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

Answer:

.30

Step-by-step explanation:

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

The height is .5

Thus .5 * .6 = .30

Answer 2

Answer:

The probability is $0.3$

Step-by-step 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 $f(x)=\frac{1}{b-a}$ when x ∈ [ a , b ] and $f(x)=0$ otherwise.

In this exercise :

$f(x)=\frac{1}{b-a}=\frac{1}{125.0-123.0}=\frac{1}{2}$

$f(x)=\frac{1}{2}$ 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 $f(x)=\frac{1}{2}$ 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 $(124.7-124.1).(\frac{1}{2})=(0.6).(\frac{1}{2})=0.3$

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

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