Download Solutions to the Central Limit Theorem Review 1. A) The proportion

Solutions to the Central Limit Theorem Review
1. A) The proportion is 68/200 = .34 or 34% .
B) The answer is
p(1  p)
.34(.66)

 .03345  3.345% .
n
200
C) The mean of these proportions should be close to the true proportion of students
attending the game, which, we hope, is about 34%.
D) This is the same question as part B). It is still .03345 or 3.345% because it doesn’t
matter how many trials you do. The only thing that matters is how many people are in your
sample.
E) This should approximate a normal curve.
F) The sample size (200) is less than 1/10 of Lassiter’s population, (at one point it was
2002) – but only barely. The number of people predicted to attend the game (np = 200.34
= 68) is larger than 10 and the number of people who won’t be attending the game (n(1-p) =
200.66 = 132) is also greater than 10. Therefore the Central Limit Theorem can be used
here.
G) We should make the sample size 4 times as large. In other words, ask 800 students
instead of just 200. (Of course, that means the Central Limit Theorem no longer works
because Lassiter’s student population is not larger than 8000 students.)
2. A) This should not be a normal distribution because, although students could text way
more than the mean of 38, they can’t text a negative number of times. This data has a
lower limit of 0 that will be reached by all students without a phone. This will skew the
data, so it shouldn’t be normal.
B) The mean of these means should be close to the actual population mean, which we
would approximate to be 38. The standard deviation of these means should be
13.5
 .6037 . Once again, the number of trials is irrelevant - as we saw in the Gettysburg
500
Address task. (Didn’t the standard deviation of all of our means come very close to the
actual standard deviation divided by 5, 10 or 25?)
C) Yes, because we have a distribution of sample means, and such distributions are
approximately normal, especially if n is large.
D) Our sample of 500 is way less than 1/10 of the number of high school students.
Although we can’t check anything with proportions, according to Ms. Linner, as long as n is
large (more than 50), the Central Limit Theorem applies.
E) Gather 4 times as many student responses – in other words, ask 2000 students.
3. A) p(3) = C(5, 3)(.72)3(.28)2  .2926  29.26% .
B) p(3) = C(10, 3)(.72)3(.28)7  .00604  .604%
C) This is p(3) + p(4) + p(5) + … + p(10) or 1 – p(0) – p(1) – p(2) = 1 – 1(.72)0(.28)10 –
10(.72)1(.28)9 – 45(.72)2(.28)8  .9990  99.9%
D) np = .72(3000) = 2160
E) It is a cluster sample or a convenience sample.
F) The standard deviation is
p(1  p)
.72(.28)

 .0081975  .8198% .
n
3000
4. A) My initial answer was “yes”, since heights are normal distributions, but that isn’t
correct. Since the mean height here is taller than the actual mean male height, we are
choosing students from the right area of the normal curve, which looks like this:
So the distribution is most likely not normal.
77  74 

B) p(h > 6’5”) = p(h > 77”) = p z 
  .208737 .
3.7 

67  74 

C) p(h < 5’7”) = p(h < 67”) = p z 
  .02925 .
3.7 

D) According to the Empirical Rule, about 68%, but by the calculator, 68.27% .
E) According to the Empirical Rule, about 95%, but the machine says 95.45% .