I have a problem I need some help on. I have answered parts. Any guidance would be appreciated. K
For many years the numbers of accidents per week at a hazardous intersection have been studied. The number of accidents per week varies with mean 2.2 and standard deveiation 1.4.
1. Could the exact distributin of the number of accidents per week be described by a normal distribution?
I've drawn the normal curve showing the mean of 2.2 in the center of the curve and deviating both directions 1.4
2. Looking at only 52 weeks we can calculate x-bar the mean number of accidents per week. What is the approximate distribution of x-bar according to the Central Limit Theorem?
N(2.2, 0.194)
3. What is the approximate probability that x-bar is greater than 2 accidents per week?
P(x-bar > 2)
4. What is the probability that there are fewer than 100 accidents at the intersection in a year?
1. Could the exact distribution of the number of accidents per week be described by a normal distribution?
No, because the number of accidents is always an integer larger or equal than zero.
But the mean number of accidents per week, computed over some time T does tend to the normal distribution for
T ---> infinity. The standard deviation of the mean over a period of T is the original standard deviation divided by sqrt(T).
Thanks, what about probability that there are fewer than 100 accidents at the intersection in a year? The hint says to "restate the event in terms of x-bar minus mean number of accidents per week.
If you have 100 accidents in a given year, then in that year the average per week was x-bar = 100/52. You know what the true average is and you know what the standard deviation of the yearly average is (sigma/sqrt[52]).
You then look up the probability that
Z < (x-bar - average)/[sigma/sqrt[52]]
from the standard normal distribution table, or using statistical software.
To calculate the probability that there are fewer than 100 accidents at the intersection in a year, we need to restate the event in terms of x-bar (the mean number of accidents per week) minus the mean number of accidents per week.
Let's call the mean number of accidents per week "μ" and the standard deviation of the yearly average "σ/√52" (where σ is the original standard deviation).
Now, we can use the standard normal distribution to calculate the probability:
P(x-bar < 100/52) = P((x-bar - μ) / (σ/√52) < (100/52 - μ) / (σ/√52))
By substituting the values of μ and σ, we can calculate the probability using the standard normal distribution table or statistical software.
To find the probability that there are fewer than 100 accidents at the intersection in a year, we need to restate the event in terms of x-bar minus the mean number of accidents per week.
Let's say x-bar represents the average number of accidents per week. We know that the mean number of accidents per week is 2.2. Therefore, to find the average number of accidents in a year, we multiply the average number of accidents per week by 52.
So, (x-bar - average) represents the difference between the average number of accidents per week and the mean number of accidents per week.
The standard deviation of the yearly average, sigma/sqrt[52], represents the variation in the average number of accidents per week over the course of a year.
To calculate the Z-score, we divide the difference between x-bar and the mean by the standard deviation: (x-bar - average) / [sigma/sqrt[52]].
We then lookup the probability that Z is less than this calculated Z-score in the standard normal distribution table or using statistical software.
Note: To calculate the value of sigma (standard deviation), we divide the given standard deviation of 1.4 by sqrt(52), as mentioned earlier.