Distributions In a continuous uniform distribution,

$\mu =\frac{minimum+maximum}{2} and\sigma =\frac{range}{\sqrt{12}}$

a. Find the mean and standard deviation for the distribution of the waiting times represented in Figure 6-2, which accompanies Exercises 5–8.

b. For a continuous uniform distribution with $\mu =0 and \sigma =1$, the minimum is$-\sqrt{3}$ and the maximum is $\sqrt{3}$. For this continuous uniform distribution, find the probability of randomly selecting a value between –1 and 1, and compare it to the value that would be obtained by incorrectly treating the distribution as a standard normal distribution. Does the distribution affect the results very much?

The mean and standard deviation of the uniform distribution is,

$\mu =\frac{\text{Minimum + Maximum}}{2} and \sigma =\frac{\text{Range}}{\sqrt{12}}$

a.

The waiting time (in minutes) follows the continuous uniform distribution which are represented in Figure 6-2.

A continuous random variable has a uniform distribution if its values are spread evenly over the range of possibilities.

Refer to Figure 6-2,

$$\begin{gathered} \text{Minimum}=0 \\ \text{Maximum} =5 \end{gathered}$$

The mean of uniform distribution is,

$$\begin{gathered} \mu =\frac{\text{Minimum + Maximum}}{2} \\ =\frac{5}{2} \\ =2.5 \end{gathered}$$

Thus, the mean is 2.5 minutes.

The range is computed as follows,

$$\begin{gathered} \text{Range}=\text{Maximum}-\text{Minimum} \\ =5-0 \\ =5 \end{gathered}$$

The standard deviation of uniform distribution is,

$$\begin{gathered} \sigma =\frac{\text{Range}}{\sqrt{12}} \\ =\frac{5}{\sqrt{12}} \\ =1.4434 \end{gathered}$$

Thus, the standard deviation is 1.44 minutes.

b.

Let X be the random variable which follows uniform distribution with the minimum value$-\sqrt{3}$ andthemaximum value $\sqrt{3}$.

Thus,

$X~U(-\sqrt{3},\sqrt{3})$

The probability of X is even, which is computed as,

$$\begin{gathered} P\left(x\right)=\frac{1}{\text{Range}} \\ =\frac{1}{\text{Maximum}-\text{Minimum}} \\ =\frac{1}{\sqrt{3}+\sqrt{3}} \\ =0.2886 \end{gathered}$$

The probability that a value falls between –1 and 1 can be calculated from the following graph,

The shaded area indicates the area between –1 and 1. The area under the graph is 1 which implies that area and probability has one-to-one correspondence.

The probability of area between –1 and 1 is equal to the area of shaded rectangle with height 0.2886 and width 2 units.

$$\begin{gathered} P\left(-1<X<1\right)=\left(\text{Height} \times \text{Width}\right) \text{of} \text{shaded} \text{area} \\ =0.2886\times 2 \\ =0.5772 \end{gathered}$$

Thus, the probability of area between –1 and 1 is 0.5772.

Let Z be the randomly selected variable which follows standard normal distribution.

Thus,

$$\begin{gathered} Z~N\left(\mu ,{\sigma }^2\right) \\ ~ N\left(0, {\text{1}}^2\right) \end{gathered}$$

The probability that a randomly selected value between –1 and 1 is expressed as,

$$\begin{gathered} P\left(-1<Z<1\right)=\text{Area}\text{less}\text{than}1-\text{Area}\text{less}\text{than}-1 \\ =P\left(Z<1\right)-P\left(Z<-1\right) ...\left(2\right) \end{gathered}$$ Refer to the standard normal table,

• The cumulative probability of –1 is obtained from the cell intersection for rows –1.0 and the column value 0.00, which is 0.1587.
• The cumulative probability of 1 is obtained from the cell intersection for rows 1.0 and the column value 0.00,which is 0.8413.

Thus,

$$\begin{gathered} P\left(Z<1\right)=0.8413 \\ P\left(Z<-1\right)=0.1587 \end{gathered}$$

Substitute in equation(2),

$$\begin{gathered} P\left(-1<Z<1\right)=P\left(Z<1\right)-P\left(Z<-1\right) \\ =0.8413-0.1587 \\ =0.6826 \end{gathered}$$

The probability that a randomly selected value falls between –1 and 1 is 0.5772 for uniform distribution, and 0.6826 for standard normal distribution.

So, the distribution affects the result significantly.