Question

2. Let Xi, X2, and X represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent normal random variables with expected values μι μ2, and ,13 and variances σ σ and σ3 respectively. a. If,-μ2-13-65 and ol-02-03-20, calculate P(Xi + X2 +X3 210). What is P(150 sXX23210)? Using the μ's and σ's given in part (a), calculate POX259)and P(62 SX568) Using the μ's and ơi's given inpart (a), calculate P(-10 Xi-5X2-5X3 5). IfAi-40, μ2-50, μ3-60, σ1-10, σ2-12, and σ)-14, calculate P(Xi +X2 + X3 160) and P(Xi +X22213). b. c.

Answer 1

We use the fact that the linear combination of independent normal random variables is also a normal random variable.  

a) Mean of $S=X_1+X_2+X_3$ is $65+65+65=195$ and variance is $20+20+20=60$

Define the standard random variable Z as $z={S-195\over \sqrt{60}}$

using normal table required probability we get

$P(S\leq 210)=P\left ( z\leq {210-195\over \sqrt{60}} \right )\\ ~~~~~~~~~~~~~~~~~~~~~=P(z\leq 1.94)\\ ~~~~~~~~~~~~~~~~~~~~~=0.9738$

$P(150\leq S\leq 210)=P\left ( {150-195\over \sqrt{60}}\leq z\leq {210-195\over \sqrt{60}} \right )\\ ~~~~~~~~~~~~~~~~~~~~~=P(-5.81\leq z\leq 1.94)\\ ~~~~~~~~~~~~~~~~~~~~~=0.9738$

b) The mean of $\bar{X}$ is 65 and variance is ${20\over 3}$

Then

Define the standard random variable Z as $z={\bar{X}-65\over \sqrt{{20\over 3}}}$

using normal table required probability we get

$P(\bar{X}\geq 59)=P\left (z\geq {\sqrt{3}(59-65)\over \sqrt{{20}}} \right )\\ ~~~~~~~~~~~~~=P(z\geq-2.32 )\\ ~~~~~~~~~~~~~=0.9898$

$P(62\leq \bar{X}\leq 68)=P\left ({\sqrt{3}(62-65)\over \sqrt{{20}}}z\leq {\sqrt{3}(68-65)\over \sqrt{{20}}} \right )\\ ~~~~~~~~~~~~~=P(-1.16\leq z\geq1.16 )\\ ~~~~~~~~~~~~~=0.754$

c) mean of $S_1=X_1-0.5X_2-0.5X_3$ is $65-0.5(65)-0.5(65)=0$ variance is $20+0.25(20)+0.25(20)=30$

Define standard random variable z as $z={S_1-0\over \sqrt{30}}$

Then using Normal table as done above gives

$P(-10\leq X_1-0.5X_2-0.5X_3\leq 5)=0.7854$

d) Using part d) gives

$P( X_1+X_2+X_3\leq 160)=P(z\leq -4.52)=0.0000$

using the same procedure we get

$P( X_1+X_2\geq 2X_3)=0.50$