The random variable Z ~ N(0, 1) A is the event Z > 1.1 B is the event Z < -1.9 C is the event -1.5 < Z < 1.5 (a) Find (i) P(A) (ii) P(B) (iii) P(C) (iv) P(A ∪ C) The random variable X has a normal distribution with mean 21 and standard deviation 5 (b) Find the value of w such that P(X > w | X > 28) = 0.625 - Edexcel - A-Level Maths Statistics - Question 6 - 2015 - Paper 1

To find P(B), we compute the probability that Z is less than -1.9:

Using the properties of the standard normal distribution:

$P(Z < -1.9)$

From the table: $P(Z < -1.9) \\approx 0.9713$

So,
$P(B) = 0.9713$

To find P(C), we calculate the probability that Z lies between -1.5 and 1.5:

$P(-1.5 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.5)$

From the normal distribution tables:

• $P(Z < 1.5) \\approx 0.9332$
• $P(Z < -1.5) \\approx 0.0668$

Therefore,
$P(C) = 0.9332 - 0.0668 = 0.8664$

To find the probability of the union of events A and C:

$P(A ∪ C) = P(A) + P(C) - P(A ∩ C)$

Since A and C are independent:

• We can compute: $P(A ∩ C) = P(A) * P(C)$

So:

• Using earlier calculations: $P(A ∪ C) = P(A) + P(C) - P(A) * P(C)$

Substituting the values:

• $P(A ∪ C) = 0.1357 + 0.8664 - (0.1357 * 0.8664)$

Calculating this: $P(A ∪ C) = 0.9332$

To solve for w, we use the formula for conditional probability:

$P(X > w | X > 28) = \frac{P(X > w \cap X > 28)}{P(X > 28)}$

This simplifies to:

$P(X > w) = P(X > 28) * 0.625$

Next, we compute: $P(X > 28) = 1 - P(X < 28)$

Using the normal distribution properties:

• Mean = 21, Standard Deviation = 5

Then calculate the Z-score: $Z = \frac{28 - 21}{5} = 1.4$

Using a standard normal table:

• $P(Z < 1.4) \approx 0.9192$
• Thus,
  $P(X > 28) = 1 - 0.9192 \\approx 0.0808$

Substituting back: $P(X > w) = 0.0808 * 0.625 = 0.0505$

We can now obtain w from: $w = 21 + 5 * Z^{-1}(0.0505)$

Finally,

• Approximate this using the Z-value for 0.0505.
• Solving gives us:
  w ≈ 20.2