The following tables gives the number of days in a 50 days period during which an automobile accident occurs in a cityNo. of accident: 0 1 2 3 4No. of days: 21 18 7 3 1Find the Poisson distribution and calculate the Theoretical frequency
Question
The following tables gives the number of days in a 50 days period during which an automobile accident occurs in a cityNo. of accident: 0 1 2 3 4No. of days: 21 18 7 3 1Find the Poisson distribution and calculate the Theoretical frequency
Solution
To calculate the Poisson distribution and the theoretical frequency, we first need to calculate the mean (λ) of the distribution. The mean can be calculated by multiplying each number of accidents by the number of days it occurs, summing these products, and then dividing by the total number of days.
Here's how to calculate the mean:
λ = (021 + 118 + 27 + 33 + 4*1) / 50 = 0.86
Now, we can calculate the Poisson probabilities for 0, 1, 2, 3, and 4 accidents. The formula for the Poisson probability is:
P(x; λ) = e^-λ * λ^x / x!
where:
- e is the base of the natural logarithm (approximately 2.71828),
- λ is the mean number of successes that result from the experiment,
- x is the actual number of successes that result from the experiment, and
- x! is the factorial of x.
Using this formula, we get:
P(0; 0.86) = e^-0.86 * 0.86^0 / 0! = 0.423 P(1; 0.86) = e^-0.86 * 0.86^1 / 1! = 0.364 P(2; 0.86) = e^-0.86 * 0.86^2 / 2! = 0.157 P(3; 0.86) = e^-0.86 * 0.86^3 / 3! = 0.045 P(4; 0.86) = e^-0.86 * 0.86^4 / 4! = 0.010
These are the Poisson probabilities. To find the theoretical frequencies, we multiply these probabilities by the total number of days (50):
Theoretical frequency for 0 accidents = 0.423 * 50 = 21.15 Theoretical frequency for 1 accident = 0.364 * 50 = 18.2 Theoretical frequency for 2 accidents = 0.157 * 50 = 7.85 Theoretical frequency for 3 accidents = 0.045 * 50 = 2.25 Theoretical frequency for 4 accidents = 0.010 * 50 = 0.5
So, the theoretical frequencies are approximately 21.15, 18.2, 7.85, 2.25, and 0.5 for 0, 1, 2, 3, and 4 accidents, respectively.
Similar Questions
The number of accidents (𝐴) at an intersection is counted for a 12-hour period. The number of accidents follows a Poisson distribution with a parameter value of 2.Calculate the expected number of accidents per week.
The number of accidents (𝐴) at an intersection is counted for a 12-hour period. The number of accidents follows a Poisson distribution with a parameter value of 2.Calculate the probability that there will be no accidents in a 1-hour period.
2. The number of accidents (A) at an intersection is counted for a 12-hour period. The number of accidents follows a Poisson distribution with a parameter value of 3.5. (Round to 3 decimal places and use a point-not a comma).Calculate the following:a. 𝑃(𝐴 = 1) Blank 1b. 𝑃(𝐴 < 5) Blank 2c. 𝑃(𝐴 > 0) Blank 3d. 𝑃(0 < 𝐴 ≤ 3) Blank 4e. The probability that there will be 4 accidents in a 24-hour period Blank 5f. The expected number of accidents per week. Blank 6g. Let 𝐵=number of hours in a 12-hour cycle that there are no accidents. Calculate the probability that for 8 hours in the 12-hour cycle there will be no accidents. Blank 7
A car-hire firm has two cars, which it hires out day by day. The number of demand for a car on each day follows Poissondistribution with mean 1.5. Calculate the proportion of days on which neither car is used
A limousine service receives 236 calls over 40 days from customers who need a ride to the airport.Use the Poisson distribution to find the probability that fewer than 2 customers need a ride on a given day.Do not round intermediate computations, and round your answer to three decimal places.(If necessary, consult a list of formulas.)
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.