Question

Suppose the time it takes a barber to complete a haircuts is uniformly distributed between 5 and 17 minutes, inclusive. Let X = the time, in minutes, it takes a barber to complete a haircut. Then X ~ U (5, 17). Find the probability that a randomly selected barber needs at least seven minutes to complete the haircut, P(x > 7) (round to 4 decimal places) Answer:

Answer 1

Answer:

$P(x>7)$

And for this case we can use the cumulative distribution given by:

$F(x) =\frac{x-a}{b-a}, a \leq x \leq b$

And if we use the complement rule and the before formula we have:

$P(x>7) = 1- P(x<7) = 1- F(7) = 1 -\frac{7-5}{17-5}= 1- 0.1667 =0.8333$

Step-by-step explanation:

For this problem we denote the random variable X as  the time, in minutes, it takes a barber to complete a haircut and the distribution for X is given by:

$X \sim Unif (a=5, b=17)$

And we want to find the following probability:

$P(x>7)$

And for this case we can use the cumulative distribution given by:

$F(x) =\frac{x-a}{b-a}, a \leq x \leq b$

And if we use the complement rule and the before formula we have:

$P(x>7) = 1- P(x<7) = 1- F(7) = 1 -\frac{7-5}{17-5}= 1- 0.1667 =0.8333$