Question

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 50.05 and 60.06 minutes. Find the probability that a given class period runs greater thangreater than 51.5 minutes.

Answer 1

Answer:

$P(X<51.5) = \frac{51.5-50.05}{10.01}=0.1449$

And then we can calculate the probability of interest using the complement rule:

$P(X>51.5) = 1-P(X<51.5) = 1-0.1449=0.8551$

Step-by-step explanation:

Let's assume that the random variable X ="Lengths of her classes", and we know that the distribution for X is given by:

$X \sim Unif (a=50.05, b= 60.06)$

The expected value is given by:

$E(X) = \frac{b+a}{2}=\frac{60.06+50.05}{2}=55.055$

And the variance  given by:

$Var(X) = \frac{(b-a)^2}{12}=\frac{(60.06-50.05)^2}{12}=8.35$

The density function is given by:

$f(x) = \frac{x}{b-a}=\frac{1}{60.06-50.05}= 0.099  , 50.05 \leq X \leq 60.06$

And 0 for other case.

And on this case we want to find the following probability $P(X>51.5)$ first we need to calculate $P(X<51.5)$, so we can calculate this with the following integral:

$P(X<51.5) = \int_{50.05}^{51.5} \frac{1}{10.01} dx = \frac{1}{10.01} x\Big|_{50.05}^{51.5}$

And after evaluate using the fundamental calculus theorem we got:

$P(X<51.5) = \frac{51.5-50.05}{10.01}=0.1449$

And then we can calculate the probability of interest using the complement rule:

$P(X>51.5) = 1-P(X<51.5) = 1-0.1449=0.8551$