Category Archives: Engineering Mathematics Blog

Chapter 1: Introduction to Statistics and Data Analysis


Problem 1.2

Problem 1.3

Problem 1.4

Problem 1.5

Problem 1.6

Problem 1.7

Problem 1.8

Problem 1.9

Problem 1.10

Problem 1.11

Problem 1.12

Problem 1.13

Problem 1.14

Problem 1.15

Problem 1.16

Problem 1.17

Problem 1.18

Problem 1.19

Problem 1.20

Problem 1.21

Problem 1.22

Problem 1.23

Problem 1.24

Problem 1.25

Problem 1.26

Problem 1.27

Problem 1.28

Problem 1.29

Problem 1.30

Problem 1.31

Problem 1.32

Problem 1.33


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Probability & Statistics for Scientists & Engineers Ninth Edition by Walpole, Myers, Myers, and Ye

Chapter 2: Probability

Chapter 3: Random Variables and Probability Distributions

Chapter 4: Mathematical Expectation

Chapter 5: Some Discrete Probability Distribution

Chapter 6: Some Continuous Probability Distribution

Chapter 7: Functions of Random Variables

Chapter 8: Fundamental Sampling Distributions and Data Descriptions

Chapter 9: One- and Two-Sample Estimation Problems

Chapter 10: One- and Two-Sample Tests of Hypotheses

Chapter 11: Simple Linear Regression and Correlation

Chapter 12: Multiple Linear Regression and Certain Nonlinear Regression Models

Chapter 13: One-Factor Experiments: General

Chapter 14: Factorial Experiments (Two or More Factors)

Chapter 15: 2k Factorial Experiments and Fractions

Chapter 16: Nonparametric Statistics

Chapter 17: Statistical Quality Control

Chapter 18: Bayesian Statistics


Textbook Solutions for Probability and Statistics



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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 7


Clear Lake has a surface area of 708,000 m2 (70.8 ha.). For a given month, the lake has an inflow of 1.5 m3/s and an outflow of 1.25 m3/s. A +1.0-m storage change or increase in lake level was recorded. If a precipitation gage recorded a total of 24 cm for this month, determine the evaporation loss (in cm) for the lake. Assume that seepage loss is negligible.


Solution:

We are given the following values:

\begin{align*}
\text{Area}, \ A&=708,000 \ \text{m}^2 \\
\text{Inflow}, \ I&=1.5 \ \text{m}^3/\text{s} \\
\text{Outflow}, \ O & = 1.25 \ \text{m}^3/\text{s} \\
\text{change in storage}, \ \Delta S & = 1.0 \ \text{m} \\
\text{Precipitation}, \ P&=24 \ \text{cm} \\
\text{time}, \ t &= 1 \ \text{month} = 30 \ \text{days}
\end{align*}

The required value is the \text{Evaporation}, \ E.

We shall use the formula

\Delta S=I+P-O-E

Solving for E in terms of the other variables, we have

E=I+P-O-\Delta S

Before we can substitute all the given values, we need to convert everything to the same unit of cm.

\begin{align*}
\text{Inflow}&=\frac{1.5\:\frac{\text{m}^3}{\text{s}}\cdot \frac{100\:\text{cm}}{1\:\text{m}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{708,000\:\text{m}^2} \\
\text{Inflow}&=549.1525 \ \text{cm}
\end{align*}
\begin{align*}
\text{Outflow}&=\frac{1.25\:\frac{\text{m}^3}{\text{s}}\cdot \frac{100\:\text{cm}}{1\:\text{m}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{708,000\:\text{m}^2} \\
\text{Outflow}&=457.6271 \ \text{cm}
\end{align*}
\Delta S=1.0 \ \text{m} \times \frac{100 \ \text{cm}}{1.0 \ \text{m}}=100 \ \text{cm}

Now, we can substitute the given values in the formula

\begin{align*}
E & =I+P-O-\Delta S \\
E& =549.1525 \ \text{cm}+24 \text{cm}-457.6271 \ \text{cm}-100 \ \text{cm} \\
E& =15.5254 \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 6


A lake with a surface area of 1050 acres was monitored over a period of time. During a one-month period, the inflow was 33 cfs, the outflow was 27 cfs, and a 1.5-in. seepage loss was measured. During the same month, the total precipitation was 4.5 in. Evaporation loss was estimated as 6.0 in. Estimate the storage change for this lake during the month.


Solution:

We are given the following values:

\begin{align*}
\text{Area}, \ A&=1050 \ \text{acres} \\
\text{Time}, \ t&=1 \ \text{month} \\
\text{Inflow}, \ I&=33 \ \text{cfs} \\
\text{Outflow}, \ O&=27 \ \text{cfs} \\
\text{Ground seepage}, \ G&=1.5 \ \text{in} \\
\text{Precipitation}, \ P&=4.5 \ \text{in} \\
\text{Evaporation}, \ E&=6.0 \ \text{in}
\end{align*}

The formula that we are going to use is:

\sum \text{Inflows}-\sum \text{Outflows}=\text{Change in Storage}, \Delta S \\
\\ 
\sum I-\sum Q=\Delta S

In this case, the inflows are \text{Inflow} \ I and \text{Precipitation}, \ P, while the others are outflows. Our formula now becomes

\color{Blue} \sum \text{Inflow}-\color{Red} \sum \text{Outflow}=\color{Green} \Delta S
\\
\color{Blue}(I+P)- \color{Red}(O+G+E)=\color{Green}\Delta S

Before substituting, we need to convert all the given to inches. More specifically the outflow O and inflow I.

The inflow and outflow, in cfs, will be divided by the given are to come up with units of inches.

The inflow is

\begin{align*}
\text{Inflow}&=\frac{33\:\frac{\text{ft}^3}{\text{s}}\cdot \frac{1\:\text{acre}}{43560\:\text{ft}^2}\cdot \frac{12\:\text{in}}{1\:\text{ft}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{1050\:\text{acres}} \\
\text{Inflow}& =22.4416 \ \text{in}
\end{align*}

The outflow is

\begin{align*}
\text{Outflow}&=\frac{27\:\frac{\text{ft}^3}{\text{s}}\cdot \frac{1\:\text{acre}}{43560\:\text{ft}^2}\cdot \frac{12\:\text{in}}{1\:\text{ft}}\cdot \frac{3600\:\text{s}}{1\:\text{hr}}\cdot \frac{24\:\text{hr}}{1\:\text{day}}\cdot \frac{30\:\text{days}}{1\:\text{month}}\cdot 1\:\text{month}}{1050\:\text{acres}} \\
\text{Outflow}& =18.3613\ \text{in}
\end{align*}

Now that everything is in inches, we can now substitute the values in the formula

\begin{align*}
\Delta S&=\left( I+P \right)-\left( O+G+E \right) \\
\Delta S&=\left( 22.4416 \ \text{in}+4.5 \ \text{in} \right)-\left( 18.3613 \ \text{in}+1.5 \ \text{in}+6 \ \text{in} \right) \\
\Delta S&=1.0803 \ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

We can also state the change in storage in terms of volume by multiplying the given area

\begin{align*}
\Delta S \ \text{in volume} & =1.0803 \ \text{in}\times 1050 \ \text{acres}\times \frac{1 \ \text{ft}}{12 \ \text{in}}\\
\Delta S \ \text{in volume} & = 94.5263 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 5


List seven major factors that determine a watershed’s response to a given rainfall.


Solution:

The seven major factors that determine a watershed’s response to a given rainfall are:

  1. Drainage Area
  2. Channel Slope
  3. Soil Types
  4. Land Use
  5. Land Cover
  6. Main Channel and tributary characteristics-channel morphology
  7. The shape, slope, and character of the floodplain

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 4


Explain how air masses are classified. Where are these types of air masses located?


Solution:

They are classified in two ways: the source from which they are generated, land (continental) or water (maritime), and the latitude of generation (polar or tropical).

These air masses are present in the United States. The Continental polar emanates from Canada and passes over the northern United States. The maritime polar air mass also comes southward from the Atlantic Coast of Canada and affects the New England states. Another maritime polar comes from the Pacific and hits the extreme northwestern states. The maritime tropical air masses come from the Pacific, the Gulf of Mexico, and the Atlantic (these affect the entire Southern United States). Continental tropical air masses form only during summer. They originate in Texas and affect the states bordering the north.


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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 3


Explain the difference between humidity and relative humidity.


Solution:

Humidity is a measure of the amount of water vapor in the atmosphere and can be expressed in several ways. Specific humidity is a mass of water vapor in a unit mass of moist air while relative humidity is a ratio of the air’s actual water vapor content compared to the amount of water vapor at saturation for that temperature.


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