CLT Application #2 : 2016 SAT scores on the Critical Reading section followed a
ID: 3242019 • Letter: C
Question
CLT Application #2: 2016 SAT scores on the Critical Reading section followed a normal distribution with a mean of 494 and a standard deviation of 117.
https://research.collegeboard.org/programs/sat/data/cb-seniors-2016 (Links to an external site.)Links to an external site.
a) If a random sample consisting of a single student were chosen, what is the probability their score would be at least 600?
b) If a random sample consisting of 5 students were chosen, what is the probability the sample mean score would be at least 600?
c) If a random sample consisting of 10 students were chosen, what is the probability the sample mean score would be at least 600?
d) If a random sample consisting of 17 students were chosen, what is the probability the sample mean score would be at least 600?
CLT Application #3: Undergraduate students at UCF have a mean age of 23 years with a standard deviation of 5 years.
a) What do you think the distribution of undergraduate student ages would look like? Guess/Sketch! Hint: It would NOT be normally distributed; why?
b) How large of a sample size is needed for the distribution of sample mean ages to be normal?
c) If a random sample of 40 students were selected, what would the z-score be for a sample mean age of 21? Would this z-score be unusual? Why or why not?
d) If a random sample of 400 students were selected, what would the z-score be for a sample mean age of 21? Would this z-score be unusual? Why or why not?
Explanation / Answer
CLT Application #2
Mean = 494 and standard deviation = 117
(a) Let x be the 2016 SAT scores on the Critical Reading section
Pr(x >= 600; 494; 117) = 1 - Pr(x <600 ; 494; 117)
Z -value = (600 - 494)/117 = 0.906
so Pr(x >= 600; 494; 117) = 1 - Pr(x <600 ; 494; 117) = 1 - (0.906)
where is normal standard cumulative distribution
Pr(x >= 600; 494; 117) = 1- 0.8175 = 0.1825
(b) sample size n = 5
standard error of the sample mean = /n = 117/5 = 52.324
so Pr(xbar >=600; 494; 52.324) = 1 - Pr(xbar < 600; 494; 52.324)
Z - value = (600 - 494)/ 52.324 = 2.026
so Pr(xbar >=600; 494; 52.324) = 1 - Pr(xbar < 600; 494; 52.324) = 1- (2.026)
where is normal standard cumulative distribution
Pr(x >= 600; 494; 52.324) = 1- 0.9786 = 0.0214
(c) sample size = 10
standard error of the sample mean = /n = 117/10 = 37.00
so Pr(xbar >=600; 494; 37.00) = 1 - Pr(xbar < 600; 494; 37.00)
Z - value = (600 - 494)/ 37.00 = 2.865
so Pr(xbar >=600; 494; 37.00) = 1 - Pr(xbar < 600; 494; 37.00) = 1- (2.865)
where is normal standard cumulative distribution
Pr(x >= 600; 494; 52.324) = 1- 0.9979= 0.0021
(d)
sample size = 17
standard error of the sample mean = /n = 117/17 = 28.38
so Pr(xbar >=600; 494; 28.38) = 1 - Pr(xbar < 600; 494; 28.38)
Z - value = (600 - 494)/ 28.38 = 3.735
so Pr(xbar >=600; 494; 28.38) = 1 - Pr(xbar < 600; 494; 28.38) = 1- (3.735)
where is normal standard cumulative distribution
Pr(x >= 600; 494; 28.38) = 1- 0.9999= 0.0001
CLT #3
(a) As the undergraduate students have students from 4 years and normally large number of students dropped out which decrease the number of students in later years so the distribution may be exponential distribution.
(b) Sample size must have be of minimum 30 for the distribution of samle mean ages to be normal.
(c) sample size = 40
so standard error of the sample mean = /n = 5/ 40 = 0.791
so Z - score = (21 -23)/ 0.791 = -2.53
yes this is unusuall and it is because it is more than 2 standard deviation away from mean value of 23.
(d) sample size = 400
so standard error of the sample mean = /n = 5/ 400 = 0.25
so Z - score = (21 -23)/ 0.25 = -8
yes this is quite unusual and it is because it is more than 2 standard deviation away from mean value of 23.