91Ó°ÊÓ

The standard deviation is a statistic that measures the amount of variation or dispersion in a set of numbers. A low standard deviation implies that the values are close to the set's mean (also known as the anticipated value), whereas a high standard deviation shows that the values are spread out over a larger range.

The random variable \({T_0}\)is defined as

\({T_0} = {X_1} + {X_2} + {X_3}\)

The mean value of the random variable \({T_0}\)is

\(\begin{aligned}E\left( {{T_0}} \right) & = E\left( {{X_1} + {X_2} + {X_3}} \right)\\ &= E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + E\left( {{X_3}} \right)\\ &= 60 + 60 + 60\\ &= 180\end{aligned}\)

the variance is

\(\begin{aligned}E\left( {{T_0}} \right) &= V\left( {{X_1} + {X_2} + {X_3}} \right)\\ &= V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + V\left( {{X_3}} \right)\\ &= 15 + 15 + 15\\ &= 45\end{aligned}\)

(1): the random variables are independent therefore the equality strands.

Finally, the standard deviation of random variable \({T_0}\)is

\({\sigma _{{T_0}}} = \sqrt {V\left( {{T_0}} \right)} = \sqrt {45} = 6.708\)

The first requested probability is

\(\begin{aligned}P\left( {{T_0} \le 200} \right) &= P\left( {{X_1} + {X_2} + {X_3} \le 200} \right)\\ &= P\left( {\frac{{{X_1} + {X_2} + {X_3} - {\mu _{{T_0}}}}}{{{\sigma _{{T_0}}}}} \le \frac{{200 - 180}}{{6.708}}} \right)\\ &= P(Z \le 2.98)\\ &= 0.9986\end{aligned}\)

(2): from the normal probability table in the appendix. The probability can also be computed with a software.

The second requested probability is

\(\begin{aligned}P\left( {150 \le {T_0} \le 200} \right) &= P\left( {\frac{{150 - 180}}{{6.708}} \le \frac{{{X_1} + {X_2} + {X_3} - {\mu _{{T_0}}}}}{{{\sigma _{{T_0}}}}} \le \frac{{200 - 180}}{{6.708}}} \right)\\ &= P( - 4.47 \le Z \le 2.98)\\ &= P(Z \le 2.98) - P(Z \le 4.47)\\ &= 0.9986\end{aligned}\)

(2): from the normal probability table in the appendix. The probability can also be computed with a software.

The mean value of sample average \(\bar X\)is

\(\mu \bar X = \mu = 60\)

and the standard deviation of sample average \(\bar X\)is

\({\sigma _{\bar X}} = \frac{1}{{\sqrt n }} \cdot \sigma = \frac{1}{{\sqrt 3 }} \cdot \sqrt {15} = 2.236\)

Using this, compute the probabilities as

\(\begin{aligned}P(\bar X \ge 55) &= P\left( {\frac{{\bar X - \mu \bar X}}{{{\sigma _{\bar X}}}} \ge \frac{{55 - 60}}{{2.236}}} \right)\\ &= P(Z \ge - 2.236)\\ &= 1 - P(Z < - 2.236)\\ &= 0.9875\end{aligned}\)

similarly, the other probability is

\(P(58 \le \bar X \le 62) = P( - 0.89 \le Z \le 0.89) = 0.6266\)

where we used the same method as earlier.

First find the expected value and the standard deviation of random variable

\({T_1} = {X_1} - 0.5{X_2} - 0.5{X_3}\)

Mean value of random variable \({T_1}\)is

\(\begin{aligned}E\left( {{T_1}} \right) &= E\left( {{X_1} - 0.5{X_2} - 0.5{X_3}} \right)\\ &= E\left( {{X_1}} \right) - 0.5E\left( {{X_2}} \right) - 0.5E\left( {{X_3}} \right)\\ &= 60 - 0.5 \cdot 60 - 0.5 \cdot 60\\ &= 0\end{aligned}\)

the variance is

\(\begin{aligned}E\left( {{T_1}} \right) &= V\left( {{X_1} - 0.5{X_2} - 0.5{X_3}} \right)\\ &= V\left( {{X_1}} \right) + {( - 0.5)^2}V\left( {{X_2}} \right) + {( - 0.5)^2}V\left( {{X_3}} \right)\\ &= 15 + {( - 0.5)^2} \cdot 15 + {( - 0.5)^2} \cdot 15\\ &= 22.5\end{aligned}\)

(1): the random variables are independent therefore the equality stands.

Finally, the standard deviation of random variable \({T_1}\)is

\({\sigma _{{T_1}}} = \sqrt {V\left( {{T_1}} \right)} = \sqrt {22.5} = 4.7434\)

Using the same method as above. subtracting the expected values and dividing with standard deviation we get

\(\begin{aligned}P\left( { - 10 \le {T_1} \le 5} \right) &= P\left( {\frac{{ - 10 - 0}}{{4.7434}} \le Z \le \frac{{5 - 0}}{{4.7434}}} \right)\\ &= P( - 2.11 \le Z \le 1.05)\\ &= P(Z \le 1.05) - P(Z \le - 2.11)\\ &= 0.8531 - 0.0174\\ &= 0.8357\end{aligned}\)

Given different expectations and variances the mean value of \({T_0}\)and standard deviation of \({T_0}\)would change.

Mean value of random variable \({T_0}\)is

\(\begin{aligned}E\left( {{T_0}} \right) &= E\left( {{X_1} + {X_2} + {X_3}} \right)\\ &= E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + E\left( {{X_3}} \right)\\ &= 40 + 50 + 60\\ &= 150\end{aligned}\)

the variance is

\(\begin{aligned}E\left( {{T_0}} \right) &= V\left( {{X_1} + {X_2} + {X_3}} \right)\\ &= V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + V\left( {{X_3}} \right)\\ &= 10 + 12 + 14\\ &= 36\end{aligned}\)

(1): the random variables are independent therefore the equality stands.

Finally, the standard deviation of random variable \({T_0}\)is

\({\sigma _{{T_0}}} = \sqrt {V\left( {{T_0}} \right)} = \sqrt {36} = 6\)

The first requested probability is

\(\begin{aligned}P\left( {{T_0} \le 160} \right) &= P\left( {{X_1} + {X_2} + {X_3} \le 160} \right)\\ &= P\left( {\frac{{{X_1} + {X_2} + {X_3} - {\mu _{{T_0}}}}}{{{\sigma _{{T_0}}}}} \le \frac{{160 - 150}}{6}} \right)\\ &= P(Z \le 1.67)\\ &= 0.9525\end{aligned}\)

(2): from the normal probability table in the appendix. The probability can also be computed with a software.

In order to compute the second probability, we shall look at even

\({X_1} + {X_2} \ge 2{X_3}\)

which can be equally written as

\({X_1} + {X_2} - 2{X_3} \ge 0\)

Denote with \({T_2}\)mentioned linear combination, hence

\({T_2} = {X_1} + {X_2} - 2{X_3}\)

Mean value of random variable \({T_2}\)is

\(\begin{aligned}E\left( {{T_2}} \right) &= E\left( {{X_1} + {X_2} - 2{X_3}} \right)\\ &= E\left( {{X_1}} \right) + E\left( {{X_2}} \right) - 2E\left( {{X_3}} \right)\\ &= 40 + 50 - 2 \cdot 60\\ &= - 30\end{aligned}\)

the variance is

\(\begin{aligned}E\left( {{T_2}} \right) &= V\left( {{X_1} + {X_2} - 2{X_3}} \right)\\ &= V\left( {{X_1}} \right) + {1^2}V\left( {{X_2}} \right) + {( - 2)^2}V\left( {{X_3}} \right)\\ &= 10 + 12 + 4 \cdot 12\\ &= 78\end{aligned}\)

(1): the random variables are independent therefore the equality stands.

Finally, the standard deviation of random variable \({T_2}\)is

\({\sigma _{{T_2}}} = \sqrt {V\left( {{T_2}} \right)} = \sqrt {78} = 8.832\)

Using the same method, the following is true

\(\begin{aligned}P\left( {{T_2} \ge 0} \right) &= P\left( {{X_1} + {X_2} - 2{X_3} \ge 0} \right)\\ &= P\left( {\frac{{{X_1} + {X_2} - 2{X_3} - {\mu _{{T_2}}}}}{{{\sigma _{{T_2}}}}} \ge \frac{{0 - ( - 30)}}{{8.832}}} \right)\\ &= P(Z \ge 3.4)\\ &= 0.0003\end{aligned}\)

(2): from the normal probability table in the appendix. The probability can also be computed with a software.