MATH 534 Final Exam

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(CD).

Solution:
P(CD) = P(D) * P(C|D)
= 0.3*0.2
= 0.06




MATH534 Final Exam Solution’s Study Guide. 1
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rev. 1/14/19


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, 5. If a binomial variable has n=34 and p=0.30, find the mean, variance,
and standard deviation.

Solution:
Mean = np = (34)(0.30) = 10.2
Variance = npq = (34)(0.30)(0.70) = 7.14
Standard deviation = sqrt(npq) = sqrt(7.14) = 2.67

6. A company claims that 61% of consumers know about their product. If
the company asks 18 consumers, what is the probability that (a)
exactly 10 will say yes, that (b) between 10 and 12 will say yes.

Solution: Can be done with Excel binomial spreadsheet. Can be done
by hand with these steps. For Part A, it would be
=BINOM.DIST(10,18,0.61,FALSE) to get 0.167 or 16.7%. For Part B, it
would be =BINOM.DIST(12,18,0.61,TRUE)-
BINOM.DIST(10,18,0.61,TRUE) to get 0.3634 or about 36%.

7. If the population has a mean of 58 and a standard deviation of 23.4,
what would be the mean and standard deviation of a sampling
distribution with a sample size of 22?

Solution:
Mean = mean of the population = 58
Standard deviation = population st dev divided by square root of the
sample size = 23.4 / sqrt(22) = 4.99

8. A restaurant asks 40 customers about their service times. If the
population mean service time is 28 minutes with a standard deviation
of 5.1 minutes, what is the probability that the mean of the sample
service times would be less than 26 minutes?

Solution:
The mean would still be 28 while the sample means standard deviation
would be 5.1/sqrt(40) = 0.806 minutes. The probability of less than 26
minutes can be found through the Excel spreadsheet for normal
distributions or with
=NORM.DIST(26, 28, 0.806, TRUE)
= 0.00654
Or less than 1%.




9. On a dry surface, the braking distance (in meters) of a certain car is a
normal distribution with mu = μ = 45.1 m and sigma = σ = 0.5.


MATH534 Final Exam Solution’s Study Guide. 2
This study source was downloaded by 100000829818949 from CourseHero.com on 03-01-2022 05:40:37 GMT -06:00
rev. 1/14/19


https://www.coursehero.com/file/42661808/MATH534-Final-Exam-Solutionsdocx/