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Week 3 Statistics 2 Questions | Prasnya
Statistics 2 > Week 3
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Type:
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Difficulty:
All
Year:
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30Q
01
A random variable
X
X
X
has the moment generating function
M
X
(
λ
)
=
e
λ
/
10
−
1
λ
/
10
,
λ
≠
0.
M_X(\lambda)=\dfrac{e^{\lambda/10}-1}{\lambda/10},\quad \lambda\ne 0.
M
X
(
λ
)
=
λ
/10
e
λ
/10
−
1
,
λ
=
0.
Find the variance of
X
X
X
. Enter the answer correct to three decimal places.
Numerical
MEDIUM
3 marks
06 April 2026
02
Using Chebyshev's inequality, find the smallest integer value
n
n
n
such that
P
(
∣
X
ˉ
n
−
μ
∣
≥
0.1
)
≤
0.2
P(|\bar{X}_n-\mu|\ge 0.1)\le 0.2
P
(
∣
X
ˉ
n
−
μ
∣
≥
0.1
)
≤
0.2
, where
σ
2
=
1
\sigma^2=1
σ
2
=
1
.
Numerical
MEDIUM
3 marks
06 April 2026
03
Suppose
X
1
,
X
2
,
X
3
∼
X_1,X_2,X_3\sim
X
1
,
X
2
,
X
3
∼
i.i.d.
X
X
X
such that
E
[
X
]
=
10
E[X]=10
E
[
X
]
=
10
and
Var
(
X
)
=
4
\operatorname{Var}(X)=4
Var
(
X
)
=
4
. Define
Y
=
X
1
−
X
2
+
X
3
Y=X_1-X_2+X_3
Y
=
X
1
−
X
2
+
X
3
and
Z
=
2
Y
Z=2Y
Z
=
2
Y
. Choose the correct option(s).
Multiple correct
EASY
3 marks
03 August 2025
04
At a customer support center, the number of service requests received in a day is modeled by a Poisson random variable
X
X
X
with mean
λ
=
18
\lambda=18
λ
=
18
. The total time spent processing requests and total words written are modeled as
T
=
7
X
+
12
T=7X+12
T
=
7
X
+
12
and
W
=
4
X
−
1
W=4X-1
W
=
4
X
−
1
. Find
Cov
(
T
,
W
)
\operatorname{Cov}(T,W)
Cov
(
T
,
W
)
.
Numerical
EASY
3 marks
03 August 2025
05
If
Y
=
X
ˉ
1
−
X
ˉ
2
Y=\bar X_1-\bar X_2
Y
=
X
ˉ
1
−
X
ˉ
2
, use Chebyshev’s inequality to find a lower bound for
P
(
∣
Y
∣
<
18
)
P(|Y|<18)
P
(
∣
Y
∣
<
18
)
. Enter the answer correct to two decimal places.
Comprehension
MEDIUM
2 marks
03 August 2025
06
Let
X
∼
B
e
r
n
o
u
l
l
i
(
0.3
)
X\sim\mathrm{Bernoulli}(0.3)
X
∼
Bernoulli
(
0.3
)
. Find the MGF of the centered version of
X
X
X
.
Single correct
EASY
3 marks
16 March 2025
07
Using Chebyshev’s inequality, find a lower bound for
P
(
∣
Y
−
0.5
∣
<
0.6
)
P(|Y-0.5|<0.6)
P
(
∣
Y
−
0.5∣
<
0.6
)
. Enter the answer correct to three decimal places.
Comprehension
MEDIUM
3 marks
16 March 2025
08
Find
E
[
Y
]
E\left[Y\right]
E
[
Y
]
for
Y
=
∑
i
=
1
4
X
i
+
2
∑
i
=
16
25
X
i
Y=\sum_{i=1}^{4}X_i+2\sum_{i=16}^{25}X_i
Y
=
∑
i
=
1
4
X
i
+
2
∑
i
=
16
25
X
i
.
Comprehension
MEDIUM
2 marks
01 December 2024
09
Using Chebyshev’s inequality, find an upper bound for
P
(
∣
X
ˉ
−
0.5
∣
>
0.25
)
P(|\bar X-0.5|>0.25)
P
(
∣
X
ˉ
−
0.5∣
>
0.25
)
, where
X
ˉ
=
(
X
1
+
⋯
+
X
16
)
/
16
\bar X=(X_1+\cdots+X_{16})/16
X
ˉ
=
(
X
1
+
⋯
+
X
16
)
/16
. Enter the answer correct to two decimal places.
Comprehension
MEDIUM
2 marks
01 December 2024
10
If
Cov
(
X
,
Y
)
=
0
\operatorname{Cov}(X,Y)=0
Cov
(
X
,
Y
)
=
0
, find
E
[
X
Y
]
E[XY]
E
[
X
Y
]
.
Comprehension
MEDIUM
3 marks
01 December 2024
11
Which option(s) is/are correct?
Comprehension
MEDIUM
1 marks
01 December 2024
12
Let
X
1
X_1
X
1
and
X
2
X_2
X
2
be i.i.d., where
X
X
X
has PMF
P
(
X
=
0
)
=
0.2
P(X=0)=0.2
P
(
X
=
0
)
=
0.2
,
P
(
X
=
1
)
=
0.4
P(X=1)=0.4
P
(
X
=
1
)
=
0.4
,
P
(
X
=
2
)
=
0.4
P(X=2)=0.4
P
(
X
=
2
)
=
0.4
. Define
Y
=
X
1
+
X
2
Y=X_1+X_2
Y
=
X
1
+
X
2
. Find the MGF of
Y
Y
Y
.
Single correct
EASY
3 marks
04 August 2024
13
Consider a random variable
X
X
X
with
E
[
X
]
=
1
E[X]=1
E
[
X
]
=
1
,
E
[
X
2
]
=
0
E[X^2]=0
E
[
X
2
]
=
0
, and
E
[
X
3
]
=
2
E[X^3]=2
E
[
X
3
]
=
2
. Define
Y
=
−
1
+
X
+
3
X
2
Y=-1+X+3X^2
Y
=
−
1
+
X
+
3
X
2
. Find
Cov
(
X
,
Y
)
\operatorname{Cov}(X,Y)
Cov
(
X
,
Y
)
.
Numerical
MEDIUM
3 marks
04 August 2024
14
Using Chebyshev’s inequality, find the least upper bound on the probability that the sample mean deviates from the population mean by more than
0.5
0.5
0.5
hours.
Comprehension
MEDIUM
3 marks
04 August 2024
15
Suppose
X
X
X
and
Y
Y
Y
are two independent random variables with probability density functions
f
(
x
)
=
8
x
3
f(x)=\frac{8}{x^3}
f
(
x
)
=
x
3
8
for
x
>
2
x>2
x
>
2
and
0
0
0
otherwise, and
g
(
y
)
=
2
y
g(y)=2y
g
(
y
)
=
2
y
for
0
<
y
<
1
0<y<1
0
<
y
<
1
and
0
0
0
otherwise. Calculate the value of
E
[
X
Y
]
E[XY]
E
[
X
Y
]
.
Single correct
MEDIUM
3 marks
24 March 2024
16
Which of the following inequalities result from Chebyshev's inequality?
Comprehension
MEDIUM
3 marks
24 March 2024
17
Find the minimum value of
n
n
n
such that the sample mean lies in
[
μ
−
5
,
μ
+
5
]
[\mu-5,\mu+5]
[
μ
−
5
,
μ
+
5
]
with probability more than
0.95
0.95
0.95
using Chebyshev's inequality.
Comprehension
MEDIUM
2 marks
24 March 2024
18
Find the value of
C
o
v
(
X
,
Y
)
\mathrm{Cov}(X,Y)
Cov
(
X
,
Y
)
.
Comprehension
MEDIUM
3 marks
24 March 2024
19
What conclusion will you make based on the obtained value in the given part?
Comprehension
EASY
2 marks
24 March 2024
20
Which of the following inequalities are true with respect to Chebyshev inequality?
Comprehension
MEDIUM
3 marks
03 December 2024
21
Find the minimum value of
n
n
n
such that the sample mean lies in
[
2.5
,
3.5
]
[2.5,3.5]
[
2.5
,
3.5
]
with probability more than
0.95
0.95
0.95
using Chebyshev inequality.
Comprehension
MEDIUM
3 marks
03 December 2024
22
Suppose
X
X
X
is a discrete random variable and has moment generating function
M
X
(
t
)
=
1
7
+
3
7
e
2
t
+
2
7
e
4
t
+
1
7
e
6
t
M_X(t)=\frac{1}{7}+\frac{3}{7}e^{2t}+\frac{2}{7}e^{4t}+\frac{1}{7}e^{6t}
M
X
(
t
)
=
7
1
+
7
3
e
2
t
+
7
2
e
4
t
+
7
1
e
6
t
. What is the PMF of
X
X
X
?
Single correct
EASY
3 marks
06 August 2023
23
Suppose
X
1
,
X
2
,
X
3
,
X
4
∼
i.i.d.
X
X_1,X_2,X_3,X_4\sim \text{i.i.d. }X
X
1
,
X
2
,
X
3
,
X
4
∼
i.i.d.
X
with
E
[
X
]
=
10
E[X]=10
E
[
X
]
=
10
and
Var
(
X
)
=
4
\operatorname{Var}(X)=4
Var
(
X
)
=
4
. Define
S
=
2
X
1
−
2
X
2
−
X
3
+
3
X
4
S=2X_1-2X_2-X_3+3X_4
S
=
2
X
1
−
2
X
2
−
X
3
+
3
X
4
. Choose the correct option(s) from below:
Multiple correct
MEDIUM
3 marks
06 August 2023
24
Find the expected value of
Y
=
∑
i
=
1
20
X
i
+
∑
i
=
11
20
X
i
Y=\sum_{i=1}^{20}X_i+\sum_{i=11}^{20}X_i
Y
=
∑
i
=
1
20
X
i
+
∑
i
=
11
20
X
i
.
Comprehension
EASY
2 marks
06 August 2023
25
Using Chebyshev's inequality, find an upper bound for
P
(
∣
X
ˉ
−
10
∣
>
2
)
P(|\bar{X}-10|>2)
P
(
∣
X
ˉ
−
10∣
>
2
)
, where
X
ˉ
=
(
X
1
+
X
2
+
⋯
+
X
20
)
/
20
\bar{X}=(X_1+X_2+\cdots+X_{20})/20
X
ˉ
=
(
X
1
+
X
2
+
⋯
+
X
20
)
/20
is the sample mean. Enter the answer correct to 3 decimal places.
Comprehension
MEDIUM
3 marks
06 August 2023
26
Let
X
1
,
X
2
,
…
,
X
n
X_1,X_2,\ldots,X_n
X
1
,
X
2
,
…
,
X
n
be i.i.d.
X
X
X
with mean
μ
=
0
\mu=0
μ
=
0
and variance
σ
2
=
1
\sigma^2=1
σ
2
=
1
. Using Chebyshev's inequality, what should be the minimum value of
n
n
n
such that the probability that the sample mean
X
1
+
X
2
+
⋯
+
X
n
n
\frac{X_1+X_2+\cdots+X_n}{n}
n
X
1
+
X
2
+
⋯
+
X
n
lies between
−
0.5
-0.5
−
0.5
and
0.5
0.5
0.5
is at least
0.95
0.95
0.95
?
Single correct
MEDIUM
3 marks
02 April 2023
27
Suppose
X
1
,
X
2
,
X
3
,
X
4
X_1,X_2,X_3,X_4
X
1
,
X
2
,
X
3
,
X
4
are i.i.d. Bernoulli
(
2
/
3
)
(2/3)
(
2/3
)
. Define
Y
=
2
X
1
+
3
X
2
+
4
X
3
+
5
X
4
Y=2X_1+3X_2+4X_3+5X_4
Y
=
2
X
1
+
3
X
2
+
4
X
3
+
5
X
4
. Find
V
a
r
(
Y
)
\mathrm{Var}(Y)
Var
(
Y
)
.
Numerical
MEDIUM
3 marks
02 April 2023
28
Let
X
1
,
X
2
,
…
,
X
n
X_1,X_2,\ldots,X_n
X
1
,
X
2
,
…
,
X
n
be i.i.d. Poisson
(
9
)
(9)
(
9
)
. Using Chebyshev's inequality, what should be the minimum value of
n
n
n
such that the probability that the sample mean
X
ˉ
\bar X
X
ˉ
lies between
8.6
8.6
8.6
and
9.4
9.4
9.4
is at least
0.95
0.95
0.95
?
Numerical
MEDIUM
3 marks
20 November 2022
29
Find the moment generating function of the random variable
Y
Y
Y
.
Comprehension
MEDIUM
3 marks
20 November 2022
30
Find the expected value of
Y
Y
Y
. Enter the answer correct to two decimal places.
Comprehension
EASY
2 marks
20 November 2022
Showing 30 questions.
Statistics 2 > Week 3
All PYQs
Topic-Wise PYQs
Start Weekly Test
Mock tests
Week mock
Topic mock
Subject mock
Type:
All
Difficulty:
All
Year:
All
30Q
01
A random variable
X
X
X
has the moment generating function
M
X
(
λ
)
=
e
λ
/
10
−
1
λ
/
10
,
λ
≠
0.
M_X(\lambda)=\dfrac{e^{\lambda/10}-1}{\lambda/10},\quad \lambda\ne 0.
M
X
(
λ
)
=
λ
/10
e
λ
/10
−
1
,
λ
=
0.
Find the variance of
X
X
X
. Enter the answer correct to three decimal places.
Numerical
MEDIUM
3 marks
06 April 2026
02
Using Chebyshev's inequality, find the smallest integer value
n
n
n
such that
P
(
∣
X
ˉ
n
−
μ
∣
≥
0.1
)
≤
0.2
P(|\bar{X}_n-\mu|\ge 0.1)\le 0.2
P
(
∣
X
ˉ
n
−
μ
∣
≥
0.1
)
≤
0.2
, where
σ
2
=
1
\sigma^2=1
σ
2
=
1
.
Numerical
MEDIUM
3 marks
06 April 2026
03
Suppose
X
1
,
X
2
,
X
3
∼
X_1,X_2,X_3\sim
X
1
,
X
2
,
X
3
∼
i.i.d.
X
X
X
such that
E
[
X
]
=
10
E[X]=10
E
[
X
]
=
10
and
Var
(
X
)
=
4
\operatorname{Var}(X)=4
Var
(
X
)
=
4
. Define
Y
=
X
1
−
X
2
+
X
3
Y=X_1-X_2+X_3
Y
=
X
1
−
X
2
+
X
3
and
Z
=
2
Y
Z=2Y
Z
=
2
Y
. Choose the correct option(s).
Multiple correct
EASY
3 marks
03 August 2025
04
At a customer support center, the number of service requests received in a day is modeled by a Poisson random variable
X
X
X
with mean
λ
=
18
\lambda=18
λ
=
18
. The total time spent processing requests and total words written are modeled as
T
=
7
X
+
12
T=7X+12
T
=
7
X
+
12
and
W
=
4
X
−
1
W=4X-1
W
=
4
X
−
1
. Find
Cov
(
T
,
W
)
\operatorname{Cov}(T,W)
Cov
(
T
,
W
)
.
Numerical
EASY
3 marks
03 August 2025
05
If
Y
=
X
ˉ
1
−
X
ˉ
2
Y=\bar X_1-\bar X_2
Y
=
X
ˉ
1
−
X
ˉ
2
, use Chebyshev’s inequality to find a lower bound for
P
(
∣
Y
∣
<
18
)
P(|Y|<18)
P
(
∣
Y
∣
<
18
)
. Enter the answer correct to two decimal places.
Comprehension
MEDIUM
2 marks
03 August 2025
06
Let
X
∼
B
e
r
n
o
u
l
l
i
(
0.3
)
X\sim\mathrm{Bernoulli}(0.3)
X
∼
Bernoulli
(
0.3
)
. Find the MGF of the centered version of
X
X
X
.
Single correct
EASY
3 marks
16 March 2025
07
Using Chebyshev’s inequality, find a lower bound for
P
(
∣
Y
−
0.5
∣
<
0.6
)
P(|Y-0.5|<0.6)
P
(
∣
Y
−
0.5∣
<
0.6
)
. Enter the answer correct to three decimal places.
Comprehension
MEDIUM
3 marks
16 March 2025
08
Find
E
[
Y
]
E\left[Y\right]
E
[
Y
]
for
Y
=
∑
i
=
1
4
X
i
+
2
∑
i
=
16
25
X
i
Y=\sum_{i=1}^{4}X_i+2\sum_{i=16}^{25}X_i
Y
=
∑
i
=
1
4
X
i
+
2
∑
i
=
16
25
X
i
.
Comprehension
MEDIUM
2 marks
01 December 2024
09
Using Chebyshev’s inequality, find an upper bound for
P
(
∣
X
ˉ
−
0.5
∣
>
0.25
)
P(|\bar X-0.5|>0.25)
P
(
∣
X
ˉ
−
0.5∣
>
0.25
)
, where
X
ˉ
=
(
X
1
+
⋯
+
X
16
)
/
16
\bar X=(X_1+\cdots+X_{16})/16
X
ˉ
=
(
X
1
+
⋯
+
X
16
)
/16
. Enter the answer correct to two decimal places.
Comprehension
MEDIUM
2 marks
01 December 2024
10
If
Cov
(
X
,
Y
)
=
0
\operatorname{Cov}(X,Y)=0
Cov
(
X
,
Y
)
=
0
, find
E
[
X
Y
]
E[XY]
E
[
X
Y
]
.
Comprehension
MEDIUM
3 marks
01 December 2024
11
Which option(s) is/are correct?
Comprehension
MEDIUM
1 marks
01 December 2024
12
Let
X
1
X_1
X
1
and
X
2
X_2
X
2
be i.i.d., where
X
X
X
has PMF
P
(
X
=
0
)
=
0.2
P(X=0)=0.2
P
(
X
=
0
)
=
0.2
,
P
(
X
=
1
)
=
0.4
P(X=1)=0.4
P
(
X
=
1
)
=
0.4
,
P
(
X
=
2
)
=
0.4
P(X=2)=0.4
P
(
X
=
2
)
=
0.4
. Define
Y
=
X
1
+
X
2
Y=X_1+X_2
Y
=
X
1
+
X
2
. Find the MGF of
Y
Y
Y
.
Single correct
EASY
3 marks
04 August 2024
13
Consider a random variable
X
X
X
with
E
[
X
]
=
1
E[X]=1
E
[
X
]
=
1
,
E
[
X
2
]
=
0
E[X^2]=0
E
[
X
2
]
=
0
, and
E
[
X
3
]
=
2
E[X^3]=2
E
[
X
3
]
=
2
. Define
Y
=
−
1
+
X
+
3
X
2
Y=-1+X+3X^2
Y
=
−
1
+
X
+
3
X
2
. Find
Cov
(
X
,
Y
)
\operatorname{Cov}(X,Y)
Cov
(
X
,
Y
)
.
Numerical
MEDIUM
3 marks
04 August 2024
14
Using Chebyshev’s inequality, find the least upper bound on the probability that the sample mean deviates from the population mean by more than
0.5
0.5
0.5
hours.
Comprehension
MEDIUM
3 marks
04 August 2024
15
Suppose
X
X
X
and
Y
Y
Y
are two independent random variables with probability density functions
f
(
x
)
=
8
x
3
f(x)=\frac{8}{x^3}
f
(
x
)
=
x
3
8
for
x
>
2
x>2
x
>
2
and
0
0
0
otherwise, and
g
(
y
)
=
2
y
g(y)=2y
g
(
y
)
=
2
y
for
0
<
y
<
1
0<y<1
0
<
y
<
1
and
0
0
0
otherwise. Calculate the value of
E
[
X
Y
]
E[XY]
E
[
X
Y
]
.
Single correct
MEDIUM
3 marks
24 March 2024
16
Which of the following inequalities result from Chebyshev's inequality?
Comprehension
MEDIUM
3 marks
24 March 2024
17
Find the minimum value of
n
n
n
such that the sample mean lies in
[
μ
−
5
,
μ
+
5
]
[\mu-5,\mu+5]
[
μ
−
5
,
μ
+
5
]
with probability more than
0.95
0.95
0.95
using Chebyshev's inequality.
Comprehension
MEDIUM
2 marks
24 March 2024
18
Find the value of
C
o
v
(
X
,
Y
)
\mathrm{Cov}(X,Y)
Cov
(
X
,
Y
)
.
Comprehension
MEDIUM
3 marks
24 March 2024
19
What conclusion will you make based on the obtained value in the given part?
Comprehension
EASY
2 marks
24 March 2024
20
Which of the following inequalities are true with respect to Chebyshev inequality?
Comprehension
MEDIUM
3 marks
03 December 2024
21
Find the minimum value of
n
n
n
such that the sample mean lies in
[
2.5
,
3.5
]
[2.5,3.5]
[
2.5
,
3.5
]
with probability more than
0.95
0.95
0.95
using Chebyshev inequality.
Comprehension
MEDIUM
3 marks
03 December 2024
22
Suppose
X
X
X
is a discrete random variable and has moment generating function
M
X
(
t
)
=
1
7
+
3
7
e
2
t
+
2
7
e
4
t
+
1
7
e
6
t
M_X(t)=\frac{1}{7}+\frac{3}{7}e^{2t}+\frac{2}{7}e^{4t}+\frac{1}{7}e^{6t}
M
X
(
t
)
=
7
1
+
7
3
e
2
t
+
7
2
e
4
t
+
7
1
e
6
t
. What is the PMF of
X
X
X
?
Single correct
EASY
3 marks
06 August 2023
23
Suppose
X
1
,
X
2
,
X
3
,
X
4
∼
i.i.d.
X
X_1,X_2,X_3,X_4\sim \text{i.i.d. }X
X
1
,
X
2
,
X
3
,
X
4
∼
i.i.d.
X
with
E
[
X
]
=
10
E[X]=10
E
[
X
]
=
10
and
Var
(
X
)
=
4
\operatorname{Var}(X)=4
Var
(
X
)
=
4
. Define
S
=
2
X
1
−
2
X
2
−
X
3
+
3
X
4
S=2X_1-2X_2-X_3+3X_4
S
=
2
X
1
−
2
X
2
−
X
3
+
3
X
4
. Choose the correct option(s) from below:
Multiple correct
MEDIUM
3 marks
06 August 2023
24
Find the expected value of
Y
=
∑
i
=
1
20
X
i
+
∑
i
=
11
20
X
i
Y=\sum_{i=1}^{20}X_i+\sum_{i=11}^{20}X_i
Y
=
∑
i
=
1
20
X
i
+
∑
i
=
11
20
X
i
.
Comprehension
EASY
2 marks
06 August 2023
25
Using Chebyshev's inequality, find an upper bound for
P
(
∣
X
ˉ
−
10
∣
>
2
)
P(|\bar{X}-10|>2)
P
(
∣
X
ˉ
−
10∣
>
2
)
, where
X
ˉ
=
(
X
1
+
X
2
+
⋯
+
X
20
)
/
20
\bar{X}=(X_1+X_2+\cdots+X_{20})/20
X
ˉ
=
(
X
1
+
X
2
+
⋯
+
X
20
)
/20
is the sample mean. Enter the answer correct to 3 decimal places.
Comprehension
MEDIUM
3 marks
06 August 2023
26
Let
X
1
,
X
2
,
…
,
X
n
X_1,X_2,\ldots,X_n
X
1
,
X
2
,
…
,
X
n
be i.i.d.
X
X
X
with mean
μ
=
0
\mu=0
μ
=
0
and variance
σ
2
=
1
\sigma^2=1
σ
2
=
1
. Using Chebyshev's inequality, what should be the minimum value of
n
n
n
such that the probability that the sample mean
X
1
+
X
2
+
⋯
+
X
n
n
\frac{X_1+X_2+\cdots+X_n}{n}
n
X
1
+
X
2
+
⋯
+
X
n
lies between
−
0.5
-0.5
−
0.5
and
0.5
0.5
0.5
is at least
0.95
0.95
0.95
?
Single correct
MEDIUM
3 marks
02 April 2023
27
Suppose
X
1
,
X
2
,
X
3
,
X
4
X_1,X_2,X_3,X_4
X
1
,
X
2
,
X
3
,
X
4
are i.i.d. Bernoulli
(
2
/
3
)
(2/3)
(
2/3
)
. Define
Y
=
2
X
1
+
3
X
2
+
4
X
3
+
5
X
4
Y=2X_1+3X_2+4X_3+5X_4
Y
=
2
X
1
+
3
X
2
+
4
X
3
+
5
X
4
. Find
V
a
r
(
Y
)
\mathrm{Var}(Y)
Var
(
Y
)
.
Numerical
MEDIUM
3 marks
02 April 2023
28
Let
X
1
,
X
2
,
…
,
X
n
X_1,X_2,\ldots,X_n
X
1
,
X
2
,
…
,
X
n
be i.i.d. Poisson
(
9
)
(9)
(
9
)
. Using Chebyshev's inequality, what should be the minimum value of
n
n
n
such that the probability that the sample mean
X
ˉ
\bar X
X
ˉ
lies between
8.6
8.6
8.6
and
9.4
9.4
9.4
is at least
0.95
0.95
0.95
?
Numerical
MEDIUM
3 marks
20 November 2022
29
Find the moment generating function of the random variable
Y
Y
Y
.
Comprehension
MEDIUM
3 marks
20 November 2022
30
Find the expected value of
Y
Y
Y
. Enter the answer correct to two decimal places.
Comprehension
EASY
2 marks
20 November 2022
Showing 30 questions.
