Question

Suppose a movie starts at 5:00 p.m. and Lindsay, a customer who is always late, arrives at the movie theater at a random time between 5:10 p.m. and 5:45 p.m. Lindsay's late arrival time, in minutes, represented by X, models a uniform distribution between 10 and 45 min. Determine the height of the uniform density curve. Provide your answer with precision to three decimal places.

height= ?

Calculate the probability that Lindsay is late by 20 min or less, P(X?20). Provide your answer with precision to three decimal places.

P(X?20) = ?

Answer 1

Answer:

a) We know that the distribution for X is given by: $X \sim Unif (a=10, b=45)$

The density function for this distirbution is given by:

$f(x) = \frac{1}{b-a}= \frac{1}{45-10}= \frac{1}{35} , 10 \leq X \leq 45$

So then the height of the uniform density curve would be given by:

$h = \frac{1}{35}= 0.029$

b) For this case we want to calculate the following probability:

$P(X\leq 20)$

And we can use the cumulative distribution function given by:

$F(X) =\frac{x-a}{b-a}=\frac{x-10}{35} , 10 \leq X \leq 45$

And we can calculate the probability like this:

$P(X \leq 20) = F(20) = \frac{20-10}{35}=0.286$

Step-by-step explanation:

For this case we define our random variable X="Lindsay's late arrival time, in minutes"

Part a

We know that the distribution for X is given by: $X \sim Unif (a=10, b=45)$

The density function for this distirbution is given by:

$f(x) = \frac{1}{b-a}= \frac{1}{45-10}= \frac{1}{35} , 10 \leq X \leq 45$

So then the height of the uniform density curve would be given by:

$h = \frac{1}{35}= 0.029$

Part b

For this case we want to calculate the following probability:

$P(X\leq 20)$

And we can use the cumulative distribution function given by:

$F(X) =\frac{x-a}{b-a}=\frac{x-10}{35} , 10 \leq X \leq 45$

And we can calculate the probability like this:

$P(X \leq 20) = F(20) = \frac{20-10}{35}=0.286$