Question

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.

Short answer

(a) Three seconds is the average time between fireworks.

(b) There's a good chance that the gap between fireworks will be longer than four seconds$P(x>4)=0.25.$

Step-by-step solution

Part (a) Step 1: Given Information

The timing between fireworks in a fireworks show is planned to be between one and five seconds, with a uniform distribution.

Part (a) Step 2:  Calculate the average interval between fireworks.

The following data is provided based on the information provided:

$\mathrm{X}~\mathrm{U}(1,5)$

Density function is given below,

Therefore,

$$\begin{gathered} a=1 \\ b=5 \end{gathered}$$

Uniform distribution the mean is calculated using the formula,

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

Therefore the average time between fireworks is 3 seconds.

Part (b) Step 1: Given Information

The timing between fireworks in a fireworks show is planned to be between one and five seconds, with a uniform distribution.

Part (b) Step 2: Determine the likelihood that the interval between fireworks is more than four seconds.

Height of probability distribution is given by $1<x<5$will be,

$$\begin{gathered} f\left(x\right)=\frac{1}{b-a} \\ f\left(x\right)=\frac{1}{5-1} \\ f\left(x\right)=\frac{1}{4} \end{gathered}$$

The required probability is,

$$\begin{gathered} P(x>4)=\text{base}\times \text{height} \\ P(x>4)=(5-4)(1/4) \\ P(x>4)=1\times (1/4) \\ P(x>4)=0.25 \end{gathered}$$

Thus , the probability that the time between fireworks is greater than four seconds is $P(x>4)=0.25$.