魅影直播

The square root of the variance is the standard deviation of a random variable, sample, statistical population, data collection, or probability distribution. It is less resilient in practice than the average absolute deviation, but it is algebraically easier.

Let \({{\rm{X}}_{\rm{1}}}\)be the speed of the first plane with a mean of \({\rm{250}}\) and a standard deviation of \({\rm{10}}\), and \({{\rm{X}}_{\rm{2}}}\) be the speed of the second plane with a mean of \({\rm{500}}\) and a standard deviation of\({\rm{10}}\)

The first plane is approximately\({\rm{10}}\) kilometers ahead of the second plane.

\(\mathop {\mathop {{\rm{2}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}}\limits_{\rm{n}} }\limits_{{\rm{distance passed/first plane\;}}} {\rm{ > }}\mathop {\mathop {{\rm{2}}{{\rm{X}}_{\rm{2}}}}\limits_{\rm{n}} }\limits_{{\rm{distance passed/second plane\;}}} \)

Obviously, the second plane had not caught up to the first plane because it had travelled a shorter distance.

Rewrite this as a linear combination of random variables \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}\) as follows:

\({{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}{\rm{ < 5}}{\rm{.}}\)

To calculate the probability of the occurrence in question, first normalize the random variable \({{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}\)

\({\rm{E}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}} \right){\rm{ = E}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ - E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ = 500 - 520 = - 20}}\)

The variance is

\({\rm{V}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}} \right){\rm{ = V}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{V}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ = 100 + 100 = 200}}\)

Because the random variables are independent, the mean values are true

The standard deviation of the data is

\({\rm{\sigma = }}\sqrt {{\rm{200}}} {\rm{ = 14}}{\rm{.14}}{\rm{.}}\)

Therefore, following is true

\(\begin{array}{*{20}{c}}{{\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}{\rm{ < 5}}} \right){\rm{ = P}}\left( {\frac{{{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}{\rm{ - ( - 20)}}}}{{{\rm{14}}{\rm{.14}}}}{\rm{ < }}\frac{{{\rm{5 - ( - 20)}}}}{{{\rm{14}}{\rm{.14}}}}} \right)}\\{{\rm{ = P(Z < 1}}{\rm{.77) = 0}}{\rm{.9616}}}\end{array}\)

\({\rm{10 + 2}}{{\rm{X}}_{\rm{1}}}\)

and the distance travelled by the second plane after \({\rm{2}}\) hours will be

\({\rm{2}}{{\rm{X}}_{\rm{2}}}\)

A random variable can be used to indicate the separation distance (the distance difference)

\(\left| {{\rm{2}}{{\rm{X}}_{\rm{2}}}{\rm{ - 10 - 2}}{{\rm{X}}_{\rm{1}}}} \right|{\rm{.\;}}\)

As a result, the chance of an event is the probability that the planes are apart by at most \({\rm{10}}\)kilometer鈥檚 after\({\rm{2}}\)hours

\(\left| {{\rm{2}}{{\rm{X}}_{\rm{2}}}{\rm{ - 10 - 2}}{{\rm{X}}_{\rm{1}}}} \right|{\rm{拢 10}}\)

or, alternatively,

\(\begin{array}{*{20}{r}}{{\rm{ - 10拢 2}}{{\rm{X}}_{\rm{2}}}{\rm{ - 10 - 2}}{{\rm{X}}_{\rm{1}}}{\rm{拢 10}}}\\{{\rm{0拢 }}{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}{\rm{拢 10}}}\end{array}\)

Finally, as in (a), the following is true:

\begin{aligned}P\left( {0拢{X_2} - {X_1}拢10} \right) &= P\left( {\frac{{0 - ( - 20)}}{{14.14}}拢\frac{{{X_2} - {X_1} - ( - 20)}}{{14.14}} < \frac{{5 - ( - 20)}}{{14.14}}} \right) \\ &= P(1.41拢Z拢2.21) = P(Z拢2.21) - P(Z拢1.41) \\ &= 0.9830 - 0.9207 = 0.0623 \\ \end{aligned}

(1): from the appendix's normal probability table. Software can also be used to calculate the likelihood.