= 0 otherwise.
Obtain the quartiles of your sample.
4. Draw a model sample of size 10 from the Discrete Uniform Distribution over the set of first 25 natural numbers. Find C.V. of your sample.
5. Draw a random sample of size 11 from discrete uniform distribution taking values 1, 2, 3, 4, 5. find coefficient of M.D. about mean of your sample.
6. Draw a model sample of size 10 from the Discrete Uniform Distribution over values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. find mean and variance of your sample.
ANSWERS Practical NO. 1.
Q. 1. Draw a model sample of size 10 from the following Discrete Uniform Distribution
P(X) = 1/8; x=1,2,3,...,8
= 0 otherwise.
Calculate A.M. and H. M. of your sample.
Answer :
first we see the discrete uniform distribution.
If X ~ DU(1,…., n ), Then the p.m.f. of discrete uniform distribution is given as,
P(X=x) = 1/n ; x=1,2,3,...,n
= 0 otherwise.
and it's mean = ( n+1)/2
& variance = (n²-1) /12.
then we, write the Aim of the question.
Aim: To Draw a model sample of size 10 from the following Discrete Uniform Distribution and Calculate A.M. and H. M. of your sample.
Statistical Formula:
A.M. = ∑Xi / n
H. M. = n / ∑(1/ Xi)
Given,
P(X) = 1/8; x=1,2,3,...,8
= 0 otherwise.
here n = 8
and P(X) = 0.125
now we draw random sample so we create observation table.
Observation Table:
|
Sr. No. |
xi |
p(xi) |
F(xi) |
Rand. No. |
xi(Random Sample) |
1/xi |
|
1 |
1 |
0.125 |
0.125 |
0.180575 |
2 |
0.5 |
|
2 |
2 |
0.125 |
0.250 |
0.005194 |
1 |
1 |
|
3 |
3 |
0.125 |
0.375 |
0.091513 |
1 |
1 |
|
4 |
4 |
0.125 |
0.500 |
0.228582 |
2 |
0.5 |
|
5 |
5 |
0.125 |
0.625 |
0.437781 |
4 |
0.25 |
|
6 |
6 |
0.125 |
0.75 |
0.900141 |
8 |
0.125 |
|
7 |
7 |
0.125 |
0.875 |
0.001238 |
1 |
1 |
|
8 |
8 |
0.125 |
1 |
0.931718 |
8 |
0.125 |
|
9 |
|
0.079296 |
1 |
1 |
||
|
10 |
0.892168 |
8 |
0.125 |
|||
|
Total |
36 |
5.625 |
||||
note: here P(X) is given for X= 1,2,3,4,5,6,7,8. and it is P(x) = 1/8 = 0.125 is same for all 8 values.
then F(X) is obtained as sum of probability, we known that F(X) = P(X≤ x )
if x =1 then F( 1) = P(X≤ 1 ) = P(X =1 )= 0.125
if x = 2 then F( 2 ) = P(X≤ 2 ) = P(X =1 )+ P(X = 2) = 0.125+0.125 = 0.250
if x = 3 then F( 3 ) = P(X≤ 3. ) = P(X =1 )+ P(X = 2)+ P(X =3) = 0.1225+0.125+0.125 = 0.375
and so on
and the Rand .No. means random number that are obtained as i. in MS- Excel write =RAND() and enter we get random number.
ii. in scientific Calculator we first press shift button then Rand# ( . button on scientific calculator) then = button.
the random number obtained as a value is slightly greater than random number should be seen in the column of F(X) and corresponding value of x in front of F(X) is selected as random sample.
i.e. first random number is 0.180 then we see the value slightly grater than 0.180 in column of F(X) that is 0.250 and corresponding value of x in front of F(X) is 2 and it is selected as random sample.
that are in column of X. and the next column is created for calculating H.M
the observation table is contains mainly five columns that are
|
xi |
p(xi) |
F(xi) |
Rand. No. |
xi |
and the next column is depended on what you are told to find.
A.M. = ∑Xi / n ; HERE Xi = total of Xi = 36 and n = total number of observations = 10
A.M. = 36/10
A.M. = 3.6
H. M. = n / ∑(1/ Xi)
H. M. = 10 / 5.625
H. m. = 1.7777
Result : Random sample of size 10 is
|
Sample |
2 |
1 |
1 |
2 |
4 |
8 |
1 |
8 |
1 |
8 |
A.M. of sample is 3.6
H. M. Of sample is 1.7777
Q. 2. Draw a model sample of size 15 from the following Discrete Uniform Distribution Taking values 10,15,20,25,30,35,40,45,50,55. Find mean deviation from mode of your sample.
Answer:
first we see the discrete uniform distribution.
If X ~ DU(1,…., n ), Then the p.m.f. of discrete uniform distribution is given as,
and it's mean = ( n+1)/2
& variance = (n²-1) /12.
here x takes values 10, 15, 20, 25, 30, 35, 40, 45, 50, 55.
then total number of observation is 10, i.e. n=10
therefore the p.m.f. of discrete uniform distribution is
P(X=x) = 1/10 = 0.1; x=10, 15, 20, 25, 30, 35, 40, 45, 50, 55.
= 0 otherwise.
then we, write the Aim of the question.
Aim: Draw a model sample of size 15 from the following Discrete Uniform Distribution. and Find mean deviation from mode of your sample.
Statistical Formula:
Mean Deviation from mode = M.D. = (1/n) ∑| Xi - Mode|
therefore the p.m.f. of discrete uniform distribution is
P(X=x) = 1/10 = 0.1; x=10, 15, 20, 25, 30, 35, 40, 45, 50, 55.
= 0 otherwise.
now we draw random sample so we create observation table.
Observation Table:
|
Sr. No. |
xi |
p(xi) |
F(xi) |
Rand. No. |
Random Sample |
|
Xi - Mode| |
|
1 |
10 |
0.1 |
0.1 |
0.235467 |
20 |
15 |
|
2 |
15 |
0.1 |
0.2 |
0.230688 |
20 |
15 |
|
3 |
20 |
0.1 |
0.3 |
0.585224 |
35 |
0 |
|
4 |
25 |
0.1 |
0.4 |
0.144258 |
15 |
20 |
|
5 |
30 |
0.1 |
0.5 |
0.574728 |
35 |
0 |
|
6 |
35 |
0.1 |
0.6 |
0.094317 |
10 |
25 |
|
7 |
40 |
0.1 |
0.7 |
0.445045 |
30 |
5 |
|
8 |
45 |
0.1 |
0.8 |
0.125616 |
15 |
20 |
|
9 |
50 |
0.1 |
0.9 |
0.872183 |
50 |
15 |
|
10 |
55 |
0.1 |
1 |
0.776099 |
45 |
10 |
|
11 |
|
0.024739 |
10 |
25 |
||
|
12 |
0.975556 |
55 |
20 |
|||
|
13 |
0.569214 |
35 |
0 |
|||
|
14 |
0.621989 |
40 |
5 |
|||
|
15 |
0.460018 |
30 |
5 |
|||
|
Total |
180 |
|||||
mode is 35 because it repeated 3 times. 35 has maximum frequency.
Calculation:
Mean Deviation from mode = M.D. = (1/n) ∑| Xi - Mode|
M.D. = (1/15) x 180
M. D. = 12
Result: Random sample of size 15 is
|
Random Sample |
20 |
20 |
35 |
15 |
35 |
10 |
30 |
15 |
50 |
45 |
10 |
55 |
35 |
40 |
30 |
Mean Deviation from mode =12.
Q. 3. Draw a model sample of size 8 from the following Discrete Uniform Distribution
P(X) = 1/13; x=1,2,3,...,13
= 0 otherwise.
Obtain the quartiles of your sample.
Answer:
here given that x has discrete uniform distribution
P(X) = 1/13; x=1,2,3,...,13
= 0 otherwise.
Aim: To Draw a model sample of size 8 and find the quartiles of your sample.
Statistical Formula:
quartiles = Qi = Size of (i(N+1)/4}th observation, for i = 1, 2, 3.
here given that x has discrete uniform distribution
P(X) = 1/13 = 0.0769; x=1,2,3,...,13
= 0 otherwise.
now we draw random sample so we create observation table.
Observation Table:
|
Sr. No. |
xi |
p(xi) |
F(xi) |
Rand. No. |
Random Sample |
ascending |
|
1 |
1 |
0.0769 |
0.0769 |
0.5956 |
8 |
2 |
|
2 |
2 |
0.0769 |
0.1538 |
0.9960 |
13 |
5 |
|
3 |
3 |
0.0769 |
0.2308 |
0.8944 |
12 |
7 |
|
4 |
4 |
0.0769 |
0.3077 |
0.3768 |
5 |
8 |
|
5 |
5 |
0.0769 |
0.3846 |
0.7382 |
10 |
10 |
|
6 |
6 |
0.0769 |
0.4615 |
0.7611 |
10 |
10 |
|
7 |
7 |
0.0769 |
0.5385 |
0.0930 |
2 |
12 |
|
8 |
8 |
0.0769 |
0.6154 |
0.4907 |
7 |
13 |
|
9 |
9 |
0.0769 |
0.6923 |
|
|
|
|
10 |
10 |
0.0769 |
0.7692 |
|
|
|
|
11 |
11 |
0.0769 |
0.8462 |
|
|
|
|
12 |
12 |
0.0769 |
0.9231 |
|
|
|
|
13 |
13 |
0.0769 |
1 |
|
|
|
Calculation:
Q1 = Size of (N+1)/4 th observation
=Size of (8+1)/4 th observation
=Size of (2.25) th observation
=5+0.25*(7-5)
Q1 = 5.5
Q2=Size of 2* (N+1)/4 th observation
=Size of 2*(8+1)/4 th observation
=Size of (4.5) th observation
=8+0.5*(10-8)
Q2 =9
Q3 =Size of 3*(N+1)/4 th observation
=Size of 3*(8+1)/4 th observation
=Size of (6.75) th observation
=10+0.75*(12-10)
Q3 =11.5
Result: The Quartiles of sample is Q1 = 5.5, Q2 =9, Q3 =11.5.
Q. 4. Draw a model sample of size 10 from the Discrete Uniform Distribution over the set of first 25 natural numbers. Find C.V. of your sample.
Answer:
here given that x has discrete uniform distribution
and the Discrete Uniform Distribution over the set of first 25 natural numbers.
i.e. x take values 1, 2, 3, 4, ......., 25
P(X) = 1/25; x=1,2,3,...,25
= 0 otherwise.
Aim: To Draw a model sample of size 10 and Find C.V. of your sample.
Statistical Formula:
C. V. = {(S.D.) / (Mean)} x 100
here Mean = ∑Xi / n
S.D. = √(Variance)
Variance = {∑X²i / n} - (x̄)²
here given that x has discrete uniform distribution
P(X) = 1/25 = 0.04 ; x=1,2,3,...,25
= 0 otherwise.
now we draw random sample so we create observation table.
Observation Table:
|
Sr.No. |
xi |
p(xi) |
F(xi) |
Rand.No. |
xi(Random sample) |
Xi^2 |
|
1 |
1 |
0.04 |
0.04 |
0.594799 |
15 |
225 |
|
2 |
2 |
0.04 |
0.08 |
0.25552 |
7 |
49 |
|
3 |
3 |
0.04 |
0.12 |
0.472988 |
12 |
144 |
|
4 |
4 |
0.04 |
0.16 |
0.62064 |
16 |
256 |
|
5 |
5 |
0.04 |
0.2 |
0.899819 |
23 |
529 |
|
6 |
6 |
0.04 |
0.24 |
0.104979 |
3 |
9 |
|
7 |
7 |
0.04 |
0.28 |
0.902166 |
23 |
529 |
|
8 |
8 |
0.04 |
0.32 |
0.178694 |
5 |
25 |
|
9 |
9 |
0.04 |
0.36 |
0.379054 |
10 |
100 |
|
10 |
10 |
0.04 |
0.4 |
0.102018 |
3 |
9 |
|
11 |
11 |
0.04 |
0.44 |
Total |
117 |
1875 |
|
12 |
12 |
0.04 |
0.48 |
|||
|
13 |
13 |
0.04 |
0.52 |
|||
|
14 |
14 |
0.04 |
0.56 |
|||
|
15 |
15 |
0.04 |
0.6 |
|||
|
16 |
16 |
0.04 |
0.64 |
|||
|
17 |
17 |
0.04 |
0.68 |
|||
|
18 |
18 |
0.04 |
0.72 |
|||
|
19 |
19 |
0.04 |
0.76 |
|||
|
20 |
20 |
0.04 |
0.8 |
|||
|
21 |
21 |
0.04 |
0.84 |
|||
|
22 |
22 |
0.04 |
0.88 |
|||
|
23 |
23 |
0.04 |
0.92 |
|||
|
24 |
24 |
0.04 |
0.96 |
|||
|
25 |
25 |
0.04 |
1 |
Calculation:
Mean = ∑Xi / n
= 117/10
Mean =11.7
Variance = {∑X²i / n} - (x̄)²
= (1875/10) -(11.7)²
Variance = 50.61
S.D. = √(Variance)
S.D. = √(50.61)
S.D.= 7.1141
C. V. = {(S.D.) / (Mean)} x 100
C. V. = {(7.1141) / (11.7)} x 100
C. V. = 60.8040%
Result:
C. V. of sample is = 60.8040%
Q. 5. Draw a random sample of size 11 from discrete uniform distribution taking values 1, 2, 3, 4, 5. find coefficient of M.D. about mean of your sample.
Answer:
here given that x has discrete uniform distribution
and the Discrete Uniform Distribution over the set of values 1, 2, 3, 4, 5.
i.e. x take values 1, 2, 3, 4, 5.
P(X) = 1/ 5 = 0.2; x=1,2,3,4, 5.
= 0 otherwise.
Aim: To Draw a model sample of size 11 and find coefficient of M.D. about mean.
Statistical Formula:
Mean = ∑Xi / n
Mean Deviation about Mean = M.D.(x̄)= 𝟏 /𝒏 ∑ |𝑿𝒊 − x̄|
Coeff. of Mean Deviation about Mean = M.D.(x̄) / x̄
now we draw random sample so we create observation table.
Observation Table:
|
Sr.No. |
xi |
p(xi) |
F(xi) |
Rand.No. |
xi(Random sample) |
| Xi -
Mean| |
|
1 |
1 |
0.2 |
0.2 |
0.357684 |
2 |
0.73 |
|
2 |
2 |
0.2 |
0.4 |
0.199077 |
1 |
1.73 |
|
3 |
3 |
0.2 |
0.6 |
0.802135 |
5 |
2.27 |
|
4 |
4 |
0.2 |
0.8 |
0.331298 |
2 |
0.73 |
|
5 |
5 |
0.2 |
1 |
0.485472 |
3 |
0.27 |
|
0.372287 |
2 |
0.73 |
||||
|
0.643617 |
4 |
1.27 |
||||
|
0.322358 |
2 |
0.73 |
||||
|
0.570065 |
3 |
0.27 |
||||
|
0.253577 |
2 |
0.73 |
||||
|
0.790086 |
4 |
1.27 |
||||
|
Total |
30 |
10.73 |
|
Random
Sample |
2 |
1 |
5 |
2 |
3 |
2 |
4 |
2 |
3 |
2 |
4 |
here given that x has discrete uniform distribution
and the Discrete Uniform Distribution over the set of values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
i.e. x take values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
P(X) = 1/ 21 = 0.0476; x= -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
= 0 otherwise.
Aim: To Draw a model sample of size 10 and find mean and variance of your sample.
Statistical Formula:
Mean = ∑Xi / n
Variance = {∑X²i / n} - (x̄)²
now we draw random sample so we create observation table.
Observation Table:
|
Sr. No. |
xi |
p(xi) |
F(xi) |
Rand. No. |
xi (Random sample) |
Xi^2 |
|
1 |
-10 |
0.0476 |
0.0476 |
0.497454 |
0 |
0 |
|
2 |
-9 |
0.0476 |
0.0952 |
0.334415 |
-3 |
9 |
|
3 |
-8 |
0.0476 |
0.1429 |
0.978027 |
10 |
100 |
|
4 |
-7 |
0.0476 |
0.1905 |
0.919356 |
9 |
81 |
|
5 |
-6 |
0.0476 |
0.2381 |
0.63215 |
3 |
9 |
|
6 |
-5 |
0.0476 |
0.2857 |
0.066648 |
-9 |
81 |
|
7 |
-4 |
0.0476 |
0.3333 |
0.373673 |
-3 |
9 |
|
8 |
-3 |
0.0476 |
0.381 |
0.235007 |
-6 |
36 |
|
9 |
-2 |
0.0476 |
0.4286 |
0.9047 |
8 |
64 |
|
10 |
-1 |
0.0476 |
0.4762 |
0.757458 |
5 |
25 |
|
11 |
0 |
0.0476 |
0.5238 |
Total |
14 |
414 |
|
12 |
1 |
0.0476 |
0.5714 |
|||
|
13 |
2 |
0.0476 |
0.619 |
|||
|
14 |
3 |
0.0476 |
0.6667 |
|||
|
15 |
4 |
0.0476 |
0.7143 |
|||
|
16 |
5 |
0.0476 |
0.7619 |
|||
|
17 |
6 |
0.0476 |
0.8095 |
|||
|
18 |
7 |
0.0476 |
0.8571 |
|||
|
19 |
8 |
0.0476 |
0.9048 |
|||
|
20 |
9 |
0.0476 |
0.9524 |
|||
|
21 |
10 |
0.0476 |
1 |
Calculation:
Mean = ∑Xi / n
=14/10
Mean =1.4
Variance = {∑X²i / n} - (x̄)²
={414/10} - (1.4)²
Variance =39.44
Result:
Random Sample is
|
Random
Sample |
0 |
-3 |
10 |
9 |
3 |
-9 |
-3 |
-6 |
8 |
5 |
The mean of sample= 1.4 and variance of sample= 39.44
Padmbhushan Vasantraodada Patil Mahavidyalaya, Kavathe Mahankal
Department of Statistics
Title: Model sampling from Binomial distribution
6. If X has binomial distribution such that P{X = 0} = P{X = 5} and n = 5 then obtain a random sample of size 11 from this distribution. Obtain sample median and sample mode.
7. If x has binomial distribution such that E(X) =2 and V(X) =1.5 then obtain a random sample of size 25 from this distribution. Obtain sample mean and sample variance. Compare these with theoretical mean and variance of the distribution.
ANSWERS Practical NO. 1.
Q1. Draw a model sample of size 10 from the binomial Distribution with parameters n = 8 and p = 0.6 also find A.M., G.M. and H. M. of your sample.
Answer :
Binomial Distribution PMF
Enter the parameters of the binomial distribution:
Enter the value of k to calculate P(X = k):
PMF for X =
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