Answer to Question #180729 in Statistics and Probability for Nathan Drake

Question

Let 𝑋~𝑁(40,144) and if 𝑌 = 2𝑋 − 1

Find the following probabilities:

(i) 𝑃(𝑋 ≤ 40)

(ii) 𝑃(𝑌 ≥ 50)

(iii)𝑃(35 ≤ 𝑋 ≤ 45)

(iv)𝑃(−45 ≤ 𝑌 ≤ 45)

(v)𝑃(−50 ≤ 𝑌 ≤ 100)

Answer 1

If 𝑋~𝑁(40,144) then the mathematical expectation is $a(X)=40$, variance is ${\sigma ^2}(X) = 144 \Rightarrow \sigma (X) = 12$ .

Then

$a(Y) = a(2X - 1) = 2a(X) - a(1) = 2 \cdot 40 - 1 = 79$

${\sigma ^2}(Y) = {\sigma ^2}(2X - 1) = {2^2}{\sigma ^2}(X) - {\sigma ^2}(1) = 4{\sigma ^2}(X) = 4 \cdot 144 = {\rm{576}} \Rightarrow \sigma (Y) = 24$

(i) Let's find

$P(X \le 40) = \Phi \left( {\frac{{\beta - a(X)}}{{\sigma (X)}}} \right) - \Phi \left( { - \infty } \right) = \Phi \left( {\frac{{40 - 40}}{{12}}} \right) + 0.5 = 0 + 0.5 = 0.5$

Answer: $P(X \le 40) = 0.5$

(ii) $P(Y \ge 50) = \Phi \left( \infty \right) - \Phi \left( {\frac{{\alpha - a(Y)}}{{\sigma (Y)}}} \right) = 0.5 - \Phi \left( {\frac{{50 - 79}}{{24}}} \right) = 0.5 + \Phi \left( {1.2} \right) = 0.5 + 0.3849 = 0.8849$

Answer: $P(Y \ge 50) = 0.8849$

(iii) $P(35 \le X \le 45) = \Phi \left( {\frac{{\beta - a(X)}}{{\sigma (X)}}} \right) - \Phi \left( {\frac{{\alpha - a(X)}}{{\sigma (X)}}} \right) = \Phi \left( {\frac{{45 - 40}}{{12}}} \right) + \Phi \left( {\frac{{35 - 40}}{{12}}} \right) = \Phi (0.42) + \Phi (0.42) = 2 \cdot 0.1628 = 0.3256$

Answer: $P(35 \le X \le 45) = 0.3256$

(iv) $P( - 45 \le Y \le 45) = \Phi \left( {\frac{{\beta - a(Y)}}{{\sigma (Y)}}} \right) - \Phi \left( {\frac{{\alpha - a(Y)}}{{\sigma (Y)}}} \right) = \Phi \left( {\frac{{45 - 79}}{{24}}} \right) - \Phi \left( {\frac{{ - 45 - 79}}{{24}}} \right) = - \Phi (1.42) + \Phi (5.17) = - 0.4222 + 0.5 = 0.0778$

Answer: $P( - 45 \le Y \le 45) = 0.0778$

(v) $P( - 50 \le Y \le 100) = \Phi \left( {\frac{{100 - 79}}{{24}}} \right) - \Phi \left( {\frac{{ - 50 - 79}}{{24}}} \right) = \Phi (0.875) + \Phi (5.375) = 0.3092 + 0.5 = 0.8092$

Answer: $P( - 50 \le Y \le 100) = 0.8092$