Supplied standard normal cumulative probabilities: Φ(z) = P(Z ≤ z)

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

These values support basic interpretation. Memorizing an extensive normal table is not required for this lesson.

Rounding convention: Use the supplied four-decimal values consistently. Results may differ slightly from calculations using more precise probabilities.

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8. Right-tail probabilities

The total area is 1.

Therefore:

P(Z > z) = 1 − Φ(z)

Worked example 4

Find P(Z > 1.5), using:

Φ(1.5) ≈ 0.9332

Then:

P(Z > 1.5) ≈ 1 − 0.9332 = 0.0668

Answer: Approximately 6.68%.

Verification

Since 1.5 is above the mean, the probability to its right must be less than 0.5.

The result 0.0668 satisfies this check.

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9. Negative z-values and symmetry

The standard normal curve is symmetric about zero.

For a > 0:

P(Z < −a) = P(Z > a)

Therefore:

Φ(−a) = 1 − Φ(a)

Worked example 5

Find P(Z < −1), using:

Φ(1) ≈ 0.8413

By symmetry:

P(Z < −1) = P(Z > 1)

Therefore:

P(Z < −1) ≈ 1 − 0.8413 = 0.1587

A related probability

To find P(Z > −1):

P(Z > −1) ≈ 1 − 0.1587 = 0.8413

This is greater than 0.5 because −1 lies to the left of the mean.

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10. Probabilities between two values

For a < b:

P(a < Z < b) = Φ(b) − Φ(a)

Subtract the area to the left of a from the area to the left of b.

Worked example 6: Both endpoints positive

Find P(0.5 < Z < 1.5).

Use:

Φ(0.5) ≈ 0.6915

Φ(1.5) ≈ 0.9332

Therefore:

P(0.5 < Z < 1.5) ≈ 0.9332 − 0.6915 = 0.2417

Worked example 7: An interval crossing zero

Find P(−1 < Z < 2).

First:

Φ(−1) ≈ 1 − 0.8413 = 0.1587

Then:

P(−1 < Z < 2) ≈ 0.9772 − 0.1587 = 0.8185

Verification by splitting at zero

Area from −1 to 0:

0.5000 − 0.1587 = 0.3413

Area from 0 to 2:

0.9772 − 0.5000 = 0.4772

Total:

0.3413 + 0.4772 = 0.8185

Both methods agree.

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11. Symmetric central regions and two tails

For a > 0:

P(−a < Z < a) = 2Φ(a) − 1

The combined probability outside that interval is:

P(Z < −a or Z > a) = 2[1 − Φ(a)]

Why these formulas work

The two tails have equal areas.

Each tail has probability:

1 − Φ(a)

Subtract both tails from 1 to obtain the central probability.

Worked example 8

Find:

1. P(−2 < Z < 2).
2. P(Z < −2 or Z > 2).

Use:

Φ(2) ≈ 0.9772

Each tail:

1 − 0.9772 = 0.0228

Both tails:

2 × 0.0228 = 0.0456

Central probability:

1 − 0.0456 = 0.9544

Answers:

• Central region: approximately 0.9544.
• Both tails: approximately 0.0456.

Rounding note

More precise probabilities give a central value of approximately 0.9545. The value 0.9544 here follows consistently from the supplied rounded Φ(2) = 0.9772.

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12. Probability-method reference

Requested region	Calculation
Left of z	Φ(z)
Right of z	1 − Φ(z)
Between a and b, where a < b	Φ(b) − Φ(a)
Left of −a, where a > 0	1 − Φ(a)
Between −a and a, where a > 0	2Φ(a) − 1
Outside −a to a, where a > 0	2[1 − Φ(a)]

Choose the region first. Then choose the calculation.

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13. Worked example 9: A probability in original units

Component lengths are modeled by:

X ~ N(100, 16)

Find the probability that a randomly selected component is longer than 106.

Step 1: Identify the parameters

μ = 100

σ = √16 = 4

Step 2: Standardize the boundary

z = (106 − 100)/4 = 1.5

Therefore:

P(X > 106) = P(Z > 1.5)

Step 3: Use the supplied probability

P(Z > 1.5) ≈ 1 − 0.9332 = 0.0668

Answer: Approximately 6.68%.

Interpretation

Under this model, approximately 6.68% of component lengths lie above 106.

This does not guarantee exactly that percentage in every finite sample.

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14. Worked example 10: An interval in original units

Suppose:

X ~ N(50, 100)

Find P(45 < X < 65).

Step 1: Find the standard deviation

σ = √100 = 10

Step 2: Standardize both endpoints

Lower:

(45 − 50)/10 = −0.5

Upper:

(65 − 50)/10 = 1.5

Therefore:

P(45 < X < 65) = P(−0.5 < Z < 1.5)

Step 3: Find the cumulative probabilities

Φ(−0.5) ≈ 1 − 0.6915 = 0.3085

Φ(1.5) ≈ 0.9332

Step 4: Subtract

P ≈ 0.9332 − 0.3085 = 0.6247

Answer: Approximately 62.47%.

The original interval is not symmetric about 50, so its outside tails are not equal.

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15. Comparing relative positions

A raw difference from the mean does not account for the dataset’s spread.

A z-score does.

Worked example 11

Two normally modeled assessments have:

• Assessment A: mean 60, standard deviation 5.
• Assessment B: mean 80, standard deviation 10.

A student scores:

• 70 on A.
• 95 on B.

Compare the student’s standardized positions.

For A:

z = (70 − 60)/5 = 2

For B:

z = (95 − 80)/10 = 1.5

The student is farther above the mean, in standard-deviation units, on Assessment A.

This compares relative positions within the two models. It does not establish that the assessments measure identical skills or have identical grading standards.

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16. Common mistakes and verification

Mistake 1: Dividing by variance

For X ~ N(50, 16), use σ = 4 in the z-score formula.

Mistake 2: Reversing the subtraction

Use:

z = (x − μ)/σ

A value below the mean should produce a negative z-score.

Mistake 3: Treating Φ(z) as a right-tail probability

In this lesson:

Φ(z) means area to the left.

Mistake 4: Forgetting to standardize one endpoint

Both interval boundaries must be converted.

Mistake 5: Subtracting cumulative probabilities in reverse order

For a < b:

Φ(b) − Φ(a)

The result must be nonnegative.

Mistake 6: Splitting unequal tails equally

Equal tails require boundaries reflected about the mean.

Mistake 7: Applying normal probabilities without a normal model

Standardizing nonnormal data gives standardized values, but does not automatically make their distribution normal.

Mistake 8: Mixing rounded reference values

Use the supplied values consistently and mark the final result as approximate.

Verification checklist

• Is normality stated?
• Have I taken the square root of the variance?
• Does the z-score sign match the value’s position?
• Am I finding a left tail, right tail, interval, or two tails?
• Have I standardized every boundary?
• Does the answer lie between 0 and 1?
• Does the region appear larger or smaller than half the curve?
• Have I used approximate notation appropriately?

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17. Independent practice

Use the supplied cumulative probabilities from Section 7.

Practice 1

For X ~ N(30, 9), find the z-scores of:

1. 36.  
2. 27.  

Practice 2

A normal variable has mean 80 and standard deviation 4.

Find x when z = −1.5.

Practice 3

For X ~ N(100, 25), express P(95 < X < 110) using Z.

Practice 4

Find P(Z > 0.5).

Practice 5

Find P(Z < −1.5).

Practice 6

Find P(0 < Z < 1).

Practice 7

Find P(−1 < Z < 1).

Practice 8

Find P(−0.5 < Z < 2).

Practice 9

For X ~ N(60, 16), find P(X > 64).

Practice 10

For X ~ N(20, 4), find P(18 < X < 23).

Practice 11

For X ~ N(50, 25), find P(X < 40 or X > 60).

Practice 12

A student obtains 84 on an assessment with mean 72 and standard deviation 6.

Another student obtains 95 on an assessment with mean 80 and standard deviation 10.

Find both z-scores and compare their positions above their respective means.

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18. Practice solutions

Solution 1

μ = 30, σ = 3

For 36:

z = (36 − 30)/3 = 2

For 27:

z = (27 − 30)/3 = −1

Solution 2

x = μ + zσ

x = 80 + (−1.5)(4) = 74

Solution 3

σ = 5

Lower boundary:

(95 − 100)/5 = −1

Upper boundary:

(110 − 100)/5 = 2

P(95 < X < 110) = P(−1 < Z < 2)

Solution 4

P(Z > 0.5) ≈ 1 − 0.6915 = 0.3085

Solution 5

By symmetry:

P(Z < −1.5) ≈ 1 − 0.9332 = 0.0668

Solution 6

P(0 < Z < 1) ≈ 0.8413 − 0.5000 = 0.3413

Solution 7

P(−1 < Z < 1) ≈ 2 × 0.8413 − 1 = 0.6826

This follows from the rounded supplied value.

Solution 8

Φ(−0.5) ≈ 1 − 0.6915 = 0.3085

P(−0.5 < Z < 2) ≈ 0.9772 − 0.3085 = 0.6687

Solution 9

σ = 4

z = (64 − 60)/4 = 1

P(X > 64) ≈ 1 − 0.8413 = 0.1587

Solution 10

σ = 2

Lower boundary:

(18 − 20)/2 = −1

Upper boundary:

(23 − 20)/2 = 1.5

P ≈ 0.9332 − 0.1587 = 0.7745

Solution 11

σ = 5

The boundaries standardize to −2 and 2.

P(X < 40 or X > 60) ≈ 2 × (1 − 0.9772)

P ≈ 0.0456

Solution 12

First student:

z = (84 − 72)/6 = 2

Second student:

z = (95 − 80)/10 = 1.5

The first student is farther above the relevant mean in standard-deviation units.

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19. Lesson assessment — Standard Normal Interpretation Quiz

Suggested time: 10–12 minutes
Questions: 6 single-answer multiple-choice questions
Marks: 1 per question
Suggested course mastery target: 5/6

Use the supplied cumulative probabilities. No extensive normal table is required.

Question 1

Given X ~ N(40, 16), what is the z-score of 48?

A. 0.5
B. 2
C. 4
D. 8

Question 2

A normal variable has mean 25 and standard deviation 3.

Which value corresponds to z = −2?

A. 31
B. 23
C. 7
D. 19

Question 3

Given Φ(1) ≈ 0.8413, find P(Z > 1).

A. 0.1587
B. 0.3413
C. 0.8413
D. 0.6826

Question 4

Given Φ(1.5) ≈ 0.9332, find P(−1.5 < Z < 1.5).

A. 0.0668
B. 0.1336
C. 0.8664
D. 0.9332

Question 5

Given X ~ N(70, 25), which standard normal event corresponds to 65 < X < 80?

A. −5 < Z < 10

B. −1 < Z < 2

C. −0.2 < Z < 0.4

D. 1 < Z < 2

Question 6

Given X ~ N(100, 100) and Φ(0.5) ≈ 0.6915, find P(X < 95).

A. 0.6915
B. 0.1915
C. 0.5000
D. 0.3085

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20. Instructor answer key

Tutor LMS placement: Put the assessment in a separate Quiz item if it will be graded. Keep the answer key and explanations out of the student-facing questions until submission.

Question	Answer	Skill assessed
1	B	Standardizing a value
2	D	Recovering an original value
3	A	Right-tail probability
4	C	Symmetric central probability
5	B	Standardizing both interval endpoints
6	D	Negative z-value and symmetry