Answer:
[tex] P(35-2 < X 35+2) = P(33< X< 37)= P(X<37) -P(X<32)[/tex]
And using the cumulative distribution function we got:
[tex] P(35-2 < X 35+2) = P(33< X< 37)= P(X<37) -P(X<32) = \frac{37-20}{50-20} -\frac{33-20}{50-20} =0.567-0.433=0.134 [/tex]
The probability that preparation is within 2 minutes of the mean time is 0.134
Step-by-step explanation:
For this case we define the following random variable X= (minutes) for a lab assistant to prepare the equipment for a certain experiment , and the distribution for X is given by:
[tex] X \sim Unif (a= 20, b =50)[/tex]
The cumulative distribution function is given by:
[tex] F(x) = \frac{x-a}{b-a} , a \leq X \leq b[/tex]
The expected value is given by:
[tex] E(X) = \frac{a+b}{2} = \frac{20+50}{2}=35[/tex]
And we want to find the following probability:
[tex] P(35-2 < X 35+2) = P(33< X< 37)[/tex]
And we can find this probability on this way:
[tex] P(35-2 < X 35+2) = P(33< X< 37)= P(X<37) -P(X<32)[/tex]
And using the cumulative distribution function we got:
[tex] P(35-2 < X 35+2) = P(33< X< 37)= P(X<37) -P(X<32) = \frac{37-20}{50-20} -\frac{33-20}{50-20} =0.567-0.433=0.134 [/tex]
The probability that preparation is within 2 minutes of the mean time is 0.134
