Chapter 5, Problem 59SE

The U.S. Coast Guard (USCG) provides a wide variety of information on boating accidents including the wind condition at the time of the accident. The following table shows the results obtained for 4401 accidents (USCG website, November 8, 2012).

Let x be a random variable reflecting the known wind condition at the time of each accident.
Set $x=0$ for noir, $x=1$ for light, $x=2$ for moderate, $x=3$ for strong, and $x=4$ for storm.
a. a probability distribution for x.
b. Compute the expected value of x.
c. Compute the variance and standard deviation for x.
d. Comment on what your results imply about the wind conditions during boating accidents.

To determine

The probability distribution of x.

Answer to Problem 59SE

The probability distribution is:

$X$	$P(X=x)$
0	0.096
1	0.57
2	0.238
3	0.077
4	0.019

Explanation of Solution

Given:

The results of wind condition at the time of 4401 accident are as given in the following table:

Wind condition	Percentage of Accident
None	9.6
Light	57
Moderate	23.8
Strong	7.7
Strom	1.9

Formula used:

The formulas are:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum _{i=1}^n{x}_i{p}_i}\\ $ V(X)=E(X^2)-{(E(X))}^2$\sigma =\sqrt{V(X)}\end{array}$

Calculation:

Consider, $X$ be a random variable reflecting the known wind condition at the time of each accident.

Set $X$ = 0 for none, $X$ = 1 for light, $X$ = 2 for moderate, $X$ = 3 for strong, and $X$ = 4 for storm.

That is, $X=\{0,1,2,3,4\}$

Thus, the required probability distribution of $X$ is −

$X$	$P(X=x)$
0	0.096
1	0.57
2	0.238
3	0.077
4	0.019

To determine

To find:The expected value of $X$

Answer to Problem 59SE

The expected value of x is 1.353.

Explanation of Solution

Calculation:

The expected value of x can be computed as:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum _{i=1}^n{x}_i{p}_i}\\ =0\times 0.096+1\times 0.238+3\times 0.077+4\times 0.019\\ =1.353\end{array}$

Hence, the expected value of x is 1.353.

To determine

To find: The standard deviation and variance of x.

Answer to Problem 59SE

The variance and standard deviation of x are 0.68839 and 0.8296 respectively.

Explanation of Solution

Calculation:

The variance and standard deviation of x can be computed as:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum _{i=1}^n{x}_i{p}_i}\\ =0\times 0.096+1\times 0.238+3\times 0.077+4\times 0.019\\ =1.353\\ E\left(X^2\right)={\displaystyle \sum _{i=1}^n{x}^2{}_i{p}_i}\\ =0^2\times 0.096+1^2\times 0.238+3^2\times 0.077+4^2\times 0.019\\ =2.519\\ V(X)=2.519-{\left[1.353\right]}^2\\ =0.68839\\ \sigma =\sqrt{0.68839}\\ =0.8296\end{array}$

Hence, the variance and standard deviation of x are 0.68839 and 0.8296 respectively.

To determine

To explain: The wind condition during boating accidents.

Explanation of Solution

The wind condition during the boating accidents can be understood from the results obtained in part b and c. That is,

On an average the wind condition during the boating accidents is 1.353 and the dispersion or the variance in the wind condition is 0.68839.

In other words, the wind condition on an average is approximately 135.3% and variability of wind condition is nearly 68.839% during the boating accidents.