Question

The police records of a metropolitan area kept over the past 300 days show the following number of fatal accidents. Number of Fatal Accidents Number of Days x f(x) 0 45 1 75 2 120 3 45 4 15 a) Develop a probability distribution for the daily fatal accidents.

Answer 1

Answer:

X     |        0    |      1       |        2      |       3      |      4    |

P(X) | 45/300 | 75/300 | 120/300 | 45/300 | 15/300|

And the mean and the variance are:

E(X)=1.7

Var(X)=1.11

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.  

Solution to the problem

For this case we have the following frequency distribution:

X     0     1      2      3     4

f(X)  45   75   120  45  15

Now we can find the probability for each value of X

P(X=0)=45/300

P(X=1)=75/300

P(X=2)=120/300

P(X=3)=45/300

P(X=4)=15/300

So then the random variable is given by this table

X     |        0    |      1       |        2      |       3      |      4    |

P(X) | 45/300 | 75/300 | 120/300 | 45/300 | 15/300|

In order to calculate the expected value we can use the following formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=(0)*(\frac{45}{300})+(1)(\frac{75}{300})+(2)(\frac{120}{300})+(3)(\frac{45}{300})+(4)(\frac{15}{300})=1.7$

In order to find the standard deviation we need to find first the second moment, given by :

$E(X^2)=\sum_{i=1}^n X^2_i P(X_i)$

And using the formula we got:

$E(X)=(0^2)*(\frac{45}{300})+(1^2)(\frac{75}{300})+(2^2)(\frac{120}{300})+(3^2)(\frac{45}{300})+(4^2)(\frac{15}{300})=4$

Then we can find the variance with the following formula:

$Var(X)=E(X^2)-[E(X)]^2 =4-(1.7)^2 =1.11$

And then the standard deviation would be given by:

$Sd(X)=\sqrt{Var(X)}=\sqrt{1.11}=1.054$

So then the distribution for the random variable is given by:

X     |        0    |      1       |        2      |       3      |      4    |

P(X) | 45/300 | 75/300 | 120/300 | 45/300 | 15/300|

And the mean and the variance are:

E(X)=1.7

Var(X)=1.11