Stats Exam 2 study guide
When we are using normal distribution we need what two things?
the mean and standard deviation mean: (n)(pi) SD: root of (n)(pi)(1-pi)
Normal Distribution: in 2020, the avg charge of tax preparation was 95$. Assume a normal distribution and a standard deviation of $10. 1. What proportion of tax fees were more than 95$? 2. What proportion of tax preparation fees were between $75 and $115? (x-mean)/standard deviation 3. If the area to the right of a positive Z is .0869, then the value of Z must be? 4. The Z-score representing the 75th percentile of the standard normal distribution is?
1. .5 because $95 IS the mean. anything above or below the mean is .5 2. (75-95)/10=-2 (115-95)/10=2 The value of 2 on the chart is .4772. Therefore the proportion of tax prep fees between $75 and $115 are .4772+.4772=.9544 3. .5-.0869=.4131; and the Z-score from the chart for .4131= 1.36
Using the normal distribution to approximate the binomial: An elementary school teacher has learned that 40% of school age children have at least 3 cavities. The teacher has 30 students in her class. 1. How many students would he expect in his class to have at least three cavities? (find the mean) 2. What is the standard deviation? (root of (n*pi)(1-pi)) 3. Using the appropriate approximation, determine P(x>19) (new K-mean)/SD= Z-score
1. (n)(pi)=(.4)(30)= 12 2. root of (12)(1-.4)= 2.683 3. (19.5-12)/2.683= 2.79= Z the probability of 2.79= .4974, therefore to find out what is above that (x>19) you must take .5-.4974= .0026 - there is a .0026 chance that more than 19 students in the class will have three cavities.
If the binomial is P(x<k) then APPROX by using
P(x<(k-1).5) Example. if P(x<3), then P(x<2.5)
If the binomial is P(x>k) then APPROX by using
P(x>k.5)
finding mean and variance of a binomial distribution. mean= n*pi variance= (n*pi)(1-pi)
example. n=75 pi=.20 Mean= 15 Variance= 12
Uniform: a bus arrives every 30 min. what is the probability that a student will wait more than 25 mins? Also, find the average wait time (b-a)/2 and standard deviation. (root of ((b-a)^2/12)
p(x)= 5*(1/30)= .1667 Avg wait time: (30-0)/2= 15 SD: root of (30-0)^2/12= 8.66