Suppose the following table shows the results obtained for 4,409 boating accidents, including the wind condition at the time of the accident.
Wind Condition |
Percentage of Accidents |
---|---|
None | 9.4 |
Light | 57.0 |
Moderate | 24.0 |
Strong | 7.7 |
Storm | 1.9 |
Let 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.
(a)Develop a probability distribution for x.
x |
f(x) |
---|---|
0 | |
1 | |
2 | |
3 | |
4 |
(b)Compute the expected value of x.
(c)Compute the variance and standard deviation for x. (Round your answers to four decimal places.)
variance
standard deviation
(d)Comment on what your results imply about the wind conditions during boating accidents.
The expected value calculated in part (b) indicates that the mean wind condition when an accident occurred is ---Select--- slightly greater than no slightly greater than light slightly greater than moderate slightly less than strong slightly less than storm wind conditions.
a)
X | P(X) |
0 | 0.094 |
1 | 0.57 |
2 | 0.24 |
3 | 0.077 |
4 | 0.019 |
b)
X | P(X) | X*P(X) | X² * P(X) |
0 | 0.094 | 0.000 | 0.0000 |
1 | 0.57 | 0.570 | 0.5700 |
2 | 0.24 | 0.480 | 0.9600 |
3 | 0.077 | 0.231 | 0.6930 |
4 | 0.019 | 0.076 | 0.3040 |
expected value of x = mean = E[X] = Σx*P(X) =
1.3570
c)
E [ X² ] = ΣX² * P(X) =
2.5270
variance = E[ X² ] - (E[ X ])² =
0.6856
std dev = √(variance) =
0.8280
d)
The expected value calculated in part (b) indicates that the mean wind condition when an accident occurred is slightly greater than light wind conditions.
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