Question

Waiting for the Train. A commuter train arrives punctually at a station every half hour. Each morning, a commuter named John leaves his house and casually strolls to the train station. The time, in minutes, that John waits for a train is a variable with density curve $y=\frac{1}{30}for0<x<30,y=0$and otherwise.

(a) Graph the density curve of this variable.

(b) Show that the area under this density curve to the left of any number x between 0 and 30equals $\frac{x}{30}$.

What percentage of all possible observations of the variable

(c) less than 5 minutes.

(d) between 10and 15 minutes.

(e) atleast20minutes.

Short answer

(a) The graph of the density function is:

(b) Thus, the area under this density curve to the left of any number x between 0 and 30equals $\frac{x}{30}$.

(c) John waits $16.67\%$ of time less than 5 minutes for a train.

(d) John waits $16.67\%$ of time between 10, 15minutes of the train.

(e) John waits 33.33%of time at least 20 minutes of the train.

Step-by-step solution

Part (a) Step 1. Given Information. 

A variable has a density curve whose equation is $y=\frac{1}{30}for0<x<30,y=0$.

Part (a) Step 2. Graph the density function. 

Graph the density function $y=\frac{1}{30}for0<x<30,y=0$.

Part (b) Step 1. The area between 0, 30. 

From part (a),

Base,$b=x$

Height, $h=\frac{1}{30}$

Thus the area of the rectangle is given by,

$$\begin{gathered} A=bh \\ =x\left(\frac{1}{30}\right) \\ =\frac{x}{30} \end{gathered}$$

Part (c) Step 1. Find the percentage. 

Area to the left of $5=\frac{5}{30}$

role="math" localid="1652600779734" $=0.1667$

John waits $16.67\%$ time of less than 5 minutes for a train.

Part (d) Step 1. Find the percentage. 

Area between $10,15=$[Area to the left of $15]-$[Area to the left of $10]$

role="math" localid="1652601050742" $$\begin{gathered} =\frac{15}{30}-\frac{10}{30} \\ =\frac{5}{30} \\ =0.1667 \end{gathered}$$

John waits $16.67\%$ of time between 10, 15 minutes of the train.

Part (e) Step 1. Find the percentage. 

[Area to the right of $20]=1-$[Area to the left of $20]$

$$\begin{gathered} =1-\frac{20}{30} \\ =\frac{10}{30} \\ =0.3333 \end{gathered}$$

John waits $33.33\%$ of time at least20 minutes of the train.