Question 2 MCQ · 0.0 marks
Useful Data has been mentioned above.
This data attachment is just for a reference & not for an evaluation.
A published solution is not available for this question yet.
Statistics for Data Science II · Quiz 1 · May 2026
Statistics for Data Science II · Quiz 1 · May 2026
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Questions and published explanations below are available without starting a test. Some questions may not have a published solution yet.
16 / 19 published solutions
Question 3 NAT · 3.0 marks
Let X be a continuous random variable with PDF as follows:
Find the value of a.
Published solution
A probability density must integrate to \(1\) over its support:
\[\int_0^a 2(1-x)\,dx=\left[2x-x^2\right]_0^a=2a-a^2=1.\]
Thus \(a^2-2a+1=(a-1)^2=0\), so \(\boxed{a=1}\). The density is nonnegative on \((0,1)\), as required.
Question 4 NAT · 3.0 marks
Let
and
be two independent random variables such that
Let another random variable
be defined as
Find the value of
Enter the answer correct to two decimal places.
Published solution
Here \(f_{X\mid Z=1}(1)=P(X=1\mid X+Y=1)\). When \(Z=1\), the possible pairs are \((1,0)\) and \((0,1)\).
By independence,
\[P(X=1,Y=0)=0.2(0.6)=0.12,\qquad P(X=0,Y=1)=0.8(0.4)=0.32.\]
Therefore
\[P(X=1\mid Z=1)=\frac{0.12}{0.12+0.32}=\frac{3}{11}\approx\boxed{0.27}.\]
Question 5 NAT · 3.0 marks
A number from 00 to 99 is selected uniformly at random.The numbers 0 to 9 are treated as two
digit numbers and denoted as 00 to 09. Let,
•
be the digit in the units place,
•
be the remainder when the number is divided by 4.
Find the value of
. Enter the answer correct to two decimal places.
Published solution
The event \(X=4\) means that the selected number ends in \(4\). Of these numbers, the multiples of \(4\) are
\[04,\;24,\;44,\;64,\;84.\]
All \(100\) numbers from \(00\) to \(99\) are equally likely. Hence
\[P(X=4,Y=0)=\frac{5}{100}=\boxed{0.05}.\]
Question 6 NAT · 3.0 marks
A random variable
has mean 10 and variance 9. Using Chebyshev's inequality, find a lower
bound on
. Enter the answer correct to two decimal places.
Published solution
The interval is centred at the mean \(\mu=10\), with distance \(6\) on each side:
\[\{4\le X\le16\}=\{|X-10|\le6\}.\]
Chebyshev's inequality gives the guaranteed lower bound
\[P(|X-10|\le6)\ge1-\frac{\operatorname{Var}(X)}{6^2}=1-\frac9{36}=\boxed{0.75}.\]
This is a bound, not necessarily the exact probability.
Question 7 NAT · 3.0 marks
A shopping mall issues 400 lucky draw coupons. Among them, one coupon is worth a gift voucher
of Rs 2000, one coupon is worth a gift voucher of Rs 1000, and all other coupons are worth Rs 0. If
a customer receives 2 coupons drawn at random from the 400, what is the expected value (in Rs)
of the prizes won by the customer?
Published solution
Let \(I_{2000}\) and \(I_{1000}\) indicate whether the two winning coupons are among the customer's two coupons. Each particular coupon is selected with probability \(2/400\).
Using linearity of expectation,
\[E[\text{prize}]=2000E[I_{2000}]+1000E[I_{1000}]=2000\frac2{400}+1000\frac2{400}=\boxed{15\text{ rupees}}.\]
The coupons need not be independent for this calculation.
Question 8 MCQ · 3.0 marks
Let
be a discrete random variable with PMF
Based on the above data, answer the given subquestions.
Find the value of
Published solution
Apply the expectation formula to \(g(X)=(X-2)^2\):
\[E[(X-2)^2]=(0-2)^2\frac12+(2-2)^2\frac13+(4-2)^2\frac16.\]
Thus
\[E[(X-2)^2]=2+0+\frac23=\boxed{\frac83}.\]
A: Incorrect: \(4/3\) is \(E[X]\), whereas the question asks for \(E[(X-2)^2]=8/3\).
B: Incorrect: the squared deviations equal \(4,0,4\), but they must be weighted by their probabilities; their expectation is not \(4\).
C: Correct: \(4(1/2)+0(1/3)+4(1/6)=8/3\).
D: Incorrect: \(6\) does not equal the probability-weighted sum \(8/3\).
Question 9 NAT · 2.0 marks
Let
be a discrete random variable with PMF
Based on the above data, answer the given subquestions.
Using Markov's inequality, obtain an upper bound for
Enter the answer correct to two
decimal places.
Published solution
Since \(X\ge0\), Markov's inequality applies:
\[P(X\ge2)\le\frac{E[X]}2.\]
The mean is
\[E[X]=0\frac12+2\frac13+4\frac16=\frac43.\]
Therefore the Markov upper bound is
\[\frac{4/3}{2}=\frac23\approx\boxed{0.67}.\]
The exact probability is \(1/3+1/6=1/2\), but the requested answer is the Markov bound.
Question 10 MCQ · 2.0 marks
Let the random variables
i.i.d Uniform{1, 2, 3, 4}.
Based on the above data, answer the given subquestions.
What is the value of
?
Published solution
Independence and uniformity give \(16\) equally likely ordered pairs \((X,Y)\). The minimum is exactly \(3\) for
\[(3,3),\;(3,4),\;(4,3).\]
Hence
\[P(\min(X,Y)=3)=\frac3{16}.\]
Equivalently, \(P(X\ge3,Y\ge3)-P(X\ge4,Y\ge4)=4/16-1/16=\boxed{3/16}\).
A: Correct: exactly three of the sixteen ordered pairs have minimum \(3\).
B: Incorrect: \(1/16\) counts only one pair; both \((3,4)\) and \((4,3)\) also qualify along with \((3,3)\).
C: Incorrect: \(1/4=P(X\ge3,Y\ge3)\) also includes \((4,4)\), whose minimum is \(4\).
D: Incorrect: \(3/4\) is much larger than the required three out of sixteen pairs.
Question 11 NAT · 3.0 marks
Let the random variables
i.i.d Uniform{1, 2, 3, 4}.
Based on the above data, answer the given subquestions.
Let
find the value of
Enter the answer correct to three decimal
places.
Published solution
Use the complementary event. Since \(W=\min(X,Y)\), the event \(W>3\) requires both \(X=4\) and \(Y=4\).
\[P(W>3)=P(X=4)P(Y=4)=\frac14\cdot\frac14=\frac1{16}.\]
Therefore
\[P(W\le3)=1-\frac1{16}=\frac{15}{16}=0.9375\approx\boxed{0.938}.\]
Question 12 MCQ · 3.0 marks
Let
be i.i.d random variables with the following probability mass function
Based on the above data, answer the given subquestions.
Find the value of
.
A published solution is not available for this question yet.
Question 13 MCQ · 2.0 marks
Let
be i.i.d random variables with the following probability mass function
Based on the above data, answer the given subquestions.
Find the value of
.
A published solution is not available for this question yet.
Question 14 NAT · 3.0 marks
Let
be i.i.d samples of a discrete random variable
whose PMF is as follows:
Based on the above data, answer the given subquestions.
What is the probability that 1 appears exactly once in the samples? Enter the answer correct to two
decimal places.
Published solution
For each independent sample, \(P(X_i=1)=1/4\) and \(P(X_i\ne1)=3/4\). The number of ones in four samples is binomial with parameters \(n=4\) and \(p=1/4\).
Therefore
\[P(\text{exactly one }1)=\binom41\frac14\left(\frac34\right)^3=\frac{27}{64}=0.421875\approx\boxed{0.42}.\]
The factor \(\binom41=4\) chooses the position of the unique \(1\).
Question 15 NAT · 2.0 marks
Let
be i.i.d samples of a discrete random variable
whose PMF is as follows:
Based on the above data, answer the given subquestions.
What is the probability that none of the four samples is equal to 0? Enter the answer correct to two
decimal places.
Published solution
Each sample avoids \(0\) with probability
\[P(X_i\ne0)=1-P(X_i=0)=1-\frac12=\frac12.\]
Since the four samples are independent,
\[P(\text{none is }0)=\left(\frac12\right)^4=\frac1{16}=0.0625\approx\boxed{0.06}.\]
Question 16 NAT · 1.0 marks
Suppose that the time
(in minutes) that a customer waits at a service counter is a random
variable with probability density function
Let
be the event that a person has to wait for more than 2 minutes and
be the event that he
has to wait for more than 1 minute. Find the value of
.
Enter the correct option number for
.
Published solution
The events are \(U=\{X>2\}\) and \(V=\{X>1\}\). Whenever \(X>2\), we automatically have \(X>1\), so \(U\subset V\).
Hence
\[P(U\cap V)=P(X>2).\]
Blank \(A\) is therefore \(P(X>2)\). Enter \(\boxed{10}\), the option number accepted by the source key.
The printed list repeats this same expression as option \(1\); the required keyed numeric entry is \(10\).
Question 17 NAT · 1.0 marks
Suppose that the time
(in minutes) that a customer waits at a service counter is a random
variable with probability density function
Let
be the event that a person has to wait for more than 2 minutes and
be the event that he
has to wait for more than 1 minute. Find the value of
.
Enter the correct option number for
.
Published solution
From \(U\cap V=\{X>2\}\) and the support \(0\le X\le4\),
\[P(U\cap V)=P(X>2)=\int_2^4\frac{x}{8}\,dx.\]
Blank \(B\) supplies the integral bounds \(\int_2^4\), which is printed as option \(4\). Therefore enter \(\boxed{4}\).
Question 18 NAT · 1.0 marks
Suppose that the time
(in minutes) that a customer waits at a service counter is a random
variable with probability density function
Let
be the event that a person has to wait for more than 2 minutes and
be the event that he
has to wait for more than 1 minute. Find the value of
.
Enter the correct option number for
.
Published solution
Blank \(C\) is the numerator probability:
\[\int_2^4\frac{x}{8}\,dx=\left[\frac{x^2}{16}\right]_2^4=\frac{16-4}{16}=\frac34.\]
In the supplied list, \(3/4\) is option \(3\). The question asks for the option number, so enter \(\boxed{3}\), not \(0.75\).
Question 19 NAT · 1.0 marks
Suppose that the time
(in minutes) that a customer waits at a service counter is a random
variable with probability density function
Let
be the event that a person has to wait for more than 2 minutes and
be the event that he
has to wait for more than 1 minute. Find the value of
.
Enter the correct option number for
.
Published solution
Blank \(D\) is the denominator \(P(V)=P(X>1)\):
\[P(V)=\int_1^4\frac{x}{8}\,dx=\left[\frac{x^2}{16}\right]_1^4=\frac{16-1}{16}=\frac{15}{16}.\]
The value \(15/16\) is option \(6\). Since an option number is requested, enter \(\boxed{6}\).
Question 20 NAT · 1.0 marks
Suppose that the time
(in minutes) that a customer waits at a service counter is a random
variable with probability density function
Let
be the event that a person has to wait for more than 2 minutes and
be the event that he
has to wait for more than 1 minute. Find the value of
.
Calculate the value for
. Enter the answer correct to one decimal place.
Published solution
Use the conditional probability formula and \(U\subset V\):
\[P(U\mid V)=\frac{P(U\cap V)}{P(V)}=\frac{P(X>2)}{P(X>1)}.\]
The numerator is \((16-4)/16=3/4\), and the denominator is \((16-1)/16=15/16\). Therefore
\[E=P(U\mid V)=\frac{3/4}{15/16}=\frac45=\boxed{0.8}.\]
Here the question asks for the numerical value of blank \(E\), not an option number.