A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform

distribution.

a. Find the average time between fireworks.

b. Find probability that the time between fireworks is greater than four seconds.

TheUniform Distribution, sometimes also known as a rectangular distribution, is a distribution that has constant probability.

It is symmetrical distribution on intervalwith probability density functionand zero everywhere else.

Let random variablehas uniform distribution on. Then.

We have that time between fireworks has uniform distribution on interval $(1,5)$which are given in seconds. Defining variable $X$as$X="thetimebetweentwofireworks"$then variable $X$also has uniform distribution on $(1,5)$Probability density function for variable$X$ is given by

(a) Calculating mean of this uniform distribution :

$$\begin{gathered} \mu =\frac{a+b}{2} \\ =\frac{1+5}{2} \\ =3 \end{gathered}$$

Hence, the mean for$U(1,5)\mathrm{is}3$

(b)

$$\begin{gathered} P(X>4)={\int }_4^{\infty }f\left(X\right).dx \\ ={\int }_4^5\frac{1}{4}.dx+{\int }_5^{\infty }0.dx \\ =\frac{1}{4}{X}_4^5+0 \\ =\frac{1}{4}(5-4) \\ =0.25 \end{gathered}$$

The probability that the time between fireworks is greater than four seconds is$0.25$