Question

The figure shows a uniform distribution of voltage levels. Find the probability a random voltage is between 124.1 volts and 124.7 volts

Answer 1

.30

Explanation:

The distance (or width) from 124.1 to 124.7 volts is .6

The height is .5

Thus .5 * .6 = .30

Answer 2

The probability is
`0.3`

Explanation:

Let's start defining the random variable.

X : '' Voltage levels ''

We know that X has an uniform distribution ⇒

X ~ U [ a , b ]

Where ''a'' and ''b'' define the interval.

In this exercise, X ~ U [ 123.0 , 125.0 ] (We can notice this in the graph of the density function).

The density function for an uniform distribution is
`f(x)=(1)/(b-a)` when x ∈ [ a , b ] and
`f(x)=0` otherwise.

In this exercise :

`f(x)=(1)/(b-a)=(1)/(125.0-123.0)=(1)/(2)`

`f(x)=(1)/(2)` when x ∈ [ 123.0 , 125.0 ]

`f(x)=0` otherwise

To find the probability
`P(124.1<X<124.7)` we need to integrate the function
`f(x)=(1)/(2)` between 124.1 and 124.7

This integrate is equal to the area below the graph of the function between 124.1 and 124.7

Given that the graph is a rectangle with height 0.5 :

This area is
`(124.7-124.1).((1)/(2))=(0.6).((1)/(2))=0.3`

`P(124.1<X<124.7)=0.3`

And that is the probability we need. We could have done the integral but it was not necessarily.