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The standard deviation is a statistical measure of a set of values' variation or dispersion. A low standard deviation implies that the values of the set tend to be close to the mean (also known as the expected value), whereas a high standard deviation shows that the values are spread out over a larger range.

Let random variables\({X_1},\;{X_2},\;{X_3}\)have mean values of\(15,\;30,\;20\)minutes, standard deviations\(1,\;2,\;1.5\)minutes, respectively. Look at random variable

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

Since,\(1{\rm{ }}hours = 60\;minutes\), the probability which needs to be obtained is\(P\left( {{T_0} \le 60} \right)\)

In order to compute it, mean values and standard deviation of random variable\({T_0}\)is required. The following holds for the mean value

\(\begin{aligned}E\left( {{T_0}} \right) &= E\left( {{X_1} + {X_2} + {X_3}} \right)\\ &= E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + E\left( {{X_3}} \right)\\ &= 15 + 30 + 20\\ &= 65\end{aligned}\)

and for the variance

\(\begin{aligned}V\left( {{T_0}} \right) &= V\left( {{X_1} + {X_2} + {X_3}} \right)\\ &= V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + V\left( {{X_3}} \right)\\ &= {1^2} + {2^2} + {1.5^2}\\ &= 7.25\end{aligned}\)

(1): the random variables are independent which implies that equality stands.

The standard deviation of random variable\({T_0}\)is

\({\sigma _{{T_0}}} = \sqrt {V\left( {{T_0}} \right)} = \sqrt {7.25} = 2.6926\)

The probability can now be computed as

\(\begin{aligned}P\left( {{T_0} \le 60} \right) &= P\left( {\frac{{{T_0} - E\left( {{T_0}} \right)}}{{{\sigma _{{T_0}}}}} \le \frac{{60 - 65}}{{2.6926}}} \right)\\ &= P(Z \le - 1.86)n\\ &= 0.0314\end{aligned}\)