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 calculate:

$P(B) = P(Z < -1.9)$ Using the standard normal distribution table, we find:

$P(Z \leq -1.9) \approx 0.0287$ Therefore,

$P(B) \approx 0.0287$

To find P(C), we can calculate:

$P(C) = P(-1.5 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.5)$ Using the standard normal distribution:

$P(Z < 1.5) \approx 0.9332$ and $P(Z < -1.5) \approx 0.0668$, Set the equation:

$P(C) \approx 0.9332 - 0.0668 = 0.8664$

To find P(A ∪ C), we use the principle of inclusion-exclusion:

$P(A ∪ C) = P(A) + P(C) - P(A ∩ C)$ We assume that A and C are independent events:

$P(A ∩ C) = P(A)P(C)$ Now substituting the values:

$P(A ∪ C) \approx 0.1357 + 0.8664 - (0.1357 \times 0.8664)$ Calculating gives us:

$P(A ∪ C) \approx 0.9332$

For a normal distribution, we can use the property of conditional probability:

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

$P(X > w | X > 28) = \frac{P(X > w)}{P(X > 28)} = 0.625$ Calculating:

1. Find P(X > 28):

   • With mean 21 and standard deviation 5:
   • Standardize: $Z = \frac{28 - 21}{5} = 1.4$
   • Obtain from Z-table, $P(Z > 1.4) \approx 0.0808$

2. Substitute into the equation to find w:

   • $P(X > w) = 0.625 \cdot P(X > 28) \approx 0.625 \cdot 0.0808 = 0.0506$
   • From the Z-table, we find w for different probabilities, concluding with:

$w = 21 + 1.64 \cdot 5 = 29.2$