The document presents a series of mathematical statistics problems and their solutions, covering topics such as probability density functions, binomial distributions, correlation coefficients, and hypothesis testing. Each section includes specific examples with corresponding calculations and interpretations of results. The document serves as a resource for understanding various statistical concepts and methodologies.

1. MATHEMATICAL STATISTICS
QUESTIONS AND ANSWERS
UNIT-1
1. A continuous random variable has the distribution function.
F(x) = {
0 𝑖𝑓 𝑥 ≤ 1
𝑘(𝑥 − 1)4
𝑖𝑓 1 < 𝑥 ≤ 3
1 𝑖𝑓 𝑥 > 3
Find (i) k (ii) p.d.f f(x)
Solution: For a probability density function, ∫ 𝑓
∞
−∞
(x) dx = 1
K=1/16
f(x) =
𝑑
𝑑𝑥
F(x)
f(x)={
0 , 𝑥 ≤ 1
(
1
4
) (𝑥 − 1)3
0 , 𝑥 > 3
, 1 < 𝑥 ≤ 3
_______________________________________________________________
2. Given the distribution function F(x) = {
0 𝑖𝑓 𝑥 < −1
𝑥+2
4
𝑖𝑓 − 1 ≤ 𝑥 < 1
1 𝑖𝑓 𝑥 ≥ 1
Find (i) P(−
1
2
< 𝑥 ≤
1
2
)
(ii) P(x = 0)
(iii) P(x = 1)
(iv) P(2 < 𝑥 ≤ 3)

2. Solution: (i) P(−
1
2
< 𝑥 ≤
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 < 𝑥 ≤ 3) = 0
_______________________________________________________
3. Let the continuous random variable X have the p.d.f
f (x) = {
2
𝑥3
𝑖𝑓 1 < 𝑥 < ∞
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find distribution function F(x).
Solution:If x < 1, F(x) = ∫ 𝑓
𝑥
−∞
(t) dt =0
If x ≥ 1, F(x) = ∫ 𝑓
𝑥
−∞
(t) dt = 1 -
1
𝑥2
F(x) ={
0, x < 1
1 −
1
𝑥2
, 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

3. (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 K2
2k2
7k2
+ k
Find (i) value of k
(ii) P(𝑥 ≥ 6)
(iii) P (𝑥 < 6)
(iv) P (1 ≤ 𝑥 < 5)
(v) E(x)
Solution: (i) value of k
k= -1 (Or) k =
1
10

4. k=-1 is not permissible. Take k =
1
10
(ii) P(𝑥 ≥ 6) =
19
100
(iii) P (𝑥 < 6) =
81
100
(iv) P (1 ≤ 𝑥 < 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) = 𝑛𝑐𝑥
(
2
3
)𝑥
(
1
3
)𝑛−𝑥
________________________________________________________________
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

5. Required Binomial distribution is:
P(x) = 𝑛𝑐𝑥
(
1
3
)𝑥
(
2
3
)𝑛−𝑥
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) = 𝑛𝑐𝑥
𝑝𝑥
𝑞𝑛−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 = 𝜇.
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 = 𝜇 is
given by
1
𝜎√2𝜋
.
(3) 𝜇 ± 𝜎 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= 𝜇 divides
the area under the normal curve into 2 equal parts.
(5) P( 𝜇 - 𝜎 < x < 𝜇 + 𝜎 ) = .6826
P( 𝜇 - 2𝜎 < x < 𝜇 + 2 𝜎 ) = .9544
P( 𝜇 - 3 𝜎 < x < 𝜇 + 3𝜎 ) = .9973
(6) Q.D :M.D : S.D =10:12:15

6. _____________________________________________________________
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 =
𝑛(∑ 𝑥𝑦)−(∑ 𝑥)(∑ 𝑦)
√𝑛 ∑ 𝑥
2
−(∑ 𝑥)
2√𝑛 ∑ 𝑦
2
−(∑ 𝑦)
2
∑ 𝑥 =21; ∑ 𝑥
2
= 91; ∑ 𝑦 =31; ∑ 𝑦
2
= 187 ; ∑ 𝑥𝑦 =105
From this, r = -0.1621
____________________________________________________________
2. Find the coefficient of correlation between X and Y from the following data
N= 10,∑ 𝑥 = 60, ∑ 𝑦 = 60, ∑ 𝑥𝑦 = 305, ∑ 𝑥2
= 400, ∑ 𝑦2
= 580.
Solution: r =
𝑁(∑ 𝑥𝑦)−(∑ 𝑥)(∑ 𝑦)
√𝑁 ∑ 𝑥
2
−(∑ 𝑥)
2√𝑁 ∑ 𝑦
2
−(∑ 𝑦)
2
By calculation, r = -.586
X 1 2 3 4 5 6
Y 7 6 5 2 3 8

7. 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:
𝜌 = 1 −
6 ∑ 𝑑2
𝑛(𝑛2 −1)
Answer: 𝜌 = 0
_____________________________________________________________
4. Obtain two regression coefficients.
x 4 5 6 8 11
y 12 10 8 7 5
Solution:
𝑏𝑦𝑥 =
𝑛(∑ 𝑥𝑦)−(∑ 𝑥)(∑ 𝑦)
𝑛 ∑ 𝑥
2
−(∑ 𝑥)
2 ; 𝑏𝑥𝑦 =
𝑛(∑ 𝑥𝑦)−(∑ 𝑥)(∑ 𝑦)
𝑛 ∑ 𝑦
2
−(∑ 𝑦)
2
𝑏𝑦𝑥 = -0.93 ; 𝑏𝑥𝑦 = -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:
𝑏𝑦𝑥 = 𝛾
𝜎𝑦
𝜎𝑥
; 𝑏𝑥𝑦 = 𝛾
𝜎𝑥
𝜎𝑦
𝑏𝑦𝑥 = 0.27 ; 𝑏𝑥𝑦 = 0.6
UNIT-4

8. 1. A coin is tossed 144 times and a person gets 80heads.Can we say that the
coin is unbiased one?
Solution: Set 𝐻0: the coin is unbiased.
Given n=144 , P = 0.5 and Q = 0.5,X=80
At 5 % level of significance, |𝑧𝛼| =1.96
Z =
𝑋−𝑛𝑝
√𝑛𝑝𝑞
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 𝐻0: 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
𝐻1 : P> 0.5(right tailed test)
At 5 % level of significance for right tailed test, |𝑧𝛼| =1.645
Z =
𝑝−𝑃
√
𝑃𝑄
𝑛
.Then By computation, z = 2.04
Calculated value =2.04>1.645=table value
The difference is significant.
____________________________________________________________

9. 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 𝐻0: 𝜇 = 𝑥̅
Given n=400 , 𝜇 = 6.48 and 𝑥̅ = 6.75, 𝜎 = 1.5
At 5 % level of significance, |𝑧𝛼| =1.96
Z =
𝑥̅−𝑛𝑝
𝜎
√𝑛
By computation, z = 3.6
Calculated value =3.6 >3
𝐻0 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 𝐻0: 𝜇 = 𝑥̅
Given n=400 , 𝜇 = 67.39 and 𝑥̅ = 67.47, 𝜎 = 1.3
At 5 % level of significance, |𝑧𝛼| =1.96
Z =
𝑥̅−𝑛𝑝
𝜎
√𝑛
By computation, z = 1.23
Calculated value |𝑧| =1.23 < 1.96
𝐻0 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

10. that the samples were drawn from the same district which has the s.d. of 11
kgs per acre?
Solution : Set 𝐻0: 𝜇1 = 𝜇2
Given 𝑛1= 100 , 𝑥̅ 1 = 210 , 𝜎 = 150, 𝑛2 = 150 , 𝑥̅ 2 =220
At 5 % level of significance, |𝑧𝛼| =1.96
Z =
𝑥̅ 1 − 𝑥̅ 2
𝜎√(
1
𝑛1
)+(
1
𝑛2
)
By computation, z = -7.04
Calculated value, |𝑧| =7.04 >3
𝐻0 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 𝐻0: 𝜇 = 𝑥̅
Calculating , 𝑥̅ =47,n =10 , 𝑠2
=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.
𝐻0 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

11. Solution: Set 𝐻0: 𝜇 = 𝑥̅
Calculating , 𝑥̅ = 97.2,n =10 , s = 14.27
Test statistic|𝑡| = 0.62
D.F=n-1=9
Table value =2.26
Calculated value =0.62
Calculated value < Table value.
𝐻0 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.
𝑋 = 50.5 S1
2
= 6.5 n1 = 15
𝑌 = 57.2 S2
2
= 5.7 n2 = 12
Solution: Set 𝐻0: 𝜎1
2
= 𝜎2
2
F=
𝑆1
2
𝑆2
2 = 1.120
Table value = 2.79
Calculated value < Table value.
𝐻0 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

12. From the above data, find out if there is any relation between literacy and
the habit of cinegores.
Solution : Set 𝐻0: there is no relation between literacy and
the habit of cinegores.
Calculation of Ψ2
:
)
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 (𝑂 − 𝐸)2
𝐸
83 70.83 2.09
57 69.17 2.14
45 57.17 2.59
68 55.83 2.65
9.47

13. Ψ
2
= ∑
(𝑂−𝐸)2
𝐸
=9.47
Table value = 3.841
Calculated value < Table value.
𝐻0 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: 𝑥1
̅̅̅ = 200 , 𝑥2
̅̅̅ = 250, 𝑠1 = 20, 𝑠2 = 25
Set 𝐻0: 𝜇1 = 𝜇2
D.F = 𝑛1 + 𝑛2 - 2 = 25+ 25 -2 = 48
t0.01=2.58
test statistic t =
𝑥̅ 1 − 𝑥̅ 2
𝑠√(
1
𝑛1
)+(
1
𝑛2
)
t = -7.65 and |𝑡| = 7.65
Calculated value > Table value.
𝐻0 is rejected.
The difference is significant.
_______________________________________________________________