Normal Distribution | Maths Teaching - Free Printable

We are given four normal distribution problems, each with a diagram and a probability condition. We need to find the value of $ x $ in each case using properties of the normal distribution.

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(a) $ X \sim N(60, 25) $, $ P(X < x) = 0.9972 $

- Mean $ \mu = 60 $
- Variance $ \sigma^2 = 25 $ → Standard deviation $ \sigma = \sqrt{25} = 5 $

We want $ P(X < x) = 0.9972 $. This means we're looking for the value $ x $ such that the cumulative probability is 0.9972.

#### Step 1: Find the z-score corresponding to $ P(Z < z) = 0.9972 $

Using standard normal tables or calculator:

$$
z \approx 2.73
$$

(You can verify: $ P(Z < 2.73) \approx 0.9968 $, $ P(Z < 2.74) \approx 0.9969 $, so slightly higher — actually, $ z \approx 2.73 $ is close; more precisely, use interpolation or calculator.)

Actually, let's use a better approximation:
- $ P(Z < 2.73) = 0.9968 $
- $ P(Z < 2.74) = 0.9969 $
- $ P(Z < 2.75) = 0.9970 $
- $ P(Z < 2.76) = 0.9971 $
- $ P(Z < 2.77) = 0.9972 $

So $ z = 2.77 $

#### Step 2: Convert back to $ x $

$$
x = \mu + z\sigma = 60 + (2.77)(5) = 60 + 13.85 = 73.85
$$

✔ Answer (a): $ x = 73.85 $

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(b) $ X \sim N(5, 4/9) $, shaded area left of $ x $ is 0.3

- $ \mu = 5 $
- $ \sigma^2 = \frac{4}{9} $ → $ \sigma = \frac{2}{3} \approx 0.6667 $

We are told $ P(X < x) = 0.3 $

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

From standard normal table:
- $ P(Z < -0.52) \approx 0.3015 $
- $ P(Z < -0.53) \approx 0.2981 $

Interpolating: $ z \approx -0.524 $

So $ z \approx -0.524 $

Now convert to $ x $:

$$
x = \mu + z\sigma = 5 + (-0.524)\left(\frac{2}{3}\right) = 5 - 0.524 \times 0.6667 \approx 5 - 0.35 = 4.65
$$

Let’s compute exactly:

$$
0.524 \times \frac{2}{3} = \frac{1.048}{3} \approx 0.3493
$$

So $ x \approx 5 - 0.3493 = 4.6507 $

✔ Answer (b): $ x \approx 4.65 $

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(c) $ X \sim N(200, 36) $, $ P(X > x) = 0.9386 $? Wait! Look at diagram

Wait — the shaded region is from $ x $ to the right, but it says 0.9386 on the shaded part.

But if it's shaded to the right, and $ P(X > x) = 0.9386 $, that would mean $ x $ is very small — but that contradicts the diagram.

Wait: look carefully.

The diagram shows:
- The shaded area is from $ x $ to the right, and labeled as 0.9386
- But this can't be correct because the total area under the curve is 1, and if the right tail is 0.9386, then $ x $ would be far to the left, but the shaded region starts at $ x $, goes right, and covers most of the curve.

Wait — actually, if the shaded area is to the right of $ x $ and is 0.9386, that means:

$$
P(X > x) = 0.9386 \Rightarrow P(X < x) = 1 - 0.9386 = 0.0614
$$

So we need $ P(X < x) = 0.0614 $

This makes sense: $ x $ is to the left of the mean, and only 6.14% of data is below it.

Now:

- $ \mu = 200 $
- $ \sigma^2 = 36 $ → $ \sigma = 6 $

Find $ z $ such that $ P(Z < z) = 0.0614 $

From standard normal table:
- $ P(Z < -1.54) = 0.0618 $
- $ P(Z < -1.55) = 0.0606 $

Interpolate:

$ 0.0614 $ is between 0.0606 and 0.0618

Difference: $ 0.0614 - 0.0606 = 0.0008 $, total range = 0.0012

So $ z \approx -1.55 + (0.0008 / 0.0012)(0.01) = -1.55 + 0.0067 \approx -1.5433 $

So $ z \approx -1.543 $

Now compute $ x $:

$$
x = \mu + z\sigma = 200 + (-1.543)(6) = 200 - 9.258 = 190.742
$$

✔ Answer (c): $ x \approx 190.74 $

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(d) $ X \sim N(0, 4) $, shaded area to the right of $ x $ is 23.15%

- $ \mu = 0 $
- $ \sigma^2 = 4 $ → $ \sigma = 2 $
- $ P(X > x) = 23.15\% = 0.2315 $

So $ P(X < x) = 1 - 0.2315 = 0.7685 $

Find $ z $ such that $ P(Z < z) = 0.7685 $

Look up in standard normal table:

- $ P(Z < 0.73) = 0.7673 $
- $ P(Z < 0.74) = 0.7704 $

Interpolate:

$ 0.7685 - 0.7673 = 0.0012 $

$ 0.7704 - 0.7673 = 0.0031 $

So $ z \approx 0.73 + (0.0012 / 0.0031)(0.01) \approx 0.73 + 0.0039 = 0.7339 $

So $ z \approx 0.734 $

Now:

$$
x = \mu + z\sigma = 0 + (0.734)(2) = 1.468
$$

✔ Answer (d): $ x \approx 1.47 $

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✔ Final Answers:

(a) $ x = 73.85 $
(b) $ x \approx 4.65 $
(c) $ x \approx 190.74 $
(d) $ x \approx 1.47 $

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Let me know if you'd like these rounded to specific decimal places or verified with exact z-table values.