MATH534 Final Exam Solution’s Study Guide
You should work each of the following on your own, then review the
solution’s guide. DO NOT look at the solution’s guide first.
1. The following numbers represent the weights in pounds of six 7-year
old children in Mrs. Jones' second grade class (25, 60, 51, 47, 49, 45).
Find the mean, median, mode, variance, and 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)
These can also be found using the MATH533 spreadsheet.
2. If the variance is 846, what is the standard deviation?
Solution: standard deviation = square root of variance = sqrt(846) =
29.086
3. 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.
4. For two events, C and D, P(C) = 0.6, P(D)=0.3, and P(C|D) = 0.2.
Find P(C𝗇D).
Solution:
P(C𝗇D) = P(D) * P(C|D)
= 0.3*0.2
= 0.06
You should work each of the following on your own, then review the
solution’s guide. DO NOT look at the solution’s guide first.
1. The following numbers represent the weights in pounds of six 7-year
old children in Mrs. Jones' second grade class (25, 60, 51, 47, 49, 45).
Find the mean, median, mode, variance, and 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)
These can also be found using the MATH533 spreadsheet.
2. If the variance is 846, what is the standard deviation?
Solution: standard deviation = square root of variance = sqrt(846) =
29.086
3. 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.
4. For two events, C and D, P(C) = 0.6, P(D)=0.3, and P(C|D) = 0.2.
Find P(C𝗇D).
Solution:
P(C𝗇D) = P(D) * P(C|D)
= 0.3*0.2
= 0.06