Question

The number of miles driven by a truck driver falls between $300\mathrm{and}700$, and follows a uniform distribution.

a. Find the probability that the truck driver goes more than $650$miles in a day.

b. Find the probability that the truck drivers goes between $400\mathrm{and}650$miles in a day.

c. At least how many miles does the truck driver travel on the furthest $10\%$of days?

Short answer

a) The probability that the truck driver goes more than $650$miles in a day is $0.125$.

b) The probability that the truck drivers goes between $400\mathrm{and}650$miles in a day is $0.625$

c) The truck driver travels at least $660$miles on the furthest $10\%$of days

Step-by-step solution

Defining the height of Probability distribution

It is given that $X~U(300,700)$

Therefore, the height or probability distribution of the data will be:

$$\begin{gathered} f\left(x\right)=\frac{1}{b-a} \\ =\frac{1}{700-300} \\ =\frac{1}{400} \end{gathered}$$

Finding the probability that the truck driver goes more than 650 miles in a day.

a) The required probability can be calculated as below:

$$\begin{gathered} P(x>650)=base\times height \\ =(700-650)\times \frac{1}{400} \\ =\frac{50}{400} \\ =0.125 \end{gathered}$$

Finding the probability that the truck drivers goes between 400 and 650 miles in a day.

b) The required probability can be calculated as below:

$$\begin{gathered} P(400<x<650)=base\times height \\ =(650-400)\times \frac{1}{400} \\ =\frac{250}{400} \\ =0.625 \end{gathered}$$

 Step 4 : Finding at least how many miles does the truck driver travel on the furthest 10% of days?

c) The required ${10}^{th}$percentile can be calculated as below:

$$\begin{gathered} P(x>k)=base\times height \\ \Rightarrow 0.10=(700-k)\times \frac{1}{400} \\ \Rightarrow 400\times 0.10=700-k \\ \Rightarrow k=660 \end{gathered}$$