# Solutions To Mathematics Textbooks/Probability and Statistics for Engineering and the Sciences (7th ed) (ISBN-10: 0-495-38217-5)/Chapter 3

## Section 3.1 - Random Variables

### Exercise 1

Let $X$ be the number of beams that failed due to shear (S). Three beams are sampled, and they will fail either from shear (S) or from flexure (F). The sample space is: $S = {SSS, SSF, SFS, SFF, FSS, FSF, FFS, FFF}.$

The random variable maps to the sample space as:

• $X = 0:\ FFF$
• $X = 1:\ SFF,\ FSF,\ FFS$
• $X = 2:\ FSS,\ SFS,\ SSF$
• $X = 3:\ SSS$

### Exercise 2

Three examples of Bernoulli random variables:

• Toss a coin once, let a success be having a the coin lands heads up. $X = \begin{cases} 1 & \text{head} \\ 0 & \text{tails} \end{cases}$
• Ask credit card users if their primary card is Visa or Mastercard. $X = \begin{cases} 1 & \text{Visa} \\ 0 & \text{Mastercard} \end{cases}$
• Inspecting pennies in a coin collection for steel pennies. $X = \begin{cases} 1 & \text{steel penny} \\ 0 & \text{copper penny} \end{cases}$

### Exercise 3

Consider an Experiment in which the number of pumps in use at each of two gas stations was determined. Define the random variables:

• $W =$ the number of pumps with a waiting line of vehicles waiting to fuel up.
• $Z =$ the number of pumps providing both Diesel and Gasoline.

### Exercise 4

Let $X$ be the number of nonzero digits in a randomly selected zip code.

$X = \begin{cases} 0 & \text{zip digits are: all zeros, e.g., 00000} \\ 1 & \text{zip digits are: 4 zero, 1 nonzero, e.g, 00100} \\ 2 & \text{zip digits are: 3 zeros, 2 nonzero, e.g., 10050} \\ 3 & \text{zip digits are: 2 zeros, 3 nonzero, e.g., 90210} \\ 4 & \text{zip digits are: 1 zero, 4 nonzero, e.g., 80223} \\ 5 & \text{zip digits are: 0 zeros, 5 nonzero, e.g., 76491} \end{cases}$

### Exercise 5

Consider a sample space for the number of rolls of a set of dice until the sum of 2 is rolled. Let $X = \begin{cases} 1 & \text{a sum of 2 is rolled before the 8th roll} \\ 0 & \text{a sum of 2 is not rolled until the 8th or later roll} \end{cases}$. In this case the random variable for mapping the sample space to the real numbers is no infinite, it is discrete and dichotomous.

### Exercise 6

$X$ is the number of cars observed at an intersection until a car turns Left ($L$). (Cars can turn right ($R$), left ($L$) or go straight ($A$)).

$X$ can take on the values 1, 2, 3, 4, ..., for example:

$X = \begin{cases} 1 & \{L\} \\ 2 & \{RL,\ AL\} \\ 3 & \{AAL,\ ARL,\ RAL,\ RRL\} \\ 4 & \{AAAL,\ AARL,\ ... \} \\ ... \end{cases}$

### Exercise 7

• Part a. Let $X$ be the number of unbroken eggs in a standard egg carton. The random $X$ is a discrete random variable.

$X \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

• Part b. Let $Y$ be the number of students on a class list who are absent on the first day of class. Say there are $n$ students on the class list. $Y$ is a discrete random variable.

$X \in \{0, 1, 2, 3, ..., n\}$

• Part c. Let $U$ be the number of swings before hitting a golf ball. $U$ is a discrete random variable.

$U \in \{1, 2, 3, 4, 5, 6, \ldots\}$

• Part d. Let $X$ be the length of a rattlesnake. $X$ is a continuous random variable.

$X \in (0, \infty)$

### Exercise 53.

Using the information in Exercise 30, the probability for any one person to be cited for $y$ moving violations while driving in the last three years is defined as:

$\begin{array}{l|cccc} y & 0 & 1 & 2 & 3 \\ \hline P(Y = y) & 0.60 & 0.25 & 0.10 & 0.05 \\ \end{array}$

Find the following probabilities for a group of 15 such persons.

• Part a. At least 10 have no citations. - Let $X$ be the number of person with no citations.
$X \sim \text{Binomial}\left(n = 15, p = 0.60\right)$
$\begin{array}{rcl} P\left(X \geq 10\right) &=& \sum_{x=10}^{15}\binom{15}{x}\left(0.60\right)^{x}\left(1-0.60\right)^{15-x} \\ &=& 0.1859 + 0.1268 + 0.0634 + 0.0219 + 0.0047 + 0.0005 \\ &=& 0.4032 \end{array}$
• Part b. Fewer than half have at least one citation. This is a little different than part a. Now the success probability is 0.40 for having at least one citation. Let $Y$ be the number of persons with at least one citation.
$Y \sim \text{Binomial}\left(n = 15, p = 0.40\right)$
$\begin{array}{rcl} P\left(Y < 15/2\right) &=& P\left( Y < 7.5\right) \\ &=& \sum_{y=0}^{7} \binom{15}{y}\left(0.40\right)^{y}\left(1-0.40\right)^{15-y} \\ &=& 0.0005 + 0.0047 + 0.0219 + 0.0634 + 0.1268 + 0.1859 + 0.2066 + 0.1771 \\ &=& 0.7869 \end{array}$
• Part c. The number of persons with at least one citation is between 5 and 10 inclusive?
$\begin{array}{rcl} P\left(5 \leq Y \leq 10\right) &=& \sum_{y=5}^{10} \binom{15}{y}\left(0.40\right)^{y}\left(1-0.40\right)^{15-y} \\ &=& 0.1859 + 0.2066 + 0.1771 + 0.1181 + 0.0612 + 0.0245 \\ &=& 0.7734 \end{array}$

### Exercise 54.

Let $X$ be the number of people who want an oversize version of a tennis racket.

$X \sim \text{Binomial}\left(n = 10, p = 0.60\right)$

• Part a. What is the probability that of ten people at least 6 will want the oversize version?
$\begin{array}{rcl} P(X \geq 6) &=& \sum_{x=6}^{10}\binom{10}{x} \left(0.60\right)^{x} \left(1-0.60\right)^{10-x} \\ &=& 0.2508 + 0.2150 + 0.1209 + 0.0403 + 0.0060 \\ &=& 0.6330 \end{array}$
• Part b. What is the the number who want the oversize version of the racket will be within one standard deviation of the mean?
First find the mean and the standard devation.
$\begin{array}{rl} E(X) = \mu &= np = 10 \times 0.60 = 6 \\ Var(X) = \sigma^2 &= np(1-p) = 2.4 \\ StdDev(X) = \sigma &= \sqrt{np(1-p)} = \sqrt{2.4} = 1.5492 \end{array}$
$\begin{array}{rcl} P\left( \left| X - \mu \right| < \sigma \right) &=& P\left( \mu - \sigma < X < \mu + \sigma \right) \\ &=& P\left(6 - 1.5492 < X < 6 + 1.5492\right) \\ &=& P\left(4.4508 < X < 7.5492\right) \\ &=& P(5 \leq X \leq 7) \\ &=& P(X = 5) + P(X = 6) + P(X = 7) \\ &=& 0.2007 + 0.2508 + 0.2150 \\ &=& 0.6665 \\ \end{array}$

• Part c. If there are 7 normal sized and 7 oversize rackets in the store, what is the probability all 10 customers will get what they want? To solve this think about what needs to happen for the stock in the store to meet the needs of the customers. If $X = 3$ then there are three people getting the oversize version and this leaves 7 to get the normal size version. If only two people wanted the oversize version then there would not be enough of the normal size rackets in stock to satisfy the 8 people who would want the normal size racket. It is easy to see the max for $X$ is 7, any more and there would not be enough oversize versions in stock.
$\begin{array}{rcl} P\left(3 \leq X \leq 7\right) &=& \sum_{x=3}^{7} \binom{10}{x}\left(0.60\right)^{x} \left( 1- 0.60 \right)^{10 - x} \\ &=& 0.0425 + 0.1115 + 0.2007 + 0.2508 + 0.2150 \\ &=& 0.8205 \end{array}$

### Exercise 55.

Note: There are two solution methods presented here. The first is a little more complex than needed, but is a good way to illustrate several concepts in probability.

We are told that 20% of all telephones of a certain type will be submitted for service under warranty. Of the telephones submitted for service 60% are repaired and 40% are replaced. If we by 10 telephones of this type, what is the probability that exactly 2 will be replaced under warranty?

Start with defining some random variables. Let $X$ be the number of phones submitted for service. Let $R$ be the number of telephones which are replaced after being submitted for service. We have some conditional probabilities here and will need to use the law of total probability to find the final answer.

We have the following distributions:

$X \sim \text{Binomial}\left(n_x = 10, p_x = 0.20\right)$

$R | X \sim \text{Binomial}\left( n_r = X, p_r = 0.40\right)$

Use the law of total probability to find the answer. Note that the first two probabilities in the sum are both zero, i.e., it is not possible to have $R = 2$ if $X = 0$ or $X = 1$.

$\begin{array}{rcl} P(R = 2) &=& P(R = 2 | X = 0)P(X=0) + P(R = 2 | X = 1)P(X = 1) \\ && + P(R = 2 | X = 2)P(X=2) + P(R=2 | X = 3)P(X=3) + \cdots \\ && + P(R = 2 | X = 10) P(X=10) \\ &=& \sum_{x=2}^{10} P\left(R = 2 | X = x\right) P\left( X = x\right) \\ &=& \sum_{x=2}^{10} \underbrace{\binom{x}{2}\left(p_r\right)^2\left(1 - p_r\right)^{x-2}}_{P(R=2|X=x)} \underbrace{\binom{10}{x} \left(p_x\right)^x\left(1-p_x\right)^{10-x}}_{P(X = x)} \\ &=& 0.1478070 \end{array}$

Now that you've seen this solution, there is another way to do this that is alot easier. Look at the probability that a telephone is replaced. Again, use the law of total probability.

$\begin{array}{rcl} P(\text{replaced}) &=& P(\text{replaced} | \text{submitted}) \times P(\text{submitted}) \\ && + P(\text{replaced} | \text{not submitted}) \times P(\text{not submitted}) \\ &=& 0.40 \times 0.20 + 0 \times 0.80 \\ &=& 0.08 \end{array}$

Now, let $W$ be the number of telephones replaced under warranty.

$W \sim \text{Binomial}\left(n = 10, p = 0.08\right)$

and the probability that 2 telephones are replaced is:

$P(W = 2) = \binom{10}{2}\left(0.08\right)^2 \left(1 - 0.08\right)^{10-2} = 0.1478070$

### Exercise 56.

Let $X$ be the number of students who receive special accommodations.

$X \sim \text{Binomial}\left( n = 25, p = 0.02 \right)$

• Part a. The probability exactly one received special accommodations?
$P(X = 1) = \binom{25}{1} 0.02^1 (1 - 0.02)^{25-1} = 0.3078902$

### Exercise 57.

We know that any one battery will have acceptable voltage with probability 0.90. A flashlight requires two batteries of acceptable voltage to work. If we assume that the batteries are independent the probability the batteries in a flashlight are acceptable is $0.90 \times 0.90 = 0.81$. Let $X$ be the number of working flashlights in a sample of size $n = 10$.

$X \sim \text{Binomial}\left(n = 10, p = 0.81\right)$

The probability at least nine of the ten flashlights will work is:

$\begin{array}{rcl} P\left( X \geq 9 \right) &=& P(X = 9) + P(X = 10) \\ &=& \binom{10}{9}\left(0.81\right)^{9} \left( 1 - 0.81 \right)^{10-9} + \binom{10}{10}\left(0.81\right)^{10} \left( 1 - 0.81 \right)^{10-10} \\ &=& 0.2852 + 0.1216 \\ &=& 0.4068 \end{array}$

### Exercise 58.

Let $X$ be the number of defective components.

$X \sim \text{Binomial}\left(n = 10, p\right)$

Find the probability the batch is accepted given the different success probabilities $p$.

$P(X \leq 2) = \begin{cases} 0.9998862 \approx 0.9999 & p = 0.01 \\ 0.9884964 \approx 0.9885 & p = 0.05 \\ 0.9298092 \approx 0.9298 & p = 0.10 \\ 0.6777995 \approx 0.6778 & p = 0.20 \\ 0.5255928 \approx 0.5256 & p = 0.25 \end{cases}$

The sampling plan which maximizes the acceptance for $p < 0.10$ and minimizes the acceptance for $p > 0.10$ is the best option. Of the three methods suggested, it appears that method d, sampling 15 and accepting if less than 2 are defective is the best call. The ideal case would be having a step function with probability 1 of acceptance for $p < 0.10$ and probability 0 of acceptance for $p \geq 0.10$.

### Exercise 59.

This exercise is a prelude to hypothesis testing in the later chapters. In this case we have

$X \sim \text{Binomial}\left( n = 25, p \right)$

where $p$ is an unknown value. We will reject the claim of $p \geq 0.80$ if we observe $X \leq 15$.

• Part a. What is the probability that the claim is rejected if the true value is $p = 0.80 :$. This is what is called a Type I Error, rejecting the claim when it is true. In this case we have a probability of rejecting the claim if $X \leq 15$ is:
$P(X \leq 15) = \sum_{x=0}^{15} \binom{25}{x} \left(p\right)^{x} \left(1 - p\right)^{25-x} = 0.01733187$.
• Part b. What is the probability of not rejecting the claim when
$p = 0.7$
$1 - P(X \leq 15) = \sum_{x=0}^{15} \binom{25}{x} \left(p\right)^{x} \left(1 - p\right)^{25-x} = 1-0.1894360 = 0.810564$.
or
$p = 0.6$.
$P(X \leq 15) = \sum_{x=0}^{15} \binom{25}{x} \left(p\right)^{x} \left(1 - p\right)^{25-x} = 1 - 0.575383 = 0.424617$.
Note: the probabilities here are also error, failure to reject the claim when it is false. These are known as Type II errors. You'll learn more about this in later chapters of this book on Hypothesis testing.
• Part c. How do the error probabilities change in the above if the rejection of the claim is done when $X \leq 14$?
• for part a: the probability is 0.00555492
• for part b: with $p = 0.7$ the probability is 0.9022
• for part b: with $p = 0.6$ the probability is 0.585775

### Exercise 60.

Let $h(X)$ be the function for the total revenue from a total of $X$ passenger cars.

$X \sim \text{Binomial}\left(n = 25, p = 0.60 \right)$

$\begin{array}{rcl} h(X) &=& 1.00 X + 2.5 \times (25 - X) \\ &=& 62.5 - 2.5 X \\ \\ E(h(X)) &=& 62.5 - 2.5 E(X) \\ &=& 62.5 - 2.5 np \\ &=& 62.5 - 2.5 \times 25 \times 0.60 \\ &=& 40.00 \end{array}$

The expected revenue of for 25 vehicles over the toll bridge is \$40.00.

### Exercise 61.

Let $A$ be the number of books received if the student writes on topic A. Let $B$ be the number of books received if the student writes on topic B.

$A \sim \text{Binomial}\left( n = 2 , p \right) \quad B \sim \text{Binomial}\left( n = 4, p \right)$

if $p = 0.9$ then the probability of having at least half of the books ordered received is

$\begin{array}{rcl} P(A \geq 1) &=& 0.99 \\ P(B \geq 2) &=& 0.9963 \end{array}$

The probability of getting at least half of the ordered books for Topic B is slightly higher than for Topic A and thus Topic B should be selected.

If the success probability for receiving a book is reduced from 0.90 to 0.50 the probabilities are:

$\begin{array}{rcl} P(A \geq 1) &=& 0.75 \\ P(B \geq 2) &=& 0.6875 \end{array}$

In this case the probability of getting at least half of the ordered books is higher for Topic A than B and Topic A should be selected.

### Exercise 62.

Let $X$ be a binomial random variable. For a fixed value of $n$ is there a value of $p$ that will allow the variance to be zero? It turns out that there are two possible values for $p$ that will allow the variance to be zero.

$Var(X) = np(1-p) = 0 \quad \text{if } p = 0 \text{ or } 1.$

This should make sense. If the success probability is 0 then the out come of the experiment will be all failures, all the values are the same and the variance, a measure of spread, is zero since the spread of the data is zero. Same if the success probability is 1, all the outcomes are success, there is no spread in the data.

To maximize the variance we need a value of $p = 1/2$. Here is the proof:

$\begin{array}{rcl} \frac{\partial}{\partial p} Var(X) &=& \frac{\partial}{\partial p} np(1-p) \\ &=& n(1-p) - np \\ &=& n - np - np \\ &=& n - 2np \\ \text{set } n - 2np &=& 0 \\ p &=& 1/2 \\ \\ \frac{\partial^2}{\partial p^2} Var(X) &=& -2n < 0 \end{array}$

Since the second derivative is less than zero for all values of $p$ we know that when the first derivative is zero that we will have a maximum. The maximum is when $p = 1/2$.

### Exercise 63.

• Part a. Show that $b(x;n, 1-p) = b(n-x;n,p)$
$\begin{array}{rcl} b(x;n, 1-p) &=& b(n-x;n,p) \\ \binom{n}{x}\left(1-p\right)^x\left(1-(1-p)\right)^{n-x} &=& \binom{n}{n-x} \left(p\right)^{n-x} \left(1 - p\right)^{n - (n-x)} \\ \frac{n!}{x! (n-x)!} \left(1-p\right)^x \left(p\right)^{n-x} &=& \frac{n!}{(n-x)!(n-(n-x))!} \left(p\right)^{n-x} (1-p)^{x} \\ \frac{n!}{x! (n-x)!} \left(p\right)^{n-x} \left(1-p\right)^x &=& \frac{n!}{(n-x)!x!} \left(p\right)^{n-x} (1-p)^{x} \end{array}$

The functions are the same.

• Part b. Show that the CDF for a binomial with $n$ trials and success probability $1 - p$, $B(x; n,1-p)$ is the same as $1 - B(n-x-1;n,p)$

To do this define another random variable $Y = n - X$. The distributions are:

$X \sim \text{Binomial}\left( n , 1-p\right)$

$Y \sim \text{Binomial}\left( n, p \right)$

$\begin{array}{rcl} B(x; n,1-p) &=& P(X \leq x) \\ &=& 1 - P(X > x) \\ &=& 1 - P(n - Y > x) \\ &=& 1 - P(Y < n - x ) \\ &=& 1 - P(Y \leq n - x - 1) \\ &=& 1 - B(n-x-1; n, p) \end{array}$

• Part c. The above two parts imply that in the Binomial probability tables in the back of the textbook, it is not necessary to publish the value for the success probability $p > 0.50$ because you can find the needed probabilities using the failure probability $1-p$ which will be less than $0.50$ when ever $p > 0.50$.

### Exercise 64.

Prove the expectation of a binomial random variable is $E(X) = np$.

$\begin{array}{rcll} E(X) &=& \sum_{x=0}^n x P(X=x) \\ &=& \sum_{x=0}^n x \binom{n}{x} p^x (1-p)^{n-x} \\ \\ &=& \sum_{x=0}^n x \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} & \text{the x=0 term in the sum will be 0,} \\ && & \text{so we can change the index on the sum} \\ \\ &=& \sum_{x=1}^n x \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} \\ \\ &=& \sum_{x=1}^n \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{n-x} & \text{move the index back to zero}\\ \\ &=& \sum_{x=0}^{n-1} \frac{n!}{x!(n - (x+1))!} p^{x+1} (1-p)^{n-(x+1)} \\ \\ &=& np \sum_{x=0}^{n-1} \underbrace{\frac{(n-1)!}{x!(n-1-x)!} p^x (1-p)^{n-1-x}}_{\text{pmf of Binomial(n-1,p)}} & \text{the sum of the pmf = l}\\ \\ &=& np \end{array}$

### Exercise 65.

• Part a. Let $X$ be the number of people who pay with a debit card.
$X \sim \text{Binomial}\left(n = 100, p = 0.2 \right)$
The expectation and variance is:
$\begin{array}{rcccl} E(X) &=& np &=& 20 \\ Var(X) &=& np(1-p) &=& 16 \end{array}$
The reasoning here, which is only in defining the random variable is the idea that a success is a customer paying with a debit card and a failure is for any other form of payment.
• Part b. Let $Y$ be the number of person who do not pay with cash. A success in this case is a customer paying with any form not cash and a failure is to paying with cash.
$Y \sim \text{Binomial}\left( n = 100, p = 0.70 \right)$
$\begin{array}{rcccl} E(Y) &=& np &=& 70 \\ Var(Y) &=& np(1-p) &=& 21 \end{array}$

### Exercise 66.

Let $X$ be the number of people with a reservation who show up for the limo.

$X \sim \text{Binomial}\left( n = 6, p = 0.80 \right)$

• Part a. What is the probability at least one person who shows up for the limo cannot be accommodated? Since the limo only holds four people we will not be able to accommodate everyone if five or six show up.
$\begin{array}{rcl} P\left(X \leq 5\right) &=& P(X=5) + P(X=6) \\ &=& 0.3932 + 0.2621 \\ &=& 0.6553 \end{array}$
• Part b. If there are six reservations made, what is the expected number of open seats in the limo? We need to define a function $h(x)$ for the number of open seats as a function of $x$ the number of people who show for the limo.
$\begin{array}{r|ccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline h(x) & 4 & 3 & 2 & 1 & 0 & 0 & 0 \end{array}$
The expectation of $h(X)$ is:
$\begin{array}{rcl} E(h(X)) &=& \sum_{x=0}^{6} h(x) P(X = x) \\ &=& (4)(0.0001) + (3)(0.0015) + (2)(0.0154) + (1)(0.0819) \\&&+ (0)(0.2458 + 0.3932 + 0.2621) \\ &=& 0.1176 \end{array}$
• Part c. If the probability distribution for the number of reservations is:
$\begin{array}{r|cccc} \text{Reservarions} & 3 & 4 & 5 & 6 \\ \hline \text{Probability} & 0.1 & 0.2 & 0.3 & 0.4 \\ \end{array}$
Then the probability mass function for the number of passengers on a randomly selected trip is found as follows.
We know that $X$ will take on the values 0, 1, 2, 3, 4. The value of 0 will happen if no one shows up for the limo for any one of the number of possible reservations totals. We use the law of total probability to find this. Let $R$ be the number of reservations made and $Y$ be the number of people who show up.
Note that the conditional distribution of $Y | R$ is:
$Y|R \sim \text{Binomial}\left(n = R, p = 0.80\right)$
$\begin{array}{rcl} P(X = 0) &=& \sum_{r=3}^{6} P(Y = 0 | R = r)P(R = r) \\ &=& \sum_{r=3}^{6} \left( \binom{r}{0} 0.80^x 0.20^{r} \right) P(R = r) \\ &=& 0.0080 \times 0.1 + 0.0016 \times 0.2 + 0.0003 \times 0.3 + 0.0001 \times 0.4 \\ &=& 0.00125 \\ \\ P(X = 1) &=& \sum_{r=3}^{6} P(Y = 1 | R = r)P(R = r) \\ &=& 0.01724 \\ \\ P(X = 2) &=& \sum_{r=3}^{6} P(Y = 2 | R = r)P(R = r) \\ &=& 0.09064 \\ \\ P(X = 3) &=& \sum_{r=3}^{6} P(Y = 3 | R = r)P(R = r) \\ &=& 0.22732 \\ \\ P(X = 4) &=& 1 - ( 0.00125 + 0.01724 + 0.09064 + 0.22732) \\ &=& 0.66355 \end{array}$

The probability $P(X = 4)$ is a little more work because you have to deal with the options of having five or six people show. The easy way to deal with this is to note that $P(X = 4) = 1 - P(X \ne 4)$.

### Exercise 67.

Using Chebyshev's Inequality find the lower bounds for $k = 2$ and $k = 3$ for

$X \sim \text{Binomial}\left(n=20, p = 0.5\right)$

The Chebyshev bounds are:

$\begin{array}{rcl} P\left( \left| X - \mu \right| > k \sigma \right) &\leq& \frac{1}{k^2} \\ &=& \frac{1}{4} \quad \text{For } k = 2 \\ \\ &=& \frac{1}{9} \quad \text{For } k = 3 \end{array}$

For the binomial distribution we have:

$\begin{array}{rcl} P\left( \left| X - \mu \right| > k \sigma \right) &=& 1 - P\left( \mu - k\sigma < X < \mu + k \sigma\right) \\ &=& 1 - P\left( np - k \sqrt{np(1-p)} < X < np + k \sqrt{np(1-p)} \right) \\ &=& 1 - P\left(5.527 < X < 14.472\right) \quad \text{For } k = 2\\ &=& 1 - P\left( 6 \leq X \leq 14 \right) \\ &=& 1 - 0.9586105 \\ &=& 0.0413895 \\ &=& 1 - P(3.29 < X < 16.708) \quad \text{For } k = 3 \\ &=& 1 - P(4 \leq X \leq 16) \\ &=& 1 - 0.9974232 \\ &=& 0.002576828 \end{array}$

For the binomial distribution

$X \sim \text{Binomial}\left(n=20,p=.75\right)$

For $k = 2$ the probability is 0.06523779 and for $k = 3$ the probability is 0.003942142.

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## Section 3.5 - Hyper geometric and Negative Binomial Distributions

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## Section 3.6 - The Poisson Probability Distribution

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## Supplementary Exercises

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