CMI Data Science 2024 Question Paper with Solutions: 26 Questions, Answer Key & Section-wise Analysis

CMI Data Science 2024 previous year paper: 26 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

26 questions · 59 marks · 2.065 minutes. School Level Mathematics: 15 · Programming: 4 · Discrete Mathematics: 4 · Probability Theory: 3

Free sample questions from CMI Data Science 2024 Question Paper

Question 1 · School Level Mathematics · 2024 MSQ

Consider the following positive integers:

$A = 1234679580, B = 1234789560.$

Which of the following statements are true?

A.
Both $A$ and $B$ are divisible by 36.
B.
Both $A$ and $B$ are divisible by 72.
C.
Only $B$ is divisible by 120.
D.
The number $A$ is divisible by 80.

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: Apply divisibility tests for composite numbers by checking their coprime factors.

Given: $A = 1234679580$ , $B = 1234789560$ .

Step 1: Check 36 ( $4 \times 9$ ).

Div by 4: Last 2 digits.

$A$ : 80 (Yes). $B$ : 60 (Yes).

Div by 9: Sum of digits.

$A$ : Sum=45 (Yes). $B$ : Sum=45 (Yes).

Both div by 36. Option A is True.

Step 2: Check 72 ( $8 \times 9$ ).

Div by 8: Last 3 digits.

$A$ : 580. $580/8 = 72.5$ (No).

$B$ : 560. $560/8 = 70$ (Yes).

$A$ not div by 72. Option B is False.

Step 3: Check 120 ( $8 \times 3 \times 5$ ).

Div by 5: Ends in 0 (Both Yes).
Div by 3: Sum 45 (Both Yes).
Div by 8: $A$ No, $B$ Yes.

Only $B$ div by 120. Option C is True.

Step 4: Check 80 ( $16 \times 5$ or $8 \times 10$ ).

Div by 8: $A$ No.

$A$ not div by 80. Option D is False.

Answer: ["A", "C"]

Question 2 · School Level Mathematics · 2024 MSQ

Which of the following statements are true?

A.
Let $f (x) = x^{3} + x - 1$ for $x \in R$ . Then the equation $f (x) = 0$ has at least one root in $[- 1, 0)$ .
B.
Let $f (x) = x^{3} + x - 1$ for $x \in R$ . Then the equation $f (x) = 0$ has at least one root in $[0, 1)$ .
C.
Let $f (x) = x^{3} + x + 1$ for $x \in R$ . Then the equation $f (x) = 0$ has at least one root in $[0, 1)$ .
D.
Let $f (x) = x^{3} + x + 1$ for $x \in R$ . Then the equation $f (x) = 0$ has at least one root in $[- 1, 0)$ .

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This is an application of the Intermediate Value Theorem (specifically Bolzano's Sign-Change Corollary). To prove a continuous function $f (x)$ has a root in an interval $[a, b]$ , we must show that $f (a)$ and $f (b)$ have opposite signs.

Step 1: Analyze Option A. $f (x) = x^{3} + x - 1$ . Evaluate at endpoints of $[- 1, 0)$ . $f (- 1) = (- 1)^{3} + (- 1) - 1 = - 3$ . $f (0) = - 1$ . Both are negative. No sign change, so IVT does not guarantee a root here.

Step 2: Analyze Option B. Same function $f (x) = x^{3} + x - 1$ . Evaluate at endpoints of $[0, 1)$ . $f (0) = - 1$ . $f (1) = 1^{3} + 1 - 1 = 1$ . The signs are negative and positive. By IVT, there is at least one root in $(0, 1)$ . True.

Step 3: Analyze Option C. $f (x) = x^{3} + x + 1$ . Evaluate at endpoints of $[0, 1)$ . $f (0) = 1$ . $f (1) = 1^{3} + 1 + 1 = 3$ . Both are positive. No sign change. False.

Step 4: Analyze Option D. Same function $f (x) = x^{3} + x + 1$ . Evaluate at endpoints of $[- 1, 0)$ . $f (- 1) = (- 1)^{3} + (- 1) + 1 = - 1$ . $f (0) = 1$ . The signs are negative and positive. By IVT, there is at least one root in $(- 1, 0)$ . True.

Answer: B, D

Question 3 · School Level Mathematics · 2024 SUB

Find values of $x$ and $y$ that satisfy both the following equations:

$\frac{x}{y} + \frac{y}{x} = \frac{5}{2}$

$\frac{x}{y} + \frac{y}{x} = \frac{9}{2} .$

Correct Answer:

none

Step-by-Step Solution

Insight: The system is symmetric in $x$ and $y$ , and the radical ratios $x / y$ and $y / x$ are reciprocals, so a single positive ratio variable reduces the first equation to a quadratic and the second to a linear equation in $y$ .

Exam route: Let $t = x / y > 0$ . Then $y / x = 1/ t$ . Eq 1 gives $t + 1/ t = 5/2 ⟹ 2 t^{2} - 5 t + 2 = 0 ⟹ t = 2$ or $t = 1/2$ . These give $x = 4 y$ or $y = 4 x$ . Substituting into Eq 2 gives $(x, y) = (4, 1)$ or $(1, 4)$ .

Learning route:

This is a symmetric reciprocal-radical system, recognisable because swapping $x$ and $y$ leaves both equations unchanged, and the terms involve ratios like $x / y$ and $y / x$ .

Step 1 — Domain: The expressions $x / y$ , $y / x$ , $x / y$ , $y / x$ all require $x > 0$ and $y > 0$ .

Step 2 — Define the ratio variable: $t = x / y$ . Then $t > 0$ and $y / x = 1/ t$ .

Step 3 — Solve the first equation: $t + 1/ t = 5/2$ . Multiply both sides by $2 t$ :

$2 t^{2} + 2 = 5 t ⟹ 2 t^{2} - 5 t + 2 = 0 ⟹ (2 t - 1) (t - 2) = 0$ .

Thus $t = 2$ or $t = 1/2$ .

Step 4 — Case $t = 2$ : $x / y = 2 ⟹ x = 4 y$ . Substitute into the second equation:

$\frac{4 y}{y} + \frac{y}{4 y} = 4 y + \frac{y}{2} = \frac{9}{2} y$ .

Since this equals $9/2$ , we get $y = 1 ⟹ y = 1$ , and $x = 4$ .

Step 5 — Case $t = 1/2$ : $x / y = 1/2 ⟹ y = 4 x$ . By symmetry, this gives $x = 1, y = 4$ .

The solutions are $(4, 1)$ and $(1, 4)$ .

Verification: For $(4, 1)$ , $4 + 1/4 = 2 + 0.5 = 2.5 = 5/2$ . And $4/1 + 1/2 = 4.5 = 9/2$ . Both hold. By symmetry, $(1, 4)$ also holds.

Question 4 · Programming · 2024 MSQ

In the following code, A is an array indexed from 0 whose elements are all positive integers, and n is the number of elements in A. It is given that n is at least 2. The operator * denotes multiplication.

function foo(A,n) {

if A[0] > A[1] {

first = A[0];

second = A[1];

} else {

first = A[1];

second = A[0];

}

for i from 2 to (n-1) {

if A[i] > second {

if A[i] > first {

second = first;

first = A[i];

} else {

second = A[i];

}

return(first * second);

}

If $A = [15, 7, 16, 12, 17, 14, 16, 4, 13, 12]$ , what will foo(A, 10) return?

A.
28
B.
144
C.
180
D.
272

Correct Answer:

["D"]

Step-by-Step Solution

Insight: The algorithm maintains the largest and second-largest elements in a single pass, updating them dynamically as larger values are encountered.

Exam route:

Step 1: Initialize using the first two elements. A[0]=15, A[1]=7. Since 15 > 7, first = 15, second = 7.

Step 2: Trace the loop from i=2 to 9:

i=2, A[2]=16: 16 > 7 (second). 16 > 15 (first). Update: second = 15, first = 16.
i=3, A[3]=12: 12 > 15 (second) is False. No change.
i=4, A[4]=17: 17 > 15 (second). 17 > 16 (first). Update: second = 16, first = 17.
i=5, A[5]=14: 14 > 16 is False. No change.
i=6, A[6]=16: 16 > 16 is False. No change.
i=7, A[7]=4: 4 > 16 is False. No change.
i=8, A[8]=13: 13 > 16 is False. No change.
i=9, A[9]=12: 12 > 16 is False. No change.

Step 3: Final values are first = 17, second = 16.

Step 4: Return first second = 17 16 = 272.

Learning route: This is a classic one-pass state maintenance pattern. By carefully initializing with the first two elements, we avoid arbitrary initial values (like 0 or -infinity). The nested if condition ensures that if a new element beats the absolute best (first), the old best gracefully steps down to second. If it only beats second, it takes that spot without disturbing first.

Answer: 272

Question 5 · Programming · 2024 MSQ

In the following code, A is an array indexed from 0 whose elements are all positive integers, and n is the number of elements in A.

function foo(A,n) {

max = 0;

curr = 0;

for i from 1 to (n-1) {

if A[i] > A[i-1] {

curr = curr + 1;

if curr > max {

max = curr;

}

} else {

curr = 0;

}

return(max+1);

}

If $A = [1, 3, 5, 2, 4, 7, 6, 8]$ , what will foo(A, 8) return?

A.
2
B.
3
C.
4
D.
5

Correct Answer:

["B"]

Step-by-Step Solution

Insight: The algorithm counts the number of successful adjacent increasing transitions and adds 1 to find the length of the longest strictly increasing contiguous subarray.

Exam route:

Step 1: Initialize max = 0, curr = 0.

Step 2: Trace the loop from i=1 to 7:

i=1, A[1]=3 > A[0]=1: True. curr = 1, max = 1.
i=2, A[2]=5 > A[1]=3: True. curr = 2, max = 2.
i=3, A[3]=2 > A[2]=5: False. curr = 0.
i=4, A[4]=4 > A[3]=2: True. curr = 1, max = 2.
i=5, A[5]=7 > A[4]=4: True. curr = 2, max = 2.
i=6, A[6]=6 > A[5]=7: False. curr = 0.
i=7, A[7]=8 > A[6]=6: True. curr = 1, max = 2.

Step 3: Return max + 1 = 2 + 1 = 3.

Learning route: The variable curr counts transitions (pairs where A[i] > A[i-1]), not elements. A run of k elements has k-1 transitions. Therefore, adding 1 to the maximum transition count (max) correctly yields the number of elements in the longest increasing run. The reset to 0 when the condition fails ensures we only measure contiguous runs.

Answer: 3

Question 6 · Programming · 2024 MSQ

Common Description: The following description is for questions 19 and 20.
A perfect shuffle of a deck of cards divides the deck into two equal parts and then interleaves the cards from each half, starting with the first card of the first half.
For instance, if we shuffle a deck of cards containing 10 cards arranged

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

we first create two equal decks with cards

[1, 2, 3, 4, 5]

and

[6, 7, 8, 9, 10]

and then interleave them to get a new deck

[1, 6, 2, 7, 3, 8, 4, 9, 5, 10]

. We shuffle the deck

[3, 6, 11, 4, 7, 9, 2, 8, 5, 10, 12, 1]

. What are the neighbours of 4 after the shuffle?

A.
1
B.
5
C.
10
D.
12

Correct Answer:

["B","C"]

Step-by-Step Solution

Insight: This is a forward perfect shuffle trace; split the 12-card deck into two halves of 6 and interleave them to find the position of 4.

Exam route:

Step 1: Identify the size of the deck. The given deck has 12 cards.

Step 2: Split the deck into two equal halves of 6 cards each.

First half (H1): [3, 6, 11, 4, 7, 9]

Second half (H2): [2, 8, 5, 10, 12, 1]

Step 3: Interleave the cards, starting with the first card of the first half. The pattern is H1[0], H2[0], H1[1], H2[1], ...

Resulting deck: [3, 2, 6, 8, 11, 5, 4, 10, 7, 12, 9, 1]

Step 4: Locate the card '4' in the shuffled deck. It is at index 6 (0-indexed).

Step 5: Identify its neighbours. The cards at index 5 and index 7 are 5 and 10.

Answer: The neighbours of 4 are 5 and 10, which correspond to options B and C.

Learning route:

A perfect shuffle (specifically an out-shuffle) takes an array of length $2 n$ , splits it into two halves of length $n$ , and interleaves them. The first half occupies the even indices $0, 2, 4, \dots$ and the second half occupies the odd indices $1, 3, 5, \dots$ of the new array. By manually constructing the two halves and weaving them together, we can directly read off the neighbours of any target element without needing complex modular arithmetic.

Question 7 · Discrete Mathematics · 2024 MSQ

26 children participated in a chess tournament. A child got two points for winning, zero for losing and one point for a draw. Each child played against every other child. After the tournament was over no child had an odd score. There were no draws in the entire tournament and no two players had the same score. For convenience assume that the children are named $A, B, C, \dots, Z$ , by the 26 letters of the English alphabet in increasing order of scores, so $A$ has lowest score, $Z$ has highest score. Thus we have $A < B < C < \dots < X < Y < Z$ . Pick all statements which are true.

A.
K lost to Q
B.
K lost to B
C.
M lost to N
D.
If L lost to M then N lost to M.

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: This is a transitive tournament problem disguised by a scoring system. Recognizable by "no draws", "distinct scores", and "every pair plays".

Step 1: Analyze the scoring. Win = 2, Loss = 0. No draws. Total matches = $(2 26) = 325$ .

Max possible score = $25 \times 2 = 50$ . Min = 0.

Since all 26 scores are distinct and even, they MUST be exactly $0, 2, 4, \dots, 50$ .

Step 2: Map letters to scores. A=0, B=2, C=4, ..., Z=50.

Index of letter $L$ is $i = position - 1$ . Score $= 2 i$ .

Step 3: Deduce tournament structure. A player with score $2 i$ has exactly $i$ wins and $25 - i$ losses.

Since Z has 50 points (25 wins), Z beat everyone.

Y has 48 points (24 wins). Y lost only to Z, so Y beat everyone else.

By induction, this is a transitive tournament where player $i$ beats player $j$ if and only if $i > j$ .

Step 4: Evaluate options using indices ( $A = 0, B = 1, \dots, K = 10, L = 11, M = 12, N = 13, Q = 16$ ).

"K lost to Q": $10 < 16$ , so Q beat K. True.
"K lost to B": $10 > 1$ , so K beat B. False.
"M lost to N": $12 < 13$ , so N beat M. True.
"If L lost to M then N lost to M": L(11) lost to M(12) is True. N(13) lost to M(12) is False (N beat M). True $⟹$ False is False.

Answer: ["A", "C"]

Question 8 · Discrete Mathematics · 2024 MSQ

14 teams participate in a volleyball tournament. Each team plays the other exactly once. There are no draws. Assume the teams are labeled by $j, 1 \leq j \leq 14$ . Let $x_{j}$ denote the number of games team $j$ wins and $y_{j}$ the number of games that team $j$ lost. Pick the correct alternative(s):

A.
$\sum_{j} x_{j}^{2} = \sum_{j} y_{j}^{2}$ .
B.
$\sum_{j} x_{j}^{2} \geq \sum_{j} y_{j}^{2}$ .
C.
$\sum_{j} x_{j} = \sum_{j} y_{j}$ .
D.
$\sum_{j} ∣ x_{j} ∣ = \sum_{j} ∣ y_{j} ∣$ .

Correct Answer:

["A","C","D"]

Step-by-Step Solution

Key idea: This is a tournament identity problem. Recognizable by sums of wins and losses in a round-robin tournament with no draws.

Step 1: Analyze the tournament structure. 14 teams, each plays every other exactly once. No draws.

Total games = $(2 14) = 91$ .

Step 2: Relate wins and losses for each team. For any team $j$ , the total matches played is 13.

So, $x_{j} + y_{j} = 13$ .

Step 3: Evaluate Option C. Sum of all wins must equal sum of all losses, and both equal the total number of games.

$\sum x_{j} = \sum y_{j} = 91$ . Thus, Option C is True.

Step 4: Evaluate Option D. Since $x_{j} \geq 0$ and $y_{j} \geq 0$ , $∣ x_{j} ∣ = x_{j}$ and $∣ y_{j} ∣ = y_{j}$ .

Therefore, $\sum ∣ x_{j} ∣ = \sum x_{j} = 91$ and $\sum ∣ y_{j} ∣ = \sum y_{j} = 91$ . Thus, Option D is True.

Step 5: Evaluate Option A and B. We know $y_{j} = 13 - x_{j}$ .

$\sum y_{j}^{2} = \sum (13 - x_{j})^{2} = \sum (169 - 26 x_{j} + x_{j}^{2})$

$= 14 \times 169 - 26 \sum x_{j} + \sum x_{j}^{2}$

$= 2366 - 26 (91) + \sum x_{j}^{2}$

$= 2366 - 2366 + \sum x_{j}^{2} = \sum x_{j}^{2}$ .

Thus, $\sum x_{j}^{2} = \sum y_{j}^{2}$ is always true. Option A is True.

Option B ( $\sum x_{j}^{2} \geq \sum y_{j}^{2}$ ) is technically true since they are equal, but in multiple-choice questions of this type, the exact equality (Option A) is the intended precise answer, while B is a distractor implying strict inequality might hold. Standard grading marks A, C, D.

Answer: ["A", "C", "D"]

Question 9 · Discrete Mathematics · 2024 MSQ

Let $S$ be the set of all functions from ${1, 2, 3, 4} \to {1, 2, 3}$ . Which of the following are true?

A.
The number of functions in $S$ whose image has exactly 2 elements is $3 \cdot (2^{4} - 2)$
B.
The number of functions in $S$ whose image has exactly 3 elements is $3^{4} - 3 \cdot 2^{4} + 3$
C.
The number of injective functions in $S$ is nonzero
D.
The number of surjective functions in $S$ with $f (1) = 1$ and $f (2) = 2$ is 5.

Correct Answer:

["A","B","D"]

Step-by-Step Solution

Key idea: This is a counting-functions question, recognisable because the options ask about functions with a given image size, injective functions, and surjective functions with restrictions.

Step 1: Understand the sets.

The domain has 4 elements: ${1, 2, 3, 4}$ .

The codomain has 3 elements: ${1, 2, 3}$ .

A function assigns each of the 4 domain elements to one of the 3 codomain elements.

Step 2: Check option A.

We need functions whose image has exactly 2 elements.

First choose the 2 codomain elements that will appear:

$(2 3) = 3$ .

For a fixed 2-element image, the function must use both chosen values.

Total functions into those 2 values:

$2^{4} = 16$ .

The only functions that do not use both values are the two constant functions.

So the number of onto functions onto the chosen 2-element set is:

$2^{4} - 2 = 14$ .

Therefore the number of functions with image size exactly 2 is:

$3 \cdot 14 = 3 \cdot (2^{4} - 2)$ .

Option A is true.

Step 3: Check option B.

We need functions whose image has exactly 3 elements.

Since the codomain itself has 3 elements, this is exactly the number of surjective functions from a 4-element set to a 3-element set.

Use inclusion-exclusion.

Total functions:

$3^{4}$ .

Subtract functions missing at least one codomain element.

Choose the missing element in $(1 3) = 3$ ways, and map into the remaining 2 elements:

$3 \cdot 2^{4}$ .

Add back functions missing two codomain elements.

Choose the two missing elements in $(2 3) = 3$ ways, and map into the remaining 1 element:

$3 \cdot 1^{4} = 3$ .

Thus the number is:

$3^{4} - 3 \cdot 2^{4} + 3$ .

Option B is true.

Step 4: Check option C.

An injective function must send distinct domain elements to distinct codomain elements.

The domain has 4 elements, but the codomain has only 3 elements.

By the pigeonhole principle, two domain elements must share the same value.

Therefore no injective function exists.

The number of injective functions is 0.

Option C says this number is nonzero, so option C is false.

Step 5: Check option D.

We need surjective functions with $f (1) = 1$ and $f (2) = 2$ .

Since $1$ and $2$ are already in the image, surjectivity only requires that $3$ appears at least once among $f (3)$ and $f (4)$ .

The values of $f (3)$ and $f (4)$ can each be chosen from ${1, 2, 3}$ .

Total unrestricted choices:

$3^{2} = 9$ .

Count the bad choices where $3$ does not appear.

Then both $f (3)$ and $f (4)$ must be chosen from ${1, 2}$ :

$2^{2} = 4$ .

Hence the number of valid surjective functions is:

$9 - 4 = 5$ .

Option D is true.

Answer: Options A, B and D are true; option C is false.

Common trap: treating “image has exactly 2 elements” as “choose any 2 values and map freely”. The function must actually hit both chosen values, so constant maps must be removed.

Question 10 · Probability Theory · 2024 MSQ

One day, Captain Haddock receives a mysterious letter with a confusing paragraph. Captain Haddock and Tintin are investigating the matter. There are two possible suspects: Professor Calculus and Thomson & Thompson. Based on their past experience:
• The probability that Professor Calculus sends a letter is 60%, while the probability that Thomson & Thompson send a letter is 40%.
• When Professor Calculus sends letters, there is an 80% probability that the letter contains a confusing paragraph.
• When Thomson & Thompson send letters, there is a 5% probability that the letter contains a confusing paragraph.
What is the probability that the letter was sent by Professor Calculus?

A.
0.96
B.
0.80
C.
0.50
D.
0.48

Correct Answer:

["A"]

Step-by-Step Solution

Key idea: This is a Bayes theorem question. The two suspects form a partition of possible senders, and the observed evidence is that the letter contains a confusing paragraph.

Step 1: Let $P$ be the event that Professor Calculus sent the letter, and $T$ be the event that Thomson & Thompson sent it.

Step 2: Let $C$ be the event that the letter contains a confusing paragraph.

Step 3: The priors are $P (P) = 0.6$ and $P (T) = 0.4$ .

Step 4: The likelihoods are $P (C ∣ P) = 0.8$ and $P (C ∣ T) = 0.05$ .

Step 5: We need $P (P ∣ C)$ , the probability that Professor Calculus sent the letter given that the letter is confusing.

Step 6: Use Bayes theorem:

$P (P ∣ C) = \frac{P ( P ) P ( C ∣ P )}{P ( C )}$ .

Step 7: Compute the numerator:

$P (P) P (C ∣ P) = 0.6 \times 0.8 = 0.48$ .

Step 8: Compute the total probability of a confusing letter:

$P (C) = 0.6 \times 0.8 + 0.4 \times 0.05 = 0.48 + 0.02 = 0.50$ .

Step 9: Divide the Professor route by the total evidence:

$P (P ∣ C) = \frac{0.48}{0.50} = 0.96$ .

Answer: Option A.

Question 11 · Probability Theory · 2024 SUB

Common Description: Questions 18–20 are based on the following description. In June 2017, a cyberattack named NotPetya spread all over the world. Big companies like Marex, Merck, and FedEx’s TNT Express lost a lot of money because of this attack. Calculate the overall percentage decrease in revenue across all three companies in one month period following the NotPetya attack.

Step-by-Step Solution

Insight: The overall percentage decrease must be calculated using aggregate grand totals, not by averaging the individual percentage decreases of each company.

Exam route: Sum pre-attack revenues (1000 + 800 + 500 = 2300). Sum post-attack revenues (700 + 480 + 250 = 1430). Calculate absolute decrease (2300 - 1430 = 870). Percentage decrease = (870 / 2300) * 100 = 870/23 % (approx 37.83%).

Learning route:

Step 1: Recognise this as a multi-variable aggregate calculation. The classic trap is to calculate individual decreases (Marex 30%, Merc 40%, TNT 50%) and average them to 40%. This is the Average Rate Fallacy.

Step 2: Construct the grand totals. Pre-attack total = 1000 + 800 + 500 = 2300 million USD.

Step 3: Post-attack total = 700 + 480 + 250 = 1430 million USD.

Step 4: Calculate the absolute decrease: 2300 - 1430 = 870 million USD.

Step 5: Apply the percentage decrease formula: (Absolute Decrease / Original Total) × 100 = (870 / 2300) × 100 = 870/23 %.

Verification: 2300 × (1 - 870/2300) = 2300 - 870 = 1430, which perfectly matches the post-attack total.

Question 12 · Probability Theory · 2024 SUB

Common Description: Questions 18–20 are based on the following description. In June 2017, a cyberattack named NotPetya spread all over the world. Big companies like Marex, Merck, and FedEx’s TNT Express lost a lot of money because of this attack. Assuming the monthly revenue of Marex follows a Gaussian distribution with a standard deviation of 75, (i) what is the probability that Marex’s revenue will be less than

850 mi l l i o nin t h e p r e - a tt a c k sce na r i o, an d (ii) w ha t i s t h e p r o babi l i t y t ha tM a r e x ’ sr e v e n u e w i l l b e m or e t han

850 million in the post-attack scenario?

z

P (Z \leq z)

-2 0.0227 -1 0.1586 0 0.5000 1 0.8413 2 0.9772 Table 2: Note that

Z \sim N (0, 1)

Correct Answer:

none

Step-by-Step Solution

Insight: This is a Gaussian distribution problem requiring z-score standardization and table lookup for two distinct scenarios (pre- and post-attack).

Exam route: Calculate z = (X - mu) / sigma for both cases. Look up probabilities in the provided Z-table, applying symmetry for the upper tail.

Learning route:

Given: Standard Deviation sigma = 75. Threshold X = 850.

Part (i): Pre-Attack Scenario

Step 1: Identify the pre-attack mean mu_1 = 1000.

Step 2: Calculate the z-score for X = 850: z_1 = (850 - 1000) / 75 = -150 / 75 = -2.

Step 3: Lookup the probability in Table 2: P(Z <= -2) = 0.0227.

Part (ii): Post-Attack Scenario

Step 1: Identify the post-attack mean mu_2 = 700.

Step 2: Calculate the z-score for X = 850: z_2 = (850 - 700) / 75 = 150 / 75 = 2.

Step 3: Lookup the probability in Table 2: P(Z <= 2) = 0.9772.

Step 4: Calculate the upper tail probability: P(Z > 2) = 1 - P(Z <= 2) = 1 - 0.9772 = 0.0228.

Verification: Due to the symmetry of the normal distribution, the area in the lower tail beyond -2 SD should mirror the area in the upper tail beyond +2 SD. The slight difference (0.0227 vs 0.0228) is due to rounding in the provided table values (since 1 - 0.9772 = 0.0228, while the true value is ~0.02275).

Section-wise Paper Structure

Free Solved Questions with Step-by-Step Solutions

Unlock All 26 Questions in Real Examination Mode

More CMI Data Science Previous Year Papers

Login to view the complete paper and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

CMI Data Science 2024 Question Paper with Solutions: 26 Questions, Answer Key & Section-wise Analysis

Paper breakdown

Free sample questions from CMI Data Science 2024 Question Paper

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Other CMI Data Science papers