Question

Major Hurricanes. The Atlantic Hurricane Database extends back to 1851, recording among other things the number of major hurricanes striking the U.S. Atlantic and Gulf Coast per year. A major hurricane is a hurricane measuring at least a Category 3 on the Saffir-Simpson hurricane wind scale (i.e., with winds of at least 110 mph). As published by the National Oceanic & Atmospheric Administration and the Atlantic Oceanographic & Meteorological Laboratory, the following table provides a probability distribution for the number of major hurricanes, Y, for a randomly selected year between 1851 and 2012.

Use random-variable notation to represent each of the following events. The year had

a. at least one major hurricane.

b. exactly three major hurricanes.

c. between 2 and 4 major hurricanes, inclusive.

Use the special addition rule and the probability distribution to determine

d. P(Y ≥ 1).

e. P(Y = 3).

f. P(2 ≤ Y≤ 4)

Short answer

Part a. Y ≥ 2

Part b. Y = 3

Part c. 2 ≤ Y ≤ 4

Part d. P(Y ≥ 1) = 0.815

Part e. P(Y = 3) = 0.093

Part f. P(2≤ Y≤ 4) = 0.408

Step-by-step solution

Part (a) Step 1. Given information

The probability distribution for the number of significant hurricanes, Y, for a year chosen at random between 1851 and 2012 is depicted below.

Y	P(Y = y)
0	0.185
1	0.296
2	0.266
3	0.093
4	0.049
5	0.056
6	0.037
7	0.012
8	0.006

Part (a) Step 3. Solution

At least one big storm is defined as one hurricane that is stronger than or equal to another hurricane.

As a result, the notation for random variables is,

Y ≥ 2

Part (b) Step 1. Solution 

Because there have been three big hurricanes, the random variable value must be three.

As a result, the notation for random variables is,

Y = 3

Part (c) Step 1. Solution 

Between two and four major hurricanes, inclusive meaning, the random variable value must include two and four major hurricanes.

As a result, the notation for random variables is,

2 ≤ Y ≤ 4

Part (d) Step 1. Solution 

$P(Y\ge 1)=P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4)+P(Y=5)+P(Y=6)+P(Y=7)+P(Y=8)$

$=1-P(X=0)$

$=1-0.185$

$=0.815$

Part (e) Step 1. Solution 

The random variable's probability of taking the value 3 is,

$P(Y=3)=0.093$

Part (f) Step 1. Solution 

$P(2\le Y\le 4)=P(Y=2)+P(Y=3)+P(Y=4)$

$=0.266+0.093+0.049$

$=0.408$