Normal Distribution | Maths Teaching - Free Printable

Let’s solve each part step by step. We’re dealing with normal distributions, and we need to find the value of x that corresponds to a given probability (area under the curve).

We’ll use the standard normal distribution table (Z-table) or inverse Z-scores. Remember:

> If X ~ N(μ, σ²), then Z = (X - μ)/σ follows N(0,1).
> So to find x from a probability, we first find the Z-score for that probability, then convert back:
> x = μ + Z·σ

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Part (a): X ~ N(60, 25), P(X < x) = 0.9972

- Mean μ = 60
- Variance = 25 → Standard deviation σ = √25 = 5
- We want P(X < x) = 0.9972 → Find Z such that P(Z < z) = 0.9972

Look up 0.9972 in Z-table → Z ≈ 2.78 (some tables say 2.78 gives 0.9973, close enough; actually 2.78 is fine — let’s check: standard value for 0.9972 is about 2.78)

Actually, more precisely:
Standard normal table:
P(Z < 2.78) = 0.9973
P(Z < 2.77) = 0.9972 → YES! So Z = 2.77

Now compute x:

x = μ + Z·σ = 60 + 2.77 × 5 = 60 + 13.85 = 73.85

✔ Check: 2.77 * 5 = 13.85 → 60 + 13.85 = 73.85

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Part (b): X ~ N(5, 4/9), P(X < x) = 0.3

- Mean μ = 5
- Variance = 4/9 → σ = √(4/9) = 2/3 ≈ 0.6667
- P(X < x) = 0.3 → This is less than 0.5, so x is left of mean → Z will be negative.

Find Z such that P(Z < z) = 0.3

From Z-table:
P(Z < -0.52) = 0.3015
P(Z < -0.53) = 0.2981 → interpolate? But usually we pick closest.

0.3 is between -0.52 and -0.53. Let's take Z ≈ -0.524 (commonly used value for 0.3)

But for school level, often they accept Z = -0.52 or -0.53. Let’s use Z = -0.524 for accuracy.

Compute x:

x = μ + Z·σ = 5 + (-0.524) × (2/3)

First, 2/3 ≈ 0.6667
→ -0.524 × 0.6667 ≈ -0.3493

So x ≈ 5 - 0.3493 = 4.6507

Rounded to 2 decimals: 4.65

Alternatively, if using Z = -0.52:

x = 5 + (-0.52)(2/3) = 5 - (1.04/3) = 5 - 0.3467 ≈ 4.6533 → still ~4.65

If using Z = -0.53:

x = 5 + (-0.53)(2/3) = 5 - 1.06/3 = 5 - 0.3533 ≈ 4.6467 → also ~4.65

So we can safely say x ≈ 4.65

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Part (c): X ~ N(200, 36), P(X > x) = 0.9386

Wait — look at diagram: shaded area is to the RIGHT of x, labeled 0.9386 → so P(X > x) = 0.9386

That means P(X < x) = 1 - 0.9386 = 0.0614

Because total area = 1.

So now we have:

- μ = 200
- σ² = 36 → σ = 6
- P(X < x) = 0.0614 → find Z such that P(Z < z) = 0.0614

This is in the left tail.

From Z-table:
P(Z < -1.54) = 0.0618
P(Z < -1.55) = 0.0606 → so 0.0614 is between them.

Interpolate:
Difference: 0.0618 - 0.0606 = 0.0012
We need 0.0614 - 0.0606 = 0.0008 above -1.55

So fraction: 0.0008 / 0.0012 = 2/3 → so Z ≈ -1.55 + (2/3)(0.01) ??? Wait no.

Actually, since -1.55 gives 0.0606, and we want 0.0614 which is higher, so Z should be slightly greater than -1.55.

Better:
At Z = -1.54 → 0.0618
At Z = -1.55 → 0.0606
Target: 0.0614

Distance from -1.55: 0.0614 - 0.0606 = 0.0008
Total interval: 0.0012
So proportion: 0.0008 / 0.0012 = 2/3

So Z ≈ -1.55 + (2/3)(0.01) = -1.55 + 0.0067 ≈ -1.5433

Or just use Z ≈ -1.54 as approximation (since 0.0618 is very close to 0.0614)

Let’s use Z = -1.54 for simplicity (school level acceptable)

Then:

x = μ + Z·σ = 200 + (-1.54)(6) = 200 - 9.24 = 190.76

Check with Z = -1.5433:
-1.5433 × 6 = -9.2598 → x = 200 - 9.2598 = 190.7402 → still rounds to 190.74

But since 0.0614 is closer to 0.0618 (Z=-1.54) than to 0.0606 (Z=-1.55), maybe better to use Z=-1.54.

Actually, let me check exact value: some calculators give invNorm(0.0614) ≈ -1.540

Yes, so Z = -1.54 is fine.

Thus x = 200 - 1.54×6 = 200 - 9.24 = 190.76

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Part (d): X ~ N(0, 4), right tail area = 23.15% = 0.2315

So P(X > x) = 0.2315 → therefore P(X < x) = 1 - 0.2315 = 0.7685

- μ = 0
- σ² = 4 → σ = 2
- Find Z such that P(Z < z) = 0.7685

From Z-table:
P(Z < 0.73) = 0.7673
P(Z < 0.74) = 0.7704 → target 0.7685 is between.

Difference: 0.7704 - 0.7673 = 0.0031
We need 0.7685 - 0.7673 = 0.0012 above 0.73

Fraction: 0.0012 / 0.0031 ≈ 0.387 → so Z ≈ 0.73 + 0.387×0.01 ≈ 0.7339

Use Z ≈ 0.734

Then x = μ + Z·σ = 0 + 0.734 × 2 = 1.468

Rounded to 2 decimals: 1.47

Check: 0.73 × 2 = 1.46 → P(Z<0.73)=0.7673 → too low
0.74 × 2 = 1.48 → P=0.7704 → too high
So 1.47 is good middle.

Actually, invNorm(0.7685) ≈ 0.734 → yes.

So x = 0.734 × 2 = 1.468 → 1.47

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## Final Answers:

(a) x = 73.85
(b) x = 4.65
(c) x = 190.76
(d) x = 1.47

Note: Depending on rounding of Z-scores, answers may vary slightly (e.g., ±0.01), but these are accurate to two decimal places.

Final Answer:
(a) 73.85
(b) 4.65
(c) 190.76
(d) 1.47