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 calculate P(B), we determine P(Z < -1.9):

Using the CDF:

$P(Z < -1.9) \approx 0.9713$ So,

$P(B) = 0.9713$

For event C, we need P(-1.5 < Z < 1.5):

Using the CDF:

$P(-1.5 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.5)$ From Z-tables:

$P(Z < 1.5) \approx 0.9332 \\ P(Z < -1.5) \approx 0.9332$ Thus,

$P(C) = 0.9332 - (1 - 0.9332) = 0.8664$

To find P(A ∪ C), we can use the formula:

$P(A ∪ C) = P(A) + P(C) - P(A ∩ C)$ Since A and C are independent, we find:

$P(A ∩ C) = P(A) * P(C)$ Calculating values:

(P(A) \approx 0.1357, \ P(C) \approx 0.8664$$
Thus,

$P(A ∪ C) = 0.1357 + 0.8664 - (0.1357 * 0.8664) \approx 0.9332$

Using the formula for conditional probability:

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

$P(X > w) = 0.625 * P(X > 28)$ Given that X is normally distributed with mean 21 and standard deviation 5, we first find:

$P(X > 28) = 1 - P(X \leq 28)$ Calculating:

$P(X \leq 28) = P(Z \leq \frac{28 - 21}{5}) = P(Z \leq 1.4) \approx 0.9192$ Thus:

$P(X > 28) \approx 0.0808$ Next, substituting back:

$P(X > w) = 0.625 * 0.0808 = 0.0500$ Now find w using the inverse CDF, leading to:

$w \approx 29.2$