MATH 221 FINAL EXAM REVIEW QUESTIONS WITH ANSWERS( DEVRY UNIVERSITY)
Final Exam
Review Questions – MATH221
You should work each of the following on your own, then
review the solutions guide. DO NOT look at the solutions guide
first.
1. Determine whether the following are nominal, ordinal, interval, and
ratio.
a. Daily temperatures in Ripon, WI
b. Test scores in statistics class
Solution: (a) would be interval as there is no zero while (b) would be ratio
as there is a zero.
2. The following numbers represent the weights in pounds of six 7-year
old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45}
Find the mean; median; mode; variance; standard deviation.
Solution: This would be a sample from the class
mean = 46.166 (=AVERAGE)
median = 48 (=MEDIAN)
mode does not exist (looking at the data)
variance = 134.5667 (=VARIANCE.S)
standard deviation =11.60029 (=STDEV.S)
3. If the variance is 846, what is the standard deviation?
Solution: standard deviation = square root of variance = sqrt(846) =
29.086
4. If we have the following data: 34, 38, 22, 21, 29, 37, 40, 41, 22, 20,
49, 47, 20, 31, 34, 66. Draw a stem and leaf. Discuss the shape of the
distribution.
Solution:
2|219200
3|48714
4|0197
5|
6|6
This distribution is right skewed (positively skewed) because the “tail”
extends to the right.
5. Find the regression equation of the following data.
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, X 6 5 7 6 5 6 8 9 4
y 14 33 43 54 21 33 43 24 28
Solution: y-hat = 1.32x + 24.32, found by using Data and then Data
Analysis menu option.
6. To predict the annual rice yield in pounds we use the equation y-hat =
859 + 5.76x1 + 3.82x2 where x1 represents the number of acres
planted and where x2 represents the number of acres harvested and
where r2 = 0.94.
a. Predict the annual yield when 3200 acres are planted and 3000
are harvested
b. Interpret the results of r2 value.
Solution:
(a) y-hat = 859 + 5.76*3200 + 3.82*3000
= 859 + 18432 + 11460
= 30751 which is 30,751 pounds of rice
(b) 94% of the variation in the annual rice yield can be explained by the
number of acres planted and harvested. The remaining 6% is unexplained
and is due to other factors or to chance.
7. The Student Services office did a survey of 500 students with the
following results:
Transfer Non-transfer Total
Part-time 100 110 210
Full-time 170 120 290
Total 270 230 500
a. Find the probabilities that a student is a transfer student.
b. Find the probability that a students is part-time.
c. Find the probability that a student is a transfer student and a
part-time student
d. Find the probability that a student is a transfer student, given
that the student is part-time, P(transfer|part-time).
Solution: (a) The total number of transfer students is 270. The total
number of students in the survey is 500. P(Transfer) = 270/500 = .54 (b)
The total number of part time students is 210. The total number of
students in the survey is 500. P(Part Time) = 210/500 = .42 (c) From the
table we see that there are 100 students which are both transfer and part
time. This is out of 500 students in the sample. So, 100/500 = .20 (d) This
is conditional probability and so we must change the denominator to the
total of what has already happened. There are 100 students which are
This study source was downloaded by 100000841990434 from CourseHero.com on 04-29-2022 22:42:55 GMT -05:00
https://www.coursehero.com/file/15953425/MATH221-Final-Exam-study-guide-solutions/
Final Exam
Review Questions – MATH221
You should work each of the following on your own, then
review the solutions guide. DO NOT look at the solutions guide
first.
1. Determine whether the following are nominal, ordinal, interval, and
ratio.
a. Daily temperatures in Ripon, WI
b. Test scores in statistics class
Solution: (a) would be interval as there is no zero while (b) would be ratio
as there is a zero.
2. The following numbers represent the weights in pounds of six 7-year
old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45}
Find the mean; median; mode; variance; standard deviation.
Solution: This would be a sample from the class
mean = 46.166 (=AVERAGE)
median = 48 (=MEDIAN)
mode does not exist (looking at the data)
variance = 134.5667 (=VARIANCE.S)
standard deviation =11.60029 (=STDEV.S)
3. If the variance is 846, what is the standard deviation?
Solution: standard deviation = square root of variance = sqrt(846) =
29.086
4. If we have the following data: 34, 38, 22, 21, 29, 37, 40, 41, 22, 20,
49, 47, 20, 31, 34, 66. Draw a stem and leaf. Discuss the shape of the
distribution.
Solution:
2|219200
3|48714
4|0197
5|
6|6
This distribution is right skewed (positively skewed) because the “tail”
extends to the right.
5. Find the regression equation of the following data.
This study source was downloaded by 100000841990434 from CourseHero.com on 04-29-2022 22:42:55 GMT -05:00
https://www.coursehero.com/file/15953425/MATH221-Final-Exam-study-guide-solutions/
, X 6 5 7 6 5 6 8 9 4
y 14 33 43 54 21 33 43 24 28
Solution: y-hat = 1.32x + 24.32, found by using Data and then Data
Analysis menu option.
6. To predict the annual rice yield in pounds we use the equation y-hat =
859 + 5.76x1 + 3.82x2 where x1 represents the number of acres
planted and where x2 represents the number of acres harvested and
where r2 = 0.94.
a. Predict the annual yield when 3200 acres are planted and 3000
are harvested
b. Interpret the results of r2 value.
Solution:
(a) y-hat = 859 + 5.76*3200 + 3.82*3000
= 859 + 18432 + 11460
= 30751 which is 30,751 pounds of rice
(b) 94% of the variation in the annual rice yield can be explained by the
number of acres planted and harvested. The remaining 6% is unexplained
and is due to other factors or to chance.
7. The Student Services office did a survey of 500 students with the
following results:
Transfer Non-transfer Total
Part-time 100 110 210
Full-time 170 120 290
Total 270 230 500
a. Find the probabilities that a student is a transfer student.
b. Find the probability that a students is part-time.
c. Find the probability that a student is a transfer student and a
part-time student
d. Find the probability that a student is a transfer student, given
that the student is part-time, P(transfer|part-time).
Solution: (a) The total number of transfer students is 270. The total
number of students in the survey is 500. P(Transfer) = 270/500 = .54 (b)
The total number of part time students is 210. The total number of
students in the survey is 500. P(Part Time) = 210/500 = .42 (c) From the
table we see that there are 100 students which are both transfer and part
time. This is out of 500 students in the sample. So, 100/500 = .20 (d) This
is conditional probability and so we must change the denominator to the
total of what has already happened. There are 100 students which are
This study source was downloaded by 100000841990434 from CourseHero.com on 04-29-2022 22:42:55 GMT -05:00
https://www.coursehero.com/file/15953425/MATH221-Final-Exam-study-guide-solutions/
