UPCAT Mathematics — Statistics & ProbabilityExam Answer Templates

Examiner tip

Always show the sum calculation explicitly - don't just write the final division

Model answer

Given: Data set = {12, 15, 18, 22, 25, 30}
To Find: Mean

Solution:
Mean = Sum of all observations / Number of observations
Mean = (12 + 15 + 18 + 22 + 25 + 30) / 6
Mean = 122 / 6
Mean = 20.33

Therefore, the mean is 20.33

Answer structure

• Line 1: Write Given and To Find [0.5 marks]
• Line 2: State the formula for mean [0.5 marks]
• Line 3: Substitute values and calculate sum [0.5 marks]
• Line 4: Final division and answer [0.5 marks]

Scoring breakdown

Common mark deductions

• Not showing the addition step
• Calculation errors
• Missing units or final statement

Key phrases to include

• Mean = Sum/Number
• Given
• Therefore

Examiner tip

Always verify your answer by checking if the median condition is satisfied

Model answer

Given: Five numbers with median = 15
Four known numbers: 8, 12, 18, 20
To Find: Fifth number

Solution:
Let the fifth number be x.
For median to be 15, when arranged in order, the middle (3rd) number must be 15.

Case 1: If x ≤ 15
Arranged order: 8, 12, x, 15, 18, 20 (not possible since we need 15 as median)

Case 2: If x > 15
Arranged order: 8, 12, 15, x, 18, 20 (not possible since 15 is not in our original set)

Therefore, x = 15

Verification: Arranged order: 8, 12, 15, 18, 20
Median = 15 ✓

Therefore, the fifth number is 15.

Answer structure

• Line 1: State given information clearly [0.5 marks]
• Line 2: Define variable for unknown [0.5 marks]
• Line 3: Apply median concept - middle value [1 mark]
• Line 4: Logical reasoning about position [1 mark]

Scoring breakdown

Common mark deductions

• Not considering different cases
• Not verifying the answer
• Incorrect understanding of median

Key phrases to include

• Let the fifth number be
• middle value
• arranged in order
• verification

Examiner tip

Always list the sample space systematically - use a tree diagram if needed

Model answer

Given: Fair coin tossed 3 times
To Find: P(exactly 2 heads)

Solution:
Sample space for 3 tosses:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8

Favorable outcomes (exactly 2 heads):
E = {HHT, HTH, THH}
n(E) = 3

P(exactly 2 heads) = n(E)/n(S) = 3/8

Therefore, the probability of getting exactly 2 heads is 3/8 or 0.375

Answer structure

• Line 1: Write given and to find [0.5 marks]
• Line 2: List complete sample space [1 mark]
• Line 3: Identify favorable outcomes [1 mark]
• Line 4: Apply probability formula [0.5 marks]

Scoring breakdown

Common mark deductions

• Incomplete sample space
• Missing favorable outcomes
• Not simplifying fraction

Key phrases to include

• Sample space
• favorable outcomes
• P(E) = n(E)/n(S)

Examiner tip

Show all calculations in a tabular format for clarity and to avoid arithmetic errors

Model answer

Given: Data set = {2, 4, 6, 8, 10}
To Find: Standard deviation (σ)

Solution:
Step 1: Calculate the mean
Mean (x̄) = (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6

Step 2: Calculate deviations from mean
(xi - x̄): (2-6), (4-6), (6-6), (8-6), (10-6)
Deviations: -4, -2, 0, 2, 4

Step 3: Calculate squared deviations
(xi - x̄)²: (-4)², (-2)², (0)², (2)², (4)²
Squared deviations: 16, 4, 0, 4, 16

Step 4: Calculate variance
Variance (σ²) = Σ(xi - x̄)²/n = (16 + 4 + 0 + 4 + 16)/5 = 40/5 = 8

Step 5: Calculate standard deviation
Standard deviation (σ) = √variance = √8 = 2√2 = 2.828

Therefore, the standard deviation is 2.828 (or 2√2).

Answer structure

• Step 1: Calculate mean [1 mark]
• Step 2: Find deviations from mean [1 mark]
• Step 3: Calculate squared deviations [1 mark]
• Step 4: Calculate variance [1 mark]
• Step 5: Find standard deviation [1 mark]

Scoring breakdown

Common mark deductions

• Calculation errors in any step
• Not showing squared deviations clearly
• Forgetting to take square root for final answer

Key phrases to include

• deviations from mean
• squared deviations
• variance
• standard deviation = √variance

Examiner tip

Count frequencies systematically and state which value appears most often

Model answer

Mode = 7 (appears 3 times, most frequent)

Answer structure

• State the mode value and frequency [1 mark]

Scoring breakdown

Common mark deductions

• Not identifying the correct frequency
• Giving multiple values without explanation

Key phrases to include

• most frequent
• appears

Examiner tip

Draw a Venn diagram to visualize the problem - it helps avoid double counting

Model answer

Given: Total students = 30
Like Math = 18, Like Physics = 12
Like both = 8
To Find: P(likes neither subject)

Solution:
Using the principle of inclusion-exclusion:
Students who like at least one subject = Like Math + Like Physics - Like both
= 18 + 12 - 8 = 22

Students who like neither subject = Total - Like at least one
= 30 - 22 = 8

P(likes neither) = Students who like neither / Total students
= 8/30 = 4/15

Therefore, the probability is 4/15 or 0.267

Answer structure

• Line 1: State given information [0.5 marks]
• Line 2: Apply inclusion-exclusion principle [1 mark]
• Line 3: Calculate students liking neither [1 mark]
• Line 4: Calculate probability [0.5 marks]

Scoring breakdown

Common mark deductions

• Double counting students
• Not using inclusion-exclusion principle
• Arithmetic errors

Key phrases to include

• inclusion-exclusion
• at least one
• neither subject

Examiner tip

List out each letter and its frequency clearly before applying the formula

Model answer

Given: Word STATISTICS
To Find: Number of arrangements

Solution:
Total letters = 10
Repeated letters:

• S appears 3 times
• T appears 3 times
• A appears 1 time
• I appears 2 times
• C appears 1 time

Number of arrangements = n!/(n₁! × n₂! × n₃! × ...)
= 10!/(3! × 3! × 1! × 2! × 1!)
= 3,628,800/(6 × 6 × 1 × 2 × 1)
= 3,628,800/72
= 50,400

Therefore, there are 50,400 different arrangements.

Answer structure

• Line 1: Count total letters and repetitions [1 mark]
• Line 2: Apply permutation formula for repeated objects [1 mark]
• Line 3: Calculate final answer [1 mark]

Scoring breakdown

Common mark deductions

• Missing repeated letters
• Wrong formula
• Calculation errors

Key phrases to include

• repeated letters
• n!/(n₁! × n₂!...)
• arrangements

Examiner tip

Always scan the data set completely to ensure you find the true maximum and minimum

Model answer

Given: Data = {15, 22, 18, 30, 25, 12, 28}
To Find: Range

Solution:
Maximum value = 30
Minimum value = 12

Range = Maximum value - Minimum value
Range = 30 - 12 = 18

Therefore, the range is 18.

Answer structure

• Line 1: Identify maximum and minimum values [1 mark]
• Line 2: Apply range formula and calculate [1 mark]

Scoring breakdown

Common mark deductions

• Incorrect max/min identification
• Arithmetic error in subtraction

Key phrases to include

• maximum value
• minimum value
• Range = Max - Min

Examiner tip

Remember that 'without replacement' changes the total number of balls for the second draw

Model answer

Given: 5 red balls, 3 blue balls (total = 8)
Two balls drawn without replacement
To Find: P(both balls are red)

Solution:
First draw: P(red) = 5/8
After drawing one red ball: 4 red balls remain out of 7 total
Second draw: P(red|first red) = 4/7

P(both red) = P(first red) × P(second red|first red)
= (5/8) × (4/7)
= 20/56
= 5/14

Therefore, the probability that both balls are red is 5/14.

Answer structure

• Line 1: State initial conditions [0.5 marks]
• Line 2: Calculate probability for first draw [0.5 marks]
• Line 3: Calculate conditional probability for second draw [1 mark]
• Line 4: Apply multiplication rule [1 mark]

Scoring breakdown

Common mark deductions

• Not accounting for replacement
• Wrong conditional probability
• Not simplifying fraction

Key phrases to include

• without replacement
• conditional probability
• multiplication rule

Examiner tip

Remember that n! means n × (n-1) × (n-2) × ... × 2 × 1

Model answer

Number of arrangements = 6! = 720

Answer structure

• Apply factorial formula and calculate [1 mark]

Scoring breakdown

Common mark deductions

• Wrong calculation of factorial
• Not using factorial notation

Key phrases to include

• 6!
• factorial

Examiner tip

Always identify the distribution type first, then set up the formula with correct parameters

Model answer

Given: n = 4 days, p = 0.3 (probability of rain), q = 1 - p = 0.7
To Find: P(exactly 2 rainy days)

Solution:
This follows binomial distribution with parameters n = 4, p = 0.3

For binomial distribution:
P(X = r) = ⁿCᵣ × pʳ × q^(n-r)

Where:
n = 4, r = 2, p = 0.3, q = 0.7

Step 1: Calculate ⁴C₂
⁴C₂ = 4!/(2! × 2!) = 24/(2 × 2) = 6

Step 2: Calculate p²
p² = (0.3)² = 0.09

Step 3: Calculate q^(n-r)
q² = (0.7)² = 0.49

Step 4: Apply the formula
P(X = 2) = 6 × 0.09 × 0.49
= 6 × 0.0441
= 0.2646

Therefore, the probability is 0.2646 or 26.46%.

Answer structure

• Step 1: Identify binomial distribution parameters [1 mark]
• Step 2: Calculate combination ⁴C₂ [1 mark]
• Step 3: Calculate p² and q² [1 mark]
• Step 4: Apply binomial formula [1 mark]
• Step 5: Final calculation and answer [1 mark]

Scoring breakdown

Common mark deductions

• Not recognizing binomial distribution
• Errors in combination calculation
• Wrong formula application

Key phrases to include

• binomial distribution
• ⁿCᵣ × pʳ × q^(n-r)
• parameters n, p, q

Examiner tip

Always create the cumulative frequency table systematically and double-check your median class identification

Model answer

Given: Frequency distribution table
To Find: Median

Solution:
Step 1: Prepare cumulative frequency table

Total frequency (N) = 40

Step 2: Find median position
Median position = N/2 = 40/2 = 20th observation

Step 3: Identify median class
The 20th observation lies in class 20-30 (since cf = 25 > 20)

Step 4: Apply median formula
Median = l + [(N/2 - cf)/f] × h
Where: l = 20, N/2 = 20, cf = 13, f = 12, h = 10

Median = 20 + [(20 - 13)/12] × 10
= 20 + (7/12) × 10
= 20 + 5.83
= 25.83

Therefore, the median is 25.83.

Answer structure

• Step 1: Create cumulative frequency table [1 mark]
• Step 2: Find N/2 position [1 mark]
• Step 3: Identify median class [1 mark]
• Step 4: Apply median formula [1 mark]

Scoring breakdown

Common mark deductions

• Errors in cumulative frequency
• Wrong median class
• Formula application errors

Key phrases to include

• cumulative frequency
• median class
• Median = l + [(N/2 - cf)/f] × h

Examiner tip

List all favorable outcomes systematically - consider (a,b) and (b,a) as different outcomes

Model answer

Given: Two dice thrown simultaneously
To Find: P(sum = 8)

Solution:
Total possible outcomes = 6 × 6 = 36

Favorable outcomes for sum = 8:
(2,6), (3,5), (4,4), (5,3), (6,2)
Number of favorable outcomes = 5

P(sum = 8) = Favorable outcomes/Total outcomes
= 5/36

Therefore, the probability is 5/36.

Answer structure

• Line 1: Calculate total possible outcomes [0.5 marks]
• Line 2: List favorable outcomes [1 mark]
• Line 3: Apply probability formula [0.5 marks]

Scoring breakdown

Common mark deductions

• Missing some favorable outcomes
• Wrong total outcomes
• Not listing outcomes systematically

Key phrases to include

• total outcomes = 36
• favorable outcomes
• ordered pairs

Examiner tip

Remember: combinations for selection (order doesn't matter), permutations for arrangement (order matters)

Model answer

Given: Total people = 10, Committee size = 4
To Find: Number of ways to select committee

Solution:
This is a combination problem (order doesn't matter)

Number of ways = ¹⁰C₄
= 10!/(4! × 6!)
= (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1)
= 5040/24
= 210

Therefore, there are 210 ways to select the committee.

Answer structure

• Line 1: Identify as combination problem [0.5 marks]
• Line 2: Apply combination formula [1 mark]
• Line 3: Calculate final answer [0.5 marks]

Scoring breakdown

Common mark deductions

• Using permutation instead of combination
• Calculation errors
• Wrong formula

Key phrases to include

• combination
• ¹⁰C₄
• order doesn't matter

Examiner tip

Always convert to standard normal first, then use the symmetry property: Φ(-z) = 1 - Φ(z)

Model answer

Given: μ = 160 cm, σ = 5 cm
Heights are normally distributed
To Find: P(155 < X < 165)

Solution:
Step 1: Standardize the values using Z = (X - μ)/σ

For X = 155:
Z₁ = (155 - 160)/5 = -5/5 = -1

For X = 165:
Z₂ = (165 - 160)/5 = 5/5 = 1

Step 2: Find the required probability
P(155 < X < 165) = P(-1 < Z < 1)
= P(Z < 1) - P(Z < -1)
= Φ(1) - Φ(-1)

Step 3: Use standard normal table
Φ(1) = 0.8413
Φ(-1) = 0.1587

Step 4: Calculate final probability
P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826

Therefore, the probability is 0.6826 or 68.26%.

Answer structure

• Step 1: Convert to standard normal using Z-scores [1 mark]
• Step 2: Set up probability expression [1 mark]
• Step 3: Use standard normal table values [1 mark]
• Step 4: Calculate final answer [1 mark]

Scoring breakdown

Common mark deductions

• Incorrect Z-score calculation
• Wrong probability setup
• Table reading errors

Key phrases to include

• standardize
• Z-score
• standard normal table
• Φ(z)

Do

• Write the final answer clearly
• Use correct mathematical notation
• Include units if applicable

Don't

• Waste time on lengthy explanations
• Show unnecessary working steps
• Leave answer unclear

Do

• State the formula used
• Show key substitution step
• Box the final answer
• Include proper mathematical notation

Don't

• Skip the formula statement
• Make arithmetic errors
• Forget units in final answer

Do

• Write 'Given' and 'To Find'
• Show all major steps
• Include intermediate calculations
• Verify answer if time permits

Don't

• Jump steps without explanation
• Make calculation errors
• Forget to state conclusion

Do

• Break into numbered steps
• Show all formula derivations
• Include diagrams where helpful
• Verify final answer
• State assumptions if any

Don't

• Rush through steps
• Skip verification
• Make multiple arithmetic errors
• Use unclear notation