You are dealing here with a distribution of means rather than raw scores.
The standard error of the mean (SEm) = SD/√(n-1)
Would n = number of weeks?
Z = (score-mean)/SEm
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.
100 accidents/year = 1.923/week
The number of accidents per week at a hazardous intersection varies with mean 2.2
and standard deviation 1.4.
a. What is the distribution of Xbar, the mean number of accidents in one year, (52
weeks)? {Note you do not have to use R for this question, write down the
distribution with parameters here and then proceed]
b. What is the probability that Xbar is less than 2?
c. What is the probability that Xbar is between 5 and 20?
d. What is the probability that there are fewer than 100 accidents in a year?
1 answer