Download 2. A university has been tracking the percentage of alumni giving to

2. A university has been tracking the percentage of alumni giving to its annual fund each year for the
past 10 years. The data is given here.
14%, 13%, 15%, 21%, 19%, 24%, 25%, 28%, 25%. 31%
Answer the following questions about this data.
a.what are its mean and median?
b.What is the procedure for using mean and median to determine whether the data is skewed, and if so,
in what direction?
c.Apply the procedure you described to the mean and median computed in part a.
a)
Solution:
Mean =( 14+13+…….31)/10 = 21.5
Ordered data: 13%,14%,15%,19%,21%,24%,25%,25%,28%,31%
Median = Average of the middle values = (21+24)/2 = 45/2 = 22.5
Answer: Mean = 21.5% , Median = 22.5%
b)
Procedure:
Decision:
Mean = Median
Not Skewed
Mean > Median
Right Skewed
Mean < Median
Left Skewed
c)
Mean = 21.5% < Median = 22.5%
Then the data is Skewed to the left
3. Assume you have a data set from a normally distributed random variable. Answer the
following questions about it.
a.Will the random variable be discrete, continuous, or neither? How do you
know?
b. Will the data be qualitative or quantitative? How do you know?
a) Continuous, because a normally distributed random variable is continuous
b) Quantitative, because the output of a normally distributed random variable are real numbers
5. The mean time for a racecar driver’s crew to perform a pit stop is 13.2 seconds, with a standard
deviation of 0.9 seconds. To maintain his current lead, the driver needs a pit stop in 12.5 seconds or less.
Assuming this random variable is normally distributed, what is the probability of the driver getting the
pit stop in a short enough time to maintain is lead?
Solution:
Let X = time to get the pit stop
X has a normal distribution with mean  = 13.2 and standard deviation =  = 0.9
Z = (X-)/(X-13.2)/0.9 has standard normal distribution
P(X<12.5) = P(Z< -9/7) = 0.099
Answer: 0.099
6. From a sample of size 175, the sample mean is x bar=54.37 and sample standard deviation is s= 7.07.
a. Construct a 95% confidence interval for the population mean and show your work.
b. Explain how your work in part a would have been different if the sample size had been only 12
instead.
a)
Solution:
Since the sample size is 175 > 30 we can use the z-distribution to find the CI
CI at 95% ( known) is; ( x-bar-z(0.025)/n ,x-bar+z(0.025)/n)
We know:
n=175
x-bar = 54.37
 = 7.07
z(0.025) = 1.96
95% CI for the population mean is: ( 54.37-1.96(/175) ,54.37+1.96(/175)) = (53.32,55.42)
Answer: (53.32,55.42)
b)
Solution:
Since the sample size is 12 < 30 we must use the t-distribution to find the CI
Formulas are the same than in part a) but z(0.025) must be changed by t(11,0.025) = 2.201
95% CI for the population mean is: ( 54.37-2.201(/12) ,54.37+2.201(/12)) = (49.88,58.86)
Answer: (49.88,58.86)
7. A random sample from the population of registered voters in California is to be taken and then
surveyed about an upcoming election. What sample size should be used to guarantee a sampling error
of 3% or less when estimating p at the 95%confidence level?
Solution:
E = z(0.025)(1/4n) ≤ 0.03
1.96/(4n) ≤ 0.03
(4n)/1.96 ≥ 1/0.03
(4n) ≥1.96/0.03
4n ≥ (1.96/0.03)2
n ≥ (1.96/0.03)2/4 = 1067.11
Rounding up a sample size of 1068 is needed
Answer: n = 1068