Mathematical Statistics-Questions and Answers
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About This Presentation
Subject: Mathematical Statistics
Unit wise questions with answers are given for practice
Slide Content
MATHEMATICAL STATISTIC S
QUESTIONS AND ANSWERS
UNIT-1
1. A continuous random variable has the distribution function.
F(x) = {
0 ??????� �≤1
??????(�−1)
4
??????� 1<�≤3
1 ??????� �>3
Find (i) k (ii) p.d.f f(x)
Solution: For a probability density function, ∫�
∞
−∞
(x) dx = 1
K=1/16
f(x) =
�
��
F(x)
f(x)={
0 , �≤1
(
1
4
)(�−1)
3
0 , �>3
, 1<�≤3
_______________________________________________________________
2. Given the distribution function F(x) = {
0 ??????� �<−1
�+2
4
??????�−1≤�<1
1 ??????� �≥1
Find (i) P(−
1
2
<�≤
1
2
)
(ii) P(x = 0)
(iii) P(x = 1)
(iv) P(2<�≤3)
QUESTIONS AND ANSWERS
UNIT-1
1. A continuous random variable has the distribution function.
F(x) = {
0 ??????� �≤1
??????(�−1)
4
??????� 1<�≤3
1 ??????� �>3
Find (i) k (ii) p.d.f f(x)
Solution: For a probability density function, ∫�
∞
−∞
(x) dx = 1
K=1/16
f(x) =
�
��
F(x)
f(x)={
0 , �≤1
(
1
4
)(�−1)
3
0 , �>3
, 1<�≤3
_______________________________________________________________
2. Given the distribution function F(x) = {
0 ??????� �<−1
�+2
4
??????�−1≤�<1
1 ??????� �≥1
Find (i) P(−
1
2
<�≤
1
2
)
(ii) P(x = 0)
(iii) P(x = 1)
(iv) P(2<�≤3)
Solution: (i) P(−
1
2
<�≤
1
2
) =
1
4
( since P [ a < X ≤ b] = F(b) –F(a) )
(ii) P(x = 0) = 0
(iii) P(x = 1) =
1
4
(iv) P(2<�≤3) = 0
_______________________________________________________
3. Let the continuous random variable X have the p.d.f
f (x) = {
2
�
3
??????� 1<�<∞
0 ��ℎ���??????��
Find distribution function F(x).
Solution:If x < 1, F(x) = ∫�
�
−∞
(t) dt =0
If x ≥ 1, F(x) = ∫�
�
−∞
(t) dt = 1 -
1
�
2
F(x) ={
0,x < 1
1−
1
�
2
, x ≥ 1
____________________________________________________________
4. A random variable X has the following probability function.
Xi -2 -1 0 1 2 3
Pi = P(xi) 0.1 k 0.2 2k 0.3 k
Find (i) the value of k
(ii) mean
(iii) variance
1
2
<�≤
1
2
) =
1
4
( since P [ a < X ≤ b] = F(b) –F(a) )
(ii) P(x = 0) = 0
(iii) P(x = 1) =
1
4
(iv) P(2<�≤3) = 0
_______________________________________________________
3. Let the continuous random variable X have the p.d.f
f (x) = {
2
�
3
??????� 1<�<∞
0 ��ℎ���??????��
Find distribution function F(x).
Solution:If x < 1, F(x) = ∫�
�
−∞
(t) dt =0
If x ≥ 1, F(x) = ∫�
�
−∞
(t) dt = 1 -
1
�
2
F(x) ={
0,x < 1
1−
1
�
2
, x ≥ 1
____________________________________________________________
4. A random variable X has the following probability function.
Xi -2 -1 0 1 2 3
Pi = P(xi) 0.1 k 0.2 2k 0.3 k
Find (i) the value of k
(ii) mean
(iii) variance
(iv) P(x 2)
(v) P(x < 2)
(vi) P(-1 < x < 3)
Solution: (i) the value of k = 0.1
(ii) mean = 0.8
(iii) variance = 2.16
(iv) P(x 2) = 0.4
(v) P(x < 2) =0.6
(vi) P(-1 < x < 3)= 0.7
5. A random variable X has the following probability function.
x 1 2 3 4 5 6 7
P(x) k 2k 2k 3k K
2
2k
2
7k
2
+ k
Find (i) value of k
(ii) P(�≥6)
(iii) P (�<6)
(iv) P (1≤�<5)
(v) E(x)
Solution: (i) value of k
k= -1 (Or) k =
1
10
(v) P(x < 2)
(vi) P(-1 < x < 3)
Solution: (i) the value of k = 0.1
(ii) mean = 0.8
(iii) variance = 2.16
(iv) P(x 2) = 0.4
(v) P(x < 2) =0.6
(vi) P(-1 < x < 3)= 0.7
5. A random variable X has the following probability function.
x 1 2 3 4 5 6 7
P(x) k 2k 2k 3k K
2
2k
2
7k
2
+ k
Find (i) value of k
(ii) P(�≥6)
(iii) P (�<6)
(iv) P (1≤�<5)
(v) E(x)
Solution: (i) value of k
k= -1 (Or) k =
1
10
k=-1 is not permissible. Take k =
1
10
(ii) P(�≥6) =
19
100
(iii) P (�<6) =
81
100
(iv) P (1≤�<5) =
8
10
(v) E(x) =
366
100
_______________________________________________________________
UNIT -2
1. Find the binomial distribution for which mean is 4 and variance is 4/3.
Solution: q =
1
3
p =
2
3
n= 6
Required Binomial distribution is:
P(x) = �
�
�
(
2
3
)
�
(
1
3
)
??????−�
________________________________________________________________
2. If X is binomially distributed random variable with mean = 2 and
variance 4/3. Find P(X = 5).
Solution: p =
1
3
q =
2
3
n= 6
1
10
(ii) P(�≥6) =
19
100
(iii) P (�<6) =
81
100
(iv) P (1≤�<5) =
8
10
(v) E(x) =
366
100
_______________________________________________________________
UNIT -2
1. Find the binomial distribution for which mean is 4 and variance is 4/3.
Solution: q =
1
3
p =
2
3
n= 6
Required Binomial distribution is:
P(x) = �
�
�
(
2
3
)
�
(
1
3
)
??????−�
________________________________________________________________
2. If X is binomially distributed random variable with mean = 2 and
variance 4/3. Find P(X = 5).
Solution: p =
1
3
q =
2
3
n= 6
Required Binomial distribution is:
P(x) = �
�
�
(
1
3
)
�
(
2
3
)
??????−�
P(x=5) = 0.0164
_______________________________________________________________
3. 10 coins are thrown simultaneously. Find the probability of getting at least 7
heads.
Solution: q =
1
2
; p =
1
2
; n= 10
P(x) = �
�
�
�
�
�
??????−�
P(x ≥ 7) =
176
1024
______________________________________________________________
4. Discuss about the properties of normal distribution.
(1) The normal probability curve is symmetrical about the ordinate at x = ??????.
The ordinate decreases rapidly as x increases. The curve extends to infinity
on either side of the mean. Then x axis is an asymptote to the curve.
(2) The mean, median and mode coincide the maximum ordinate at x = ?????? is
given by
1
??????√2??????
.
(3) ?????? ± ?????? are the points of inflection of the normal curve and hence the points
of inflection are also equidistance from the median.
(4) The area under the normal curve is the unity. The ordinate at x= ?????? divides
the area under the normal curve into 2 equal parts.
(5) P( ?????? - ?????? < x < ?????? + ?????? ) = .6826
P( ?????? - 2?????? < x < ?????? + 2 ?????? ) = .9544
P( ?????? - 3 ?????? < x < ?????? + 3?????? ) = .9973
(6) Q.D :M.D : S.D =10:12:15
P(x) = �
�
�
(
1
3
)
�
(
2
3
)
??????−�
P(x=5) = 0.0164
_______________________________________________________________
3. 10 coins are thrown simultaneously. Find the probability of getting at least 7
heads.
Solution: q =
1
2
; p =
1
2
; n= 10
P(x) = �
�
�
�
�
�
??????−�
P(x ≥ 7) =
176
1024
______________________________________________________________
4. Discuss about the properties of normal distribution.
(1) The normal probability curve is symmetrical about the ordinate at x = ??????.
The ordinate decreases rapidly as x increases. The curve extends to infinity
on either side of the mean. Then x axis is an asymptote to the curve.
(2) The mean, median and mode coincide the maximum ordinate at x = ?????? is
given by
1
??????√2??????
.
(3) ?????? ± ?????? are the points of inflection of the normal curve and hence the points
of inflection are also equidistance from the median.
(4) The area under the normal curve is the unity. The ordinate at x= ?????? divides
the area under the normal curve into 2 equal parts.
(5) P( ?????? - ?????? < x < ?????? + ?????? ) = .6826
P( ?????? - 2?????? < x < ?????? + 2 ?????? ) = .9544
P( ?????? - 3 ?????? < x < ?????? + 3?????? ) = .9973
(6) Q.D :M.D : S.D =10:12:15
_____________________________________________________________
5. Explain the mode of binomial distribution.
Since mode is the value of x for which p(x) is maximum,
we have p(x) ≥ p(x+1) and p(x) ≥ p(x-1)
(n+1)p -1 ≤ x ≤ (n+1)p
Case(i):If (n+1)p is not an integer, clearly mode is the integral part of (n+1)p and
the distribution is unimodal.
Case(ii):If (n+1)p is an integer ,both (n+1)p and (n+1)p will represent mode and
the distribution is bimodal.
________________________________________________________________
UNIT-3
1. Calculate Karl Pearson’s co-efficient of correlation.
Solution: r =
??????(∑��)−(∑�)(∑�)
√??????∑�
2
−(∑�)
2
√??????∑�
2
−(∑�)
2
∑� =21; ∑�
2
=91;∑� =31; ∑�
2
=187 ; ∑�� =105
From this, r = -0.1621
____________________________________________________________
2. Find the coefficient of correlation between X and Y from the following data
N= 10,∑�=60,∑�=60,∑��=305,∑�
2
=400,∑�
2
=580.
Solution: r =
�(∑��)−(∑�)(∑�)
√�∑�
2
−(∑�)
2
√�∑�
2
−(∑�)
2
By calculation, r = -.586
X 1 2 3 4 5
6
Y 7 6 5 2 3
8
5. Explain the mode of binomial distribution.
Since mode is the value of x for which p(x) is maximum,
we have p(x) ≥ p(x+1) and p(x) ≥ p(x-1)
(n+1)p -1 ≤ x ≤ (n+1)p
Case(i):If (n+1)p is not an integer, clearly mode is the integral part of (n+1)p and
the distribution is unimodal.
Case(ii):If (n+1)p is an integer ,both (n+1)p and (n+1)p will represent mode and
the distribution is bimodal.
________________________________________________________________
UNIT-3
1. Calculate Karl Pearson’s co-efficient of correlation.
Solution: r =
??????(∑��)−(∑�)(∑�)
√??????∑�
2
−(∑�)
2
√??????∑�
2
−(∑�)
2
∑� =21; ∑�
2
=91;∑� =31; ∑�
2
=187 ; ∑�� =105
From this, r = -0.1621
____________________________________________________________
2. Find the coefficient of correlation between X and Y from the following data
N= 10,∑�=60,∑�=60,∑��=305,∑�
2
=400,∑�
2
=580.
Solution: r =
�(∑��)−(∑�)(∑�)
√�∑�
2
−(∑�)
2
√�∑�
2
−(∑�)
2
By calculation, r = -.586
X 1 2 3 4 5
6
Y 7 6 5 2 3
8
3. Calculate Spearman’s rank correlation to the following data
X 1 3 4 5 7 2 6
Y 2 6 5 7 1 3 4
Solution:
?????? =1−
6 ∑�
2
??????(??????
2
−1)
Answer: ?????? =0
_____________________________________________________________
4. Obtain two regression coefficients.
x 4 5 6 8 11
y 12 10 8 7 5
Solution:
??????
�� =
??????(∑��)−(∑�)(∑�)
??????∑�
2
−(∑�)
2;??????
��
=
�(∑��)−(∑�)(∑�)
�∑�
2
−(∑�)
2
??????
�� = -0.93 ; ??????
�� = -0.98
5. Obtain two regression coefficients to the following data;
Mean of marks in mathematics = 80; Mean of marks in English = 50;
S.D of marks in mathematics = 15; S.D. of marks in English = 10;
Coefficient of correlation= 0. 4
Solution:
??????
�� = ??????
??????
�
??????
�
; ??????
�� = ??????
??????
�
??????
�
??????
�� = 0.27 ; ??????
�� = 0.6
UNIT-4
X 1 3 4 5 7 2 6
Y 2 6 5 7 1 3 4
Solution:
?????? =1−
6 ∑�
2
??????(??????
2
−1)
Answer: ?????? =0
_____________________________________________________________
4. Obtain two regression coefficients.
x 4 5 6 8 11
y 12 10 8 7 5
Solution:
??????
�� =
??????(∑��)−(∑�)(∑�)
??????∑�
2
−(∑�)
2;??????
��
=
�(∑��)−(∑�)(∑�)
�∑�
2
−(∑�)
2
??????
�� = -0.93 ; ??????
�� = -0.98
5. Obtain two regression coefficients to the following data;
Mean of marks in mathematics = 80; Mean of marks in English = 50;
S.D of marks in mathematics = 15; S.D. of marks in English = 10;
Coefficient of correlation= 0. 4
Solution:
??????
�� = ??????
??????
�
??????
�
; ??????
�� = ??????
??????
�
??????
�
??????
�� = 0.27 ; ??????
�� = 0.6
UNIT-4
1. A coin is tossed 144 times and a person gets 80heads.Can we say that the
coin is unbiased one?
Solution: Set ??????
0: the coin is unbiased.
Given n=144 , P = 0.5 and Q = 0.5,X=80
At 5 % level of significance, |�
??????| =1.96
Z =
??????−??????�
√??????��
By computation, z = 1.33
Calculated value =1.33 < 1.96=table value
The difference is not significant.
Hence the coin is unbiased.
____________________________________________________________
2. In a big city 325 men out of 600 men were found to be smokers. Does this
information support the conclusion that the majority of men in this city are
smokers?
Solution:
Set ??????
0: the number of smokers and non smokers are equal in the city.
Given n=600 and p=0.5417
P = 0.5 and Q = 0.5
??????
1 : P> 0.5(right tailed test)
At 5 % level of significance for right tailed test, |�
??????| =1.645
Z =
�−�
√
��
??????
.Then By computation, z = 2.04
Calculated value =2.04>1.645=table value
The difference is significant.
____________________________________________________________
coin is unbiased one?
Solution: Set ??????
0: the coin is unbiased.
Given n=144 , P = 0.5 and Q = 0.5,X=80
At 5 % level of significance, |�
??????| =1.96
Z =
??????−??????�
√??????��
By computation, z = 1.33
Calculated value =1.33 < 1.96=table value
The difference is not significant.
Hence the coin is unbiased.
____________________________________________________________
2. In a big city 325 men out of 600 men were found to be smokers. Does this
information support the conclusion that the majority of men in this city are
smokers?
Solution:
Set ??????
0: the number of smokers and non smokers are equal in the city.
Given n=600 and p=0.5417
P = 0.5 and Q = 0.5
??????
1 : P> 0.5(right tailed test)
At 5 % level of significance for right tailed test, |�
??????| =1.645
Z =
�−�
√
��
??????
.Then By computation, z = 2.04
Calculated value =2.04>1.645=table value
The difference is significant.
____________________________________________________________
3. A normal population has a mean of 6.48 and S.D. of 1.5.In a sample of 400
members mean is 6.75 . Is the difference significant?
Solution: Set ??????
0: ?????? = �̅
Given n=400 , ?????? = 6.48 and �̅ = 6.75, ??????= 1.5
At 5 % level of significance, |�
??????| =1.96
Z =
�̅−??????�
??????
√??????
By computation, z = 3.6
Calculated value =3.6 >3
??????
0 is rejected.
The difference is significant.
____________________________________________________________
4. A sample of 400 individuals is found to have a mean of 67.47.Can it be
reasonably regarded as a sample from a large population with 67.39 and
S.D.1.3?
Solution: Set ??????
0: ?????? = �̅
Given n=400 , ?????? = 67.39 and �̅ = 67.47, ??????= 1.3
At 5 % level of significance, |�
??????| =1.96
Z =
�̅−??????�
??????
√??????
By computation, z = 1.23
Calculated value |�| =1.23 < 1.96
??????
0 is accepted.
The difference is not significant.
___________________________________________________________
5. The mean yields of rice from two places in a district were 210 kgs and 220
kgs per acre from 100 acres and 150 acres respectively. Can it be regarded
members mean is 6.75 . Is the difference significant?
Solution: Set ??????
0: ?????? = �̅
Given n=400 , ?????? = 6.48 and �̅ = 6.75, ??????= 1.5
At 5 % level of significance, |�
??????| =1.96
Z =
�̅−??????�
??????
√??????
By computation, z = 3.6
Calculated value =3.6 >3
??????
0 is rejected.
The difference is significant.
____________________________________________________________
4. A sample of 400 individuals is found to have a mean of 67.47.Can it be
reasonably regarded as a sample from a large population with 67.39 and
S.D.1.3?
Solution: Set ??????
0: ?????? = �̅
Given n=400 , ?????? = 67.39 and �̅ = 67.47, ??????= 1.3
At 5 % level of significance, |�
??????| =1.96
Z =
�̅−??????�
??????
√??????
By computation, z = 1.23
Calculated value |�| =1.23 < 1.96
??????
0 is accepted.
The difference is not significant.
___________________________________________________________
5. The mean yields of rice from two places in a district were 210 kgs and 220
kgs per acre from 100 acres and 150 acres respectively. Can it be regarded
that the samples were drawn from the same district which has the s.d. of 11
kgs per acre?
Solution : Set ??????
0: ??????
1 = ??????
2
Given �
1= 100 , �̅
1 =210 , ??????= 150, �
2 = 150 , �̅
2 =220
At 5 % level of significance, |�
??????| =1.96
Z =
�̅
1 − �̅
2
??????√(
1
??????1
)+(
1
??????2
)
By computation, z = -7.04
Calculated value, |�| =7.04 >3
??????
0 is rejected.
The difference is highly significant.
UNIT-5
1. Weights in kilograms of 10 students are given below: 38,40,45,53,47,43,55,
48,52, 49. Can we say that the mean weight of the distribution of all students from
which the above sample was drawn is equal to 45.
Solution: Set ??????
0: ?????? = �̅
Calculating ,�̅ =47,n =10 , �
2
=28,s =5.292
Test statistic t =1.154
D.F=n-1=9
Table value =2.26
Calculated value =1.154
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
________________________________________________________________
2. The A random sample of 10 boys has the following I.Q(Intelligent Quotients).
70,120,110,101,88,83,95,98,107,100.Do these supports the assumption of a
population mean I.Q. of 100? )26.2
05.0
( =t )26.2
05.0
( =t
kgs per acre?
Solution : Set ??????
0: ??????
1 = ??????
2
Given �
1= 100 , �̅
1 =210 , ??????= 150, �
2 = 150 , �̅
2 =220
At 5 % level of significance, |�
??????| =1.96
Z =
�̅
1 − �̅
2
??????√(
1
??????1
)+(
1
??????2
)
By computation, z = -7.04
Calculated value, |�| =7.04 >3
??????
0 is rejected.
The difference is highly significant.
UNIT-5
1. Weights in kilograms of 10 students are given below: 38,40,45,53,47,43,55,
48,52, 49. Can we say that the mean weight of the distribution of all students from
which the above sample was drawn is equal to 45.
Solution: Set ??????
0: ?????? = �̅
Calculating ,�̅ =47,n =10 , �
2
=28,s =5.292
Test statistic t =1.154
D.F=n-1=9
Table value =2.26
Calculated value =1.154
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
________________________________________________________________
2. The A random sample of 10 boys has the following I.Q(Intelligent Quotients).
70,120,110,101,88,83,95,98,107,100.Do these supports the assumption of a
population mean I.Q. of 100? )26.2
05.0
( =t )26.2
05.0
( =t
Solution: Set ??????
0: ?????? = �̅
Calculating ,�̅ = 97.2,n =10 , s = 14.27
Test statistic|�| = 0.62
D.F=n-1=9
Table value =2.26
Calculated value =0.62
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
________________________________________________________________
3. Two methods of performing a certain operation compared. The following are
obtained. Test the significant difference between two variances.
� = 50.5 S1
2
= 6.5 n1 = 15
� = 57.2 S2
2
= 5.7 n2 = 12
Solution: Set ??????
0: ??????
1
2
= ??????
2
2
F=
??????
1
2
??????
2
2 = 1.120
Table value = 2.79
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
_______________________________________________________________
4. The following data is collected on two characters: )79.2)11,14(( =F
Cinegoers Non-Cinegoers
Literate 83 57
0: ?????? = �̅
Calculating ,�̅ = 97.2,n =10 , s = 14.27
Test statistic|�| = 0.62
D.F=n-1=9
Table value =2.26
Calculated value =0.62
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
________________________________________________________________
3. Two methods of performing a certain operation compared. The following are
obtained. Test the significant difference between two variances.
� = 50.5 S1
2
= 6.5 n1 = 15
� = 57.2 S2
2
= 5.7 n2 = 12
Solution: Set ??????
0: ??????
1
2
= ??????
2
2
F=
??????
1
2
??????
2
2 = 1.120
Table value = 2.79
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
_______________________________________________________________
4. The following data is collected on two characters: )79.2)11,14(( =F
Cinegoers Non-Cinegoers
Literate 83 57
From the above data, find out if there is any relation between literacy and
the habit of cinegores.
Solution : Set ??????
0: there is no relation between literacy and
the habit of cinegores.
Calculation of Ψ
2
:
)841.3( 05.0
2
=
Illiterate 45 68
Cinegoers Non-
Cinegoers
Total
Literate 70.83 69.17 140
Illiterate 57.17 55.83 113
Total 128 125 253
O E (??????−??????)
2
??????
83 70.83 2.09
57 69.17 2.14
45 57.17 2.59
68 55.83 2.65
9.47
the habit of cinegores.
Solution : Set ??????
0: there is no relation between literacy and
the habit of cinegores.
Calculation of Ψ
2
:
)841.3( 05.0
2
=
Illiterate 45 68
Cinegoers Non-
Cinegoers
Total
Literate 70.83 69.17 140
Illiterate 57.17 55.83 113
Total 128 125 253
O E (??????−??????)
2
??????
83 70.83 2.09
57 69.17 2.14
45 57.17 2.59
68 55.83 2.65
9.47
Ψ
2
= ∑
(�−??????)
2
??????
=9.47
Table value = 3.841
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
________________________________________________________________
5. The average number of articles produced by two machines per day are 200 & 250
with S.D 20 &25respectively. On the basis of records of 25 days production, Can
you regard both the machine equally efficient at 1% level of significance?(
t0.01=2.58)
Solution: �
1̅̅̅ = 200 , �
2̅̅̅ = 250, �
1 = 20, �
2 = 25
Set ??????
0: ??????
1 = ??????
2
D.F = �
1 + �
2 - 2 = 25+ 25 -2 = 48
t0.01=2.58
test statistic t =
�̅
1 − �̅
2
??????√(
1
??????1
)+(
1
??????2
)
t = -7.65 and |�| = 7.65
Calculated value > Table value.
??????
0 is rejected.
The difference is significant.
_______________________________________________________________
2
= ∑
(�−??????)
2
??????
=9.47
Table value = 3.841
Calculated value < Table value.
??????
0 is accepted.
The difference is not significant.
________________________________________________________________
5. The average number of articles produced by two machines per day are 200 & 250
with S.D 20 &25respectively. On the basis of records of 25 days production, Can
you regard both the machine equally efficient at 1% level of significance?(
t0.01=2.58)
Solution: �
1̅̅̅ = 200 , �
2̅̅̅ = 250, �
1 = 20, �
2 = 25
Set ??????
0: ??????
1 = ??????
2
D.F = �
1 + �
2 - 2 = 25+ 25 -2 = 48
t0.01=2.58
test statistic t =
�̅
1 − �̅
2
??????√(
1
??????1
)+(
1
??????2
)
t = -7.65 and |�| = 7.65
Calculated value > Table value.
??????
0 is rejected.
The difference is significant.
_______________________________________________________________
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