MATH201_Worksheet2 (1)
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Feb 20, 2024
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MATH 201 Statistics for Environmental Professionals ● Worksheet 2
Name: Amanda Perry
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.)
P(T)
=
Travel
−
Associated
Total
=
4,897
5,121
=
0.956 or 95.6%
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(F AND T)
= FemaleTravel
−
Associated
Total
=
3,163
5,121
=
0.618 or 61.8%
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(F OR T)
=
P(F)
+
P(T) -
P(F
–
T)
P(F)
= 3,266
5,121
=
0.638 P(T)
=
4,897
5,121
=
0.956 or 95.6%
P(F OR T)
=
P(F)
+
P(T)
- P(F AND T)
P(F OR T) = 0.638
+
0.956
-
0.618 =
0.976
P(F OR T)
= 0.976 or 97.6%
Question 4. Find the conditional probability that a randomly selected case from all cases represented in the table is travel-associated given the person is female. Note this is P(T | F). Show your work.
(10 pts.)
P(T | F)
=
P
(
F
∧
T
)
P
(
F
)
=
0.618
0.638
=
0.969 or 96.9%
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.)
Females are more susceptible to contracting the Zika virus through travel-related routes compared to males.
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.)
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
=
69
−
138
46
=−
1.5
x
=
69
z
=−
1.5
p
(
x
<
z
)
=
0.066807
∨
6.68%
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
=
161
−
138
46
=
0.5
p
(
x
>
161
)
=
1
−
p
(
x
<
161
)
=
0.308
∨
30.8%
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.)
69
gallons
:
Z
=−
1.5
161
gallons
:
Z
=
0.5
P
(
−
1.5
<
x
<
0.5
)
=
0.62466
or 62.46%
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.) z
=
1.282
μ
=
138
σ
=
46
z
=
1.282
x
−
138
46
46
(
1.282
)
=
x
−
138
x
=
138
+
46
(
1.282
)
=
196.97
gallons
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