Answer to Question #146720 in Statistics and Probability for shaoib

Question

On the basis of the previous record, an average of 4 accidents in one day occur on the superhighway during peak rush timings. Calculate the following if possible. If any part of question (you feel) is not possible to calculate give reason:
a. Evaluate the probability that on any Particular day there will be fewer than 3 accidents on this superhighway during the peak rush timings.
b. Compute the probability that on any Particular day there will be more than 6 accidents on this Super highway during the peak rush timings.
c. Find the Expected number of accidents per day and variance.

Answer 1

Let $X$ be the random variable representing the number of accidents in a day: $X\sim Po(\lambda t)$

$$P(X=x)=\dfrac{e^{-\lambda t}(\lambda t)^x}{x!}$$

Average number of accidents per day, $\lambda=4, t=1.$

Then $\lambda t=4(1)=4$ and

$$P(X=x)=\dfrac{e^{-4}(4)^x}{x!}$$

b)

$$P( X>6)=1-P(X=0)-P(X=1)-P(X=2)-$$

$$=\dfrac{e^{-4}(4)^0}{0!}+\dfrac{e^{-4}(4)^1}{1!}+\dfrac{e^{-4}(4)^2}{2!}=$$ $$=13e^{-4}\approx0.238103$$

The probability that on any Particular day there will be fewer than 3 accidents on this superhighway during the peak rush timings is $13e^{-4}\approx0.238103.$

a)

$$P( X<3)=P(X=0)+P(X=1)+P(X=2)$$

$$-P(X=3)-P(X=4)-P(X=5)-P(X=6)$$

$$=1-\dfrac{e^{-4}(4)^0}{0!}-\dfrac{e^{-4}(4)^1}{1!}-\dfrac{e^{-4}(4)^2}{2!}$$

$$-\dfrac{e^{-4}(4)^3}{3!}-\dfrac{e^{-4}(4)^4}{4!}-\dfrac{e^{-4}(4)^5}{5!}-\dfrac{e^{-4}(4)^6}{6!}$$

$$\approx0.110674$$

The probability that on any Particular day there will be more than 6 accidents on this superhighway during the peak rush timings is $\approx0.110674.$

c)

$$E(X)=\lambda t=Var(X)$$

Therefore

$$E(X)=Var(X)=4$$