Question

Two airplanes are flying in the same direction in adjacent parallel corridors. At time \({\rm{t = 10}}\), the first airplane is \({\rm{10}}\)km ahead of the second one. Suppose the speed of the first plane (km/hr.) is normally distributed with mean \({\rm{520\;}}\)and standard deviation \({\rm{10}}\) and the second plane’s speed is also normally distributed with mean and standard deviation \({\rm{500\; and\; 10}}\), respectively.

a. What is the probability that after \({\rm{2hr}}{\rm{. }}\)of flying, the second plane has not caught up to the first plane?

b. Determine the probability that the planes are separated by at most \({\rm{10km\; after\; 2hr}}{\rm{. }}\)

Short answer

a\({\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}{\rm{ < 5}}} \right){\rm{ = 0}}{\rm{.9616;}}\)

b. \({\rm{P}}\left( {{\rm{0£ }}{{\rm{X}}_{\rm{2}}}{\rm{ - }}{{\rm{X}}_{\rm{1}}}{\rm{£ 10}}} \right){\rm{ = 0}}{\rm{.0623}}\)

Step-by-step solution

definition of standard deviation

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.

Determining the probability that after \({\rm{2hr}}{\rm{. }}\)of flying, the second plane has not caught up to the first plane

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.

(a) The lengths travelled by the planes are \({\rm{2}}{{\rm{X}}_{\rm{1}}}\)and \({\rm{2}}{{\rm{X}}_{\rm{2}}}\),respectively, implying that the event of the second plane not catching up to the first plane may be represented with

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

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

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

Determining the probability that the planes are separated by at most \({\rm{10km\; after\; 2hr}}{\rm{. }}\) 

(b) After \({\rm{2}}\) hours, the distance travelled by the first plane will be

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

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’s after\({\rm{2}}\)hours

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

or, alternatively,

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

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

\begin{aligned}P\left( {0£{X_2} - {X_1}£10} \right) &= P\left( {\frac{{0 - ( - 20)}}{{14.14}}£\frac{{{X_2} - {X_1} - ( - 20)}}{{14.14}} < \frac{{5 - ( - 20)}}{{14.14}}} \right) \\ &= P(1.41£Z£2.21) = P(Z£2.21) - P(Z£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.