Be sure to show all your work for problems requiring calculations in order to earn
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1. (1 point) Determine which of the four levels of measurement:
Voltage measurements from my house Ratio level data
Book genre Nominal Level Data
Outside temperature in winter in Silver Spring Interval Level Data
Restaurant ratings on scale of 0-4 stars Ordinal Level Data
2. (1 point) Determine type of sampling:
I collected shoe sizes from my family Complete Census
1003 adults were called after their phone numbers were
randomly generated by computer Simple Random Sampling
Police stopped every sixth driver for sobriety test Systematic Sampling
U.S. D.O.C. collected data by randomly selecting 8
Federal prisons and surveying every prisoner Multistage Sampling
3. (2 points) I collected radon samples from basements in
my neighborhood and obtained the following:
|
Radon in cPi |
Frequency |
|
2-5 |
2 |
|
6-9 |
2 |
|
10-13 |
6 |
|
14-17 |
15 |
Identify class width Class width = 5-2 = 3
Identify class midpoints Class Midpoint =
and consecutively 7.5,11.5 and 15.5
Identify class boundaries Class boundaries = [1.5, 5.5], [5.5,9.5], [9.5, 13.5], [13.5, 17.5]
Create histogram
The required Histogram Using Excel is given below,
4. (1 point) A simple random sample of pages of my dictionary was collected. These are the number of words defined on each page. Find the mean, median, mode, and midrange.
51 63 36 43 34 62 73 39 53 79
I put the data in Excel after that Used Excel functions to get the values,
|
Statistic |
Formula |
Answer |
|
Mean |
AVERAGE() |
53.3 |
|
Median |
MEDIAN() |
52 |
|
Mode |
MODE() |
#N/A |
|
Mid Range |
(MAX()+MIN())/2 |
56.5 |
There are 1459 pages. Estimate the total number of words in the dictionary. Is the estimate likely to be an accurate estimate of the total number of words in the English language?
If there are 1459 pages an estimate of the total number of words in the dictionary would be = 1459*53.3 = 77764.7 = 77765
Yes, as we draw random sample of pages so this is expected to be an accurate estimate.
5. (1 point) Environmental scientists measured CO2 emissions of a sample of cars in tons. Is the value of 9.3 unusual? Base your answers on the standard deviation, which must be included in your answer.
7.2 7.1 7.4 7.9 6.5 7.2 8.2 9.3
Again Using Excel we have,
|
Mean |
7.6 |
|
Std. Dev |
0.85857 |
Note that the upper 3 sigma limit is = 7.6+2*0.85857 = 9.3171
So the value 9.3 is inside the 2 sigma limitand hence it is not an unusual value.
6. (2 points) A simple random sample of starting weekly salaries for college graduates was obtained. What is the 5-number summary? Construct a boxplot.
664 693 698 714 751 753 779 789 802 818 834 836
Using MINITAB the obtained table and graph is given below,
Descriptive Statistics: Values
Variable Minimum Q1 Median Q3 Maximum
Values 664.0 702.0 766.0 814.0 836.0
7. (1 point) What is the probability of selecting a person who has high blood pressure or being male?
Has High Blood Pressure
|
|
Yes |
No |
|
Male |
201 |
288 |
|
Female |
126 |
224 |
Required Probability =
What is the probability of selecting a woman?
Required Probability =
8. (1 point) If 2 out of 100 second graders are randomly selected, find the probability that they are both boys and like spring:
|
|
Spring |
Summer |
Fall |
Winter |
|
Boys |
39 |
35 |
8 |
4 |
|
Girls |
6 |
5 |
2 |
1 |
a) With replacement?
Required Probability =
b) Without replacement?
Required Probability =
9. (1 point) Human body temperatures have a mean of 98.200 F and a standard deviation of 0.620 F. Convert each of the following to a z-score and determine if it is usual or unusual.
a. 96.900 F
Z Score = ; Unusual
b. 96.980 F
Z Score = ; Usual
10. (1 point) My parents lived in Florida and wanted to know the probability of winning the Florida Fantasy 5 if they bought one ticket. They had to pick the five winning numbers from 1–36. Each randomly selected number is different and order doesn’t matter.
The required Probability =
11. (2 points) Groups of five babies are randomly selected. In each group, the random variable x is the number of babies with green eyes.
|
x |
P(x) |
|
0 |
0.528 |
|
1 |
0.360 |
|
2 |
0.098 |
|
3 |
0.013 |
|
4 |
0.001 |
|
5 |
0+ |
Is this a probability table? If not, why not? If it is, what is the mean and standard deviation?
As all the probabilities is non-negative and the total probability is 1 so this is a probability distribution table.
The mean and SD can be found using the formula,
Mean = and SD =
The detailed calculation was done by Excel and the table is given below.
|
x |
P(x) |
xP(x) |
x^2*P(x) |
|
0 |
0.528 |
0 |
0 |
|
1 |
0.36 |
0.36 |
0.36 |
|
2 |
0.098 |
0.196 |
0.392 |
|
3 |
0.013 |
0.039 |
0.117 |
|
4 |
0.001 |
0.004 |
0.016 |
|
5 |
0 |
0 |
0 |
|
|
|
|
|
|
Total |
1 |
0.599 |
0.885 |
|
|
|
|
|
|
|
Mean = |
0.599 |
|
|
|
SD = |
0.725 |
|
12. (1 points) Ten peas are generated from parents having the green/yellow pair of genes so there is a 75% probability that an individual pea will have a green pod. Find the probability that among the 10 offspring peas, at least 9 have green pods. Is this unusual? Why or why not?
Here we can use the Binomial distribution as all conditions of the Binomial distribution is followed by the variable. So we have,
X = Number of green pods out 10 offspring peas ~Binomial(10,0.75)
Required Probability = P(X ≥ 9) = 0.2440 (Using Excel)
As we can see that the required probability is quite high so this is not unusual.
13. (1 point) Find the critical value:
a. Z0.02= 2.0537 (Excel formula: Norm.S.Inv(0.98))
b. Z0.01= 2.3263(Excel formula: Norm.S.Inv(0.99))
c. Z0.10= 1.2816(Excel formula: Norm.S.Inv(0.90))
14. (2 points) Assume adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15.
Find the probability that a randomly selected adult has IQ < 115
Required Probability = 0.8413 (Excel Formula: NORM.DIST(115,100,15,TRUE))
Find the probability that a randomly selected adult has IQ > 70
Required Probability = 0.9772 (Excel Formula: 1-NORM.DIST(70,100,15,TRUE))
Find P30, which is the IQ separating the bottom 30% from the top 70%
P30 = 92.1340 (Excel Formula: NORM.INV(0.30,100,15))
15. (2 points) Assume women’s ring sizes are normally distributed with a mean of 6 and a standard deviation of 1.0.
a. If one woman is randomly selected, find the probability that her ring size is < 6.2
Required Probability = 0.5793 (Excel Formula: NORM.DIST(6.2,6,1,TRUE))
b. If 100 women are randomly selected, find the probability that the mean ring size is < 6.2
Required Probability = 0.9772 (Excel Formula: NORM.DIST(6.2,6,1/SQRT(100),TRUE))
c. A jeweler sees part b. and decides that all rings should be made < 6.2 because they would fit all but a few women with a ring guard. What’s wrong with this reasoning
The 2nd part is the probability of the average ring size, but as the ring would be worn by one individual so he should use the answer of part a.
2.4.16
It suggests that USA has the lowest weekly instructions times for school children.
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