Week 2 Worksheet_ Probability and Normal Distribution
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Jan 9, 2024
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MATH 201 Statistics for Environmental Professionals ● Worksheet 2
Name: Orissa Pollard
Section 1. Suppose that you work to control the spread of infectious disease. One of your tasks is to
understand the pathway and sex differences in Zika virus infections. The following table represents
the confirmed and probable non-congenital cases of Zika virus disease in 50 U.S. states and the DC
in 2016 by mode of infection and sex. Refer to the textbook 3.2 – 3.4.
Female
(F)
Male
(M)
Total
Travel-associated
(T)
3,163
1,734
4,897
Local mosquito
borne (L)
103
121
224
Total
3,266
1,855
5,121
Source: Centers for Disease Control and Prevention
https://www.cdc.gov/mmwr/volumes/67/wr/mm6709a1.htm
Note: Data collection has often categorized findings by binary sex.
Question 1. Find the probability that a randomly selected case from all cases represented in the
table is travel-associated. Note this is P(T). Show your work. (10 pts.)
The probability that a randomly selected case is travel-associated P (T) = Number of Travel-associated
cases / Total number of cases
P (T) = 3,163/5121 + 1,734/5121 =(3163+1734)/5,121
P (T) = 4,897 / 5,121
P (T) = 0.956
Question 2. Find the probability that a randomly selected person represented in this table is female
AND travel associated. Note this is P(F AND T).
Show your work. (10 pts.)
P (FnT) = (intersection of F and T which is equal to 3163)/(total number = 5121)
P (FnT) = 3163/5121
P (FnT) = 0.618
Question 3. Find the probability that a randomly selected case from all cases represented in the
table is a female OR travel-associated. Note this is P(F OR T). Show your work. (10 pts.)
P (FuT) = P (F) + P (T) - P (F n T) where P (F u T) is the union and P (F n T) is the intersection
P (F) = 3266/5121= 0.6378
P (T) = 4897/5121=0.9563
P (FnT) =3163/5121= 0.6177
P (FuT) = 0.6378+0.9563-0.6177
P (FuT) = 0.9764
Question 4. Find the conditional probability that a randomly selected case from all cases
represented in the table is a travel-associated given the person is female. Note this is P(T | F). Show
your work. (10 pts.)
P (T | F) = P (TnF) / P (F)
From question 1, P (TnF) = 0.6177, and P (F) = 0.6378
P (T | F) = 0.6177/ 0.6378
P (T | F) = 0.9684
Question 5. Based on the probabilities you have calculated for Questions 1 to 4. What is your
conclusion concerning the pathway and sex differences in Zika virus infection? (10 pts.)
Since P(T | F) is greater than P(T) and P(F), we can conclude that there is a strong association between
being female and being travel-associated in the Zika virus infection cases represented in this table.
Section 2. Suppose that you work for a government sector for water resources management. Your
section is responsible for estimating the required amount of water for household use. Assume that
the volume of daily water use per household in North America follows a normal distribution. The
mean volume of daily water use is 138 gallons per household and the standard deviation is 46
gallons. Refer to the textbook 6.1 and 6.2.
Question 1. Sketch (by hand or digitally) the distribution of daily water use (in gallon) per
household in North America. Clearly label x and y-axis. Refer to the textbook 6.1. Figure 6.3. (10
pts.)
0.008
0.006
probability
0.004
0.002
0
0
0
50 100 150 200 250
Question 2. Find the probability that a randomly selected household in North America uses less
than 69 gallons of water per day. Calculate the z-value using the formula and find the probability
using the z-table or other technology (e.g. calculator). Show your work. (10 pts.)
z = (x - μ) / σ
z = (69 - 138) / 46
z = -1.5
From the z-table, the probability of a z-score of -1.5 or less is 0.0668.
Therefore, the probability that a randomly selected household in North America uses less than 69 gallons
of water per day is 0.0668.
Question 3. Find the probability that a randomly selected household in North America uses greater
than 161 gallons of water per day. Calculate the z-value using the formula and find the probability
using the z-table or other technology (e.g. calculator). Show your work. (10 pts.)
z = (x - μ) / σ
z = (161 - 138) / 46
z = 0.5
Reading from the z-table, the probability of a z-score of 0.5 or more is 0.3085. Therefore, the probability
that a randomly selected household in North America utilizes more than 161 gallons of water per day is
0.3085.
Question 4. Find the probability that a randomly selected household in North America uses between
69 and 161 gallons of water per day. Show your work. (10 pts.)
P (69 < X < 161) = P (Z < (161-138)/46) – P (Z < (69-138)/46)
P (69 < X < 161) = P (Z < 0.5) – P (Z < -1.5)
P (69 < X < 161) = 0.6915 - 0.0668
P (69 < X < 161) = 0.6247
Question 5. Find the 90
th
percentile of the value of daily water use per household in gallon. Use the
formula of z-score and either z-table or other technology (e.g. calculator). Show your work. (10 pts.)
Utilizing the z-score formula: z = (x - μ) / σ
P (Z <= z) = 0.90
Reading the value in the z-table we find that the z-value for the 90th percentile is approximately 1.28.
From the z-score formula we have daily water use per household (x) = z * σ + μ
Substituting the values:
x = 1.28 * 46 + 138 = 196.88 gallons
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