Question structure	Useful method	Essential check
Finite equally likely outcomes	Favorable count divided by total count	Equal likelihood is justified.
Several stages or positions	Systematic listing or multiplication counting	Respect order and replacement.
At least one occurrence	Consider the complement	The complement means no occurrences.
Either of two conditions	Add counts and subtract overlap	Count shared outcomes once.
Mean or missing observation	Use total = count × mean	Count repeated observations.
Median or mode	Order observations or count frequencies	Distinguish values from positions and frequencies.
Variance or standard deviation	Average squared deviations, then take a square root if needed	Divide by n and distinguish variance from standard deviation.
Normal probability	Identify the region, standardize, and use supplied probabilities	Use σ, not σ², in standardization.

3. Essential relationships

Classical probability

P(A) = favorable equally likely outcomes ÷ total equally likely outcomes

P(Aᶜ) = 1 − P(A)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Mean and combined mean

Mean = total ÷ count

Total = count × mean

Combined mean = combined total ÷ combined count

Dispersion

Range = maximum − minimum

Variance = sum of squared deviations from the mean ÷ n

Standard deviation = √variance

Alternatively:

Variance = mean of squares − square of mean

For y = kx + c:

• New mean = k × original mean + c.
• New variance = k² × original variance.
• New standard deviation = |k| × original standard deviation.

Normal distribution

For X ~ N(μ, σ²):

z = (x − μ)/σ

Φ(z) = P(Z ≤ z)

P(Z > z) = 1 − Φ(z)

Φ(−z) = 1 − Φ(z)

P(a < Z < b) = Φ(b) − Φ(a)

For a continuous normal variable, including or excluding a finite endpoint does not change the probability.

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4. Supplied normal probabilities

Use these values wherever required in this lesson.

Φ(z) = P(Z ≤ z), for Z ~ N(0, 1)

z	Φ(z), approximately
0	0.5000
0.5	0.6915
1	0.8413
1.5	0.9332
2	0.9772

Rounding rule: Calculate consistently from these supplied values. Small differences from more precise normal probabilities are expected.

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5. Worked example 1: Selecting the correct counting model

A bag contains three red balls and two blue balls.

Two balls are drawn successively without replacement. At each draw, every remaining ball is equally likely to be selected.

Find the probability that the colors differ.

Step 1: Identify elementary outcomes

Label the physical balls:

R₁, R₂, R₃, B₁, B₂

Use ordered pairs of distinct balls.

Step 2: Count all outcomes

Total = 5 × 4 = 20

Step 3: Count favorable outcomes

Red then blue:

3 × 2 = 6

Blue then red:

2 × 3 = 6

The cases do not overlap.

Favorable = 12

Step 4: Calculate

P(different colors) = 12/20 = 3/5

Verification

Same-color outcomes:

• Both red: 3 × 2 = 6.
• Both blue: 2 × 1 = 2.

Different colors = 20 − 8 = 12 outcomes

Both methods agree.

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6. Worked example 2: Combining groups correctly

Group A contains six observations with mean 12.

Group B contains four observations with mean 17.

Find the combined mean.

Step 1: Recover totals

Group A total = 6 × 12 = 72

Group B total = 4 × 17 = 68

Step 2: Combine totals and counts

Combined mean = (72 + 68)/(6 + 4)

Combined mean = 140/10 = 14

Why not average the two means?

(12 + 17)/2 = 14.5

This gives the groups equal weight, although they have different sizes.

Verification: The combined mean 14 lies between 12 and 17 and is closer to 12, the mean of the larger group.

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7. Worked example 3: One dataset, several characteristics

Find the mean, median, mode, variance, and standard deviation of:

2, 4, 4, 6

Mean

Mean = 16/4 = 4

Median

The data is ordered. The middle two observations are 4 and 4.

Median = 4

Mode

Mode = 4

Variance

Deviations from the mean:

−2, 0, 0, 2

Squared deviations:

4, 0, 0, 4

Variance = 8/4 = 2

Standard deviation

Standard deviation = √2

Interpretation

The center measures all equal 4, but the positive variance shows that the observations are not all identical.

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8. Worked example 4: Understanding a transformation

A dataset has:

• Mean = 10.
• Variance = 9.
• Standard deviation = 3.

Every value is transformed using:

y = −2x + 5

Find the new mean, variance, and standard deviation.

Mean

New mean = −2 × 10 + 5 = −15

Variance

New variance = (−2)² × 9 = 36

Standard deviation

New standard deviation = |−2| × 3 = 6

Interpretation

• Multiplication by −2 reverses order and doubles distances.
• Adding 5 shifts the data.
• The shift does not change dispersion.

Check: 6² = 36.

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9. Worked example 5: Normal probability in original units

Measurements follow:

X ~ N(80, 16)

Find P(76 < X < 86).

Step 1: Identify the standard deviation

σ = √16 = 4

Step 2: Standardize both boundaries

Lower:

(76 − 80)/4 = −1

Upper:

(86 − 80)/4 = 1.5

Therefore:

P(76 < X < 86) = P(−1 < Z < 1.5)

Step 3: Use cumulative probabilities

Φ(−1) ≈ 1 − 0.8413 = 0.1587

P ≈ 0.9332 − 0.1587 = 0.7745

Answer: Approximately 77.45%.

Verification

The interval contains the mean and extends substantially on both sides. A probability greater than 0.5 is reasonable.

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10. Worked example 6: The model determines how to calculate

Compare these questions:

Question A

One integer is selected uniformly from:

{1, 2, 3, 4, 5, 6}

Find the probability of exactly 3.

There are six equally likely outcomes.

P(result = 3) = 1/6

Question B

X is a continuous normal random variable.

Find P(X = 3).

A single exact point has zero area under the density curve.

P(X = 3) = 0

Key lesson

Do not transfer a rule from a finite equally likely model directly to a continuous normal model.

A probability-zero exact value in a continuous model is not the same as an outcome absent from a finite sample space.

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11. Independent mixed practice

Suggested time: 20–25 minutes.

Choose the method yourself. Show enough working to justify each answer.

Practice 1

One integer is selected uniformly from 1 to 18.

Find the probability that it is divisible by 2 or by 3.

Practice 2

Two fair dice are rolled independently.

Find the probability that their sum is 11.

Practice 3

Two fair dice are rolled independently.

Find the probability of at least one 5.

Practice 4

A two-person team is selected uniformly from four girls and three boys.

Find the probability that it contains one girl and one boy.

Practice 5

Find the mean, median, and mode of:

3, 7, 5, 7, 8, 12

Practice 6

The mean of:

4, 8, 10, x

is 9.

Find x.

Practice 7

Five observations have mean 6. Three other observations have mean 10.

Find the combined mean.

Practice 8

Find the mean, variance, and standard deviation of:

1, 3, 5, 7

Practice 9

A dataset has mean 8, range 12, and variance 16.

Every value is changed using:

y = 3x − 5

Find the new mean, range, variance, and standard deviation.

Practice 10

Use the frequency table to find the mean, median, mode, and variance.

Value	Frequency
2	1
4	2
6	1

Practice 11

Given:

X ~ N(40, 9)

Find:

1. The standard deviation.
2. The z-score of 46.
3. P(X > 46), using the supplied table.

Practice 12

A normal variable has mean 100.

Given:

P(X < 92) = 0.15

Find:

1. P(X > 108).
2. P(92 < X < 108).

Practice 13

Given:

X ~ N(60, 25)

Find P(55 < X < 65), using the supplied table.

Practice 14

A student claims:

Any dataset with mean 0 and standard deviation 1 follows the standard normal distribution.

Explain why the claim is incorrect.

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12. Independent practice solutions

Solution 1

Multiples of 2: 9.

Multiples of 3: 6.

Multiples of both, meaning multiples of 6: 3.

Favorable = 9 + 6 − 3 = 12

P = 12/18 = 2/3

Solution 2

Favorable outcomes:

(5,6), (6,5)

P = 2/36 = 1/18

Solution 3

No 5:

5 × 5 = 25 outcomes

At least one 5:

36 − 25 = 11 outcomes

P = 11/36

Solution 4

Total unordered teams:

7 × 6 ÷ 2 = 21

Mixed teams:

4 × 3 = 12

P = 12/21 = 4/7

Solution 5

Ordered data:

3, 5, 7, 7, 8, 12

Sum = 42.

Mean = 42/6 = 7

Middle values are 7 and 7.

Median = 7

Mode = 7

Solution 6

Required total:

4 × 9 = 36

Known total:

4 + 8 + 10 = 22

x = 36 − 22 = 14

Solution 7

Combined total:

5 × 6 + 3 × 10 = 60

Combined count:

8

Combined mean = 60/8 = 7.5

Solution 8

Mean = 16/4 = 4

Squared deviations:

9, 1, 1, 9

Variance = 20/4 = 5

Standard deviation = √5

Solution 9

Original standard deviation:

√16 = 4

Therefore:

• New mean = 3 × 8 − 5 = 19.
• New range = 3 × 12 = 36.
• New variance = 3² × 16 = 144.
• New standard deviation = 3 × 4 = 12.

Solution 10

Expanded data:

2, 4, 4, 6

Mean = 4

Median = 4

Mode = 4

Squared deviations:

4, 0, 0, 4

Variance = 8/4 = 2

Solution 11

σ = √9 = 3

z = (46 − 40)/3 = 2

P(X > 46) ≈ 1 − 0.9772 = 0.0228

Solution 12

92 and 108 are equally far from the mean 100.

P(X > 108) = 0.15

P(92 < X < 108) = 1 − 0.15 − 0.15 = 0.70

Solution 13

σ = 5

The boundaries standardize to −1 and 1.

P(55 < X < 65) ≈ 2 × 0.8413 − 1 = 0.6826

The result follows from the supplied rounded probabilities.

Solution 14

The mean and standard deviation describe center and spread, but do not determine the distribution’s shape.

Standardization does not automatically create normality.

For example, a variable taking −1 and 1 with probability 1/2 each has mean 0 and standard deviation 1, but is discrete, not normally distributed.

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13. Integrated assessment — Probability & Statistics Mixed Quiz

Suggested time: 15–20 minutes
Questions: 12 single-answer multiple-choice questions
Marks: 1 per question
Suggested course mastery target: 10/12

Use variance with denominator n and the supplied normal probabilities.

Question 1

One card is selected uniformly from cards numbered 1 to 10.

What is the probability that the number is at least 8?

A. 1/5
B. 3/10
C. 4/5
D. 7/10

Question 2

A bag contains two red balls and three blue balls.

Two balls are drawn successively without replacement, uniformly from those remaining.

What is the probability that both are red?

A. 4/25
B. 2/5
C. 1/5
D. 1/10

Question 3

Two fair dice are rolled independently.

What is the probability that their sum is even?

A. 1/2
B. 1/3
C. 1/6
D. 2/3

Question 4

Find the median of:

8, 2, 10, 4, 6, 12

A. 6
B. 8
C. 7
D. 9

Question 5

Four observations have mean 7. A fifth observation of 12 is added.

What is the new mean?

A. 7.5
B. 8
C. 9.5
D. 12

Question 6

Find the variance of:

1, 3, 3, 5

A. 8
B. √2
C. 4
D. 2

Question 7

A dataset has variance 4. Every observation is transformed using:

y = −3x + 2

What is the new standard deviation?

A. 6
B. 12
C. 36
D. −6

Question 8

Group A contains two observations with mean 10.

Group B contains six observations with mean 18.

What is the combined mean?

A. 14
B. 15
C. 16
D. 28

Question 9

Given:

X ~ N(50, 16)

what is the z-score of 44?

A. −0.375
B. −1.5
C. 1.5
D. −6

Question 10

Given X ~ N(50, 16) and Φ(1.5) ≈ 0.9332, find P(X < 44).

A. 0.9332
B. 0.4332
C. 0.8664
D. 0.0668

Question 11

A normal distribution has mean 30.

Given P(24 < X < 36) = 0.80, find P(X > 36).

A. 0.10
B. 0.20
C. 0.40
D. 0.80

Question 12

Which statement is correct?

A. Every symmetric distribution is normal.

B. A dataset with a larger mean must have a larger variance.

C. Adding the same constant to every observation leaves the variance unchanged.

D. A normal variable has positive probability of equaling its mean exactly.

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14. Instructor answer key

Tutor LMS placement: Put the graded assessment in a separate Quiz item. Keep the answer key and explanations separate from student-facing questions until submission.

Question	Answer	Skill assessed
1	B	Classical probability and boundary wording
2	D	Sampling without replacement
3	A	Counting disjoint cases
4	C	Median
5	B	Updating a mean
6	D	Variance
7	A	Scaling standard deviation
8	C	Combined mean
9	B	Standardization
10	D	Negative normal boundary
11	A	Symmetry and tails
12	C	Conceptual interpretation