Question

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 45.045.0 and 55.055.0 minutes. Find the probability that a given class period runs between 50.550.5 and 51.2551.25 minutes.

Answer 1

Answer:

$P(50.5 < X< 51.25)= P(X<51.25)-P(X<50.5) = \frac{51.25-45}{10}- \frac{50.5-45}{10}= 0.625-0.55=0.075$

Step-by-step explanation:

For this case we define X as our random variable representing the lenghts of her classes in minutes.

We know that the distribution for X is uniform and is given by:

$X \sim Unif (a=45.0, b=55.0)$

The density function is given by:

$f(x) = \frac{1}{55-45}= \frac{1}{10}, 45 \leq X \leq 55$

$f(x)=0$ for other case

The cumulative distirbution function is given by:

$F(x) = \frac{x-45}{55-45}= \frac{x-45}{10} , 45 \leq x \leq 55$

And we want to find this probability:

$P(50.5 < X< 51.25)$

We can find this probability using the cumulative distribution function and we got:

$P(50.5 < X< 51.25)= P(X<51.25)-P(X<50.5) = \frac{51.25-45}{10}- \frac{50.5-45}{10}= 0.625-0.55=0.075$