To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

Question

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Answer 1

Question

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

a.

Expert Solution

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Explanation of Solution

Given:

Total number of pathogenic isolates extracted from patients n=7Percentage of pathogenic isolates that are resistant to antibiotic vancomycin p = 30%=0.30

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

P(X=x)=nxpx1−pn−x

The addition principle in binomial distribution X is shown as:

P(X≥x)=P(X=x)+P(X=x+1)+...+P(X=n)

The assumptions of the binomial distribution are:

There are only two outcomes for each trial, which are treated as a success and a failure.
There are a fixed number of trials and each trial is independent of the other.
The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

b.

Expert Solution

Answer to Problem 1PP

The probability of success p=0.3

The total number of trails n=7

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin =30% and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin p=0.30 .

The total number of pathogenic isolates extracted from patients n=7 .

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

c.

Expert Solution

Answer to Problem 1PP

The probability is 0.025.

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

PX=5=750.351−0.37−5=0.025

The required probability is 0.025.

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

d.

Expert Solution

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

P(X=x)=nxpx1−pn−xwhere x=6,n=7,p=0.3P(X=6)=760.361−0.37−6P(X=6)=7×0.360.77−6=0.0036

Now,

P(X=x)=nxpx1−pn−xwhere x=7,n=7,p=0.3P(X=7)=770.371−0.37−7P(X=7)=1×0.370.77−7=0.0002

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

e.

Expert Solution

Answer to Problem 1PP

The probability is 0.0288.

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

P(X≥5)=P(X=5)+P(X=6)+P(X=7)

It is known that,

P(X=5)=0.025P(X=6)=0.0036P(X=7)=0.0002

Now,

P(X≥5)=P(X=5)+P(X=6)+P(X=7)=0.025+0.0036+0.0002=0.0288

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.

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Answer 2

Question

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

a.

Expert Solution

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Explanation of Solution

Given:

Total number of pathogenic isolates extracted from patients n=7Percentage of pathogenic isolates that are resistant to antibiotic vancomycin p = 30%=0.30

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

P(X=x)=nxpx1−pn−x

The addition principle in binomial distribution X is shown as:

P(X≥x)=P(X=x)+P(X=x+1)+...+P(X=n)

The assumptions of the binomial distribution are:

There are only two outcomes for each trial, which are treated as a success and a failure.
There are a fixed number of trials and each trial is independent of the other.
The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

b.

Expert Solution

Answer to Problem 1PP

The probability of success p=0.3

The total number of trails n=7

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin =30% and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin p=0.30 .

The total number of pathogenic isolates extracted from patients n=7 .

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

c.

Expert Solution

Answer to Problem 1PP

The probability is 0.025.

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

PX=5=750.351−0.37−5=0.025

The required probability is 0.025.

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

d.

Expert Solution

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

P(X=x)=nxpx1−pn−xwhere x=6,n=7,p=0.3P(X=6)=760.361−0.37−6P(X=6)=7×0.360.77−6=0.0036

Now,

P(X=x)=nxpx1−pn−xwhere x=7,n=7,p=0.3P(X=7)=770.371−0.37−7P(X=7)=1×0.370.77−7=0.0002

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

e.

Expert Solution

Answer to Problem 1PP

The probability is 0.0288.

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

P(X≥5)=P(X=5)+P(X=6)+P(X=7)

It is known that,

P(X=5)=0.025P(X=6)=0.0036P(X=7)=0.0002

Now,

P(X≥5)=P(X=5)+P(X=6)+P(X=7)=0.025+0.0036+0.0002=0.0288

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.

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Answer 3

Question

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

a.

Expert Solution

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Explanation of Solution

Given:

Total number of pathogenic isolates extracted from patients n=7Percentage of pathogenic isolates that are resistant to antibiotic vancomycin p = 30%=0.30

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

P(X=x)=nxpx1−pn−x

The addition principle in binomial distribution X is shown as:

P(X≥x)=P(X=x)+P(X=x+1)+...+P(X=n)

The assumptions of the binomial distribution are:

There are only two outcomes for each trial, which are treated as a success and a failure.
There are a fixed number of trials and each trial is independent of the other.
The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

b.

Expert Solution

Answer to Problem 1PP

The probability of success p=0.3

The total number of trails n=7

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin =30% and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin p=0.30 .

The total number of pathogenic isolates extracted from patients n=7 .

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

c.

Expert Solution

Answer to Problem 1PP

The probability is 0.025.

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

PX=5=750.351−0.37−5=0.025

The required probability is 0.025.

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

d.

Expert Solution

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

P(X=x)=nxpx1−pn−xwhere x=6,n=7,p=0.3P(X=6)=760.361−0.37−6P(X=6)=7×0.360.77−6=0.0036

Now,

P(X=x)=nxpx1−pn−xwhere x=7,n=7,p=0.3P(X=7)=770.371−0.37−7P(X=7)=1×0.370.77−7=0.0002

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

e.

Expert Solution

Answer to Problem 1PP

The probability is 0.0288.

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

P(X≥5)=P(X=5)+P(X=6)+P(X=7)

It is known that,

P(X=5)=0.025P(X=6)=0.0036P(X=7)=0.0002

Now,

P(X≥5)=P(X=5)+P(X=6)+P(X=7)=0.025+0.0036+0.0002=0.0288

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.

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Answer 4

Question

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

a.

Expert Solution

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Explanation of Solution

Given:

Total number of pathogenic isolates extracted from patients n=7Percentage of pathogenic isolates that are resistant to antibiotic vancomycin p = 30%=0.30

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

P(X=x)=nxpx1−pn−x

The addition principle in binomial distribution X is shown as:

P(X≥x)=P(X=x)+P(X=x+1)+...+P(X=n)

The assumptions of the binomial distribution are:

There are only two outcomes for each trial, which are treated as a success and a failure.
There are a fixed number of trials and each trial is independent of the other.
The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

b.

Expert Solution

Answer to Problem 1PP

The probability of success p=0.3

The total number of trails n=7

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin =30% and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin p=0.30 .

The total number of pathogenic isolates extracted from patients n=7 .

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

c.

Expert Solution

Answer to Problem 1PP

The probability is 0.025.

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

PX=5=750.351−0.37−5=0.025

The required probability is 0.025.

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

d.

Expert Solution

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

P(X=x)=nxpx1−pn−xwhere x=6,n=7,p=0.3P(X=6)=760.361−0.37−6P(X=6)=7×0.360.77−6=0.0036

Now,

P(X=x)=nxpx1−pn−xwhere x=7,n=7,p=0.3P(X=7)=770.371−0.37−7P(X=7)=1×0.370.77−7=0.0002

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

e.

Expert Solution

Answer to Problem 1PP

The probability is 0.0288.

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

P(X≥5)=P(X=5)+P(X=6)+P(X=7)

It is known that,

P(X=5)=0.025P(X=6)=0.0036P(X=7)=0.0002

Now,

P(X≥5)=P(X=5)+P(X=6)+P(X=7)=0.025+0.0036+0.0002=0.0288

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.

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Answer 5

Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

a.

Expert Solution

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Explanation of Solution

Given:

Total number of pathogenic isolates extracted from patients n=7Percentage of pathogenic isolates that are resistant to antibiotic vancomycin p = 30%=0.30

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

P(X=x)=nxpx1−pn−x

The addition principle in binomial distribution X is shown as:

P(X≥x)=P(X=x)+P(X=x+1)+...+P(X=n)

The assumptions of the binomial distribution are:

There are only two outcomes for each trial, which are treated as a success and a failure.
There are a fixed number of trials and each trial is independent of the other.
The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

b.

Expert Solution

Answer to Problem 1PP

The probability of success p=0.3

The total number of trails n=7

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin =30% and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin p=0.30 .

The total number of pathogenic isolates extracted from patients n=7 .

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

c.

Expert Solution

Answer to Problem 1PP

The probability is 0.025.

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

PX=5=750.351−0.37−5=0.025

The required probability is 0.025.

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

d.

Expert Solution

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

P(X=x)=nxpx1−pn−xwhere x=6,n=7,p=0.3P(X=6)=760.361−0.37−6P(X=6)=7×0.360.77−6=0.0036

Now,

P(X=x)=nxpx1−pn−xwhere x=7,n=7,p=0.3P(X=7)=770.371−0.37−7P(X=7)=1×0.370.77−7=0.0002

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

e.

Expert Solution

Answer to Problem 1PP

The probability is 0.0288.

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

P(X≥5)=P(X=5)+P(X=6)+P(X=7)

It is known that,

P(X=5)=0.025P(X=6)=0.0036P(X=7)=0.0002

Now,

P(X≥5)=P(X=5)+P(X=6)+P(X=7)=0.025+0.0036+0.0002=0.0288

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.

Answer 6

Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

a.

Expert Solution

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Explanation of Solution

Given:

Total number of pathogenic isolates extracted from patients n=7Percentage of pathogenic isolates that are resistant to antibiotic vancomycin p = 30%=0.30

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

P(X=x)=nxpx1−pn−x

The addition principle in binomial distribution X is shown as:

P(X≥x)=P(X=x)+P(X=x+1)+...+P(X=n)

The assumptions of the binomial distribution are:

There are only two outcomes for each trial, which are treated as a success and a failure.
There are a fixed number of trials and each trial is independent of the other.
The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

Answer 7

Chapter 7, Problem 1PP

a.

To determine

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

Answer 8

a.

Expert Solution

Answer to Problem 1PP

Yes, it does satisfy the assumptions of the binomial distribution.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Total number of pathogenic isolates extracted from patients n=7Percentage of pathogenic isolates that are resistant to antibiotic vancomycin p = 30%=0.30

Formula used:The binomial distribution probability density function for n trials and p probability of success for x number of success is shown below:

P(X=x)=nxpx1−pn−x

The addition principle in binomial distribution X is shown as:

P(X≥x)=P(X=x)+P(X=x+1)+...+P(X=n)

The assumptions of the binomial distribution are:

There are only two outcomes for each trial, which are treated as a success and a failure.
There are a fixed number of trials and each trial is independent of the other.
The probability of success is the same for each trial.

In this case, the pathogenic isolates which are resistant are considered a success, and which are not resistant are considered a failure. There are a fixed number of trails or a fixed number of pathogenic isolates that are independent of each other.

The given example satisfies all the assumptions; so, the given example can be represented by binomial distribution.

Answer 13

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

b.

Expert Solution

Answer to Problem 1PP

The probability of success p=0.3

The total number of trails n=7

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin =30% and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin p=0.30 .

The total number of pathogenic isolates extracted from patients n=7 .

Answer 14

b.

To determine

To find: The probability of success for the given example. Also, compute the value of n .

Answer 15

b.

Expert Solution

Answer to Problem 1PP

The probability of success p=0.3

The total number of trails n=7

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

From the above part, it is known that the percentage of pathogenic isolates that are resistant to antibiotic vancomycin =30% and the total number of pathogenic = 7.

The probability of success or probability that a randomly selected pathogenic isolate is resistant to the antibiotic vancomycin p=0.30 .

The total number of pathogenic isolates extracted from patients n=7 .

Answer 20

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

c.

Expert Solution

Answer to Problem 1PP

The probability is 0.025.

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

PX=5=750.351−0.37−5=0.025

The required probability is 0.025.

Answer 21

c.

To determine

To calculate: The probability that exactly five pathogenic isolates are resistant.

Answer 22

c.

Expert Solution

Answer to Problem 1PP

The probability is 0.025.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

The probability that exactly 5 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

PX=5=750.351−0.37−5=0.025

The required probability is 0.025.

Answer 27

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

d.

Expert Solution

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

P(X=x)=nxpx1−pn−xwhere x=6,n=7,p=0.3P(X=6)=760.361−0.37−6P(X=6)=7×0.360.77−6=0.0036

Now,

P(X=x)=nxpx1−pn−xwhere x=7,n=7,p=0.3P(X=7)=770.371−0.37−7P(X=7)=1×0.370.77−7=0.0002

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

Answer 28

d.

To determine

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

Answer 29

d.

Expert Solution

Answer to Problem 1PP

The probabilities are 0.0036 and 0.0002 respectively.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

The probability that exactly 6 pathogenic isolates show resistance to antibiotic vancomycin is calculated as shown below:

P(X=x)=nxpx1−pn−xwhere x=6,n=7,p=0.3P(X=6)=760.361−0.37−6P(X=6)=7×0.360.77−6=0.0036

Now,

P(X=x)=nxpx1−pn−xwhere x=7,n=7,p=0.3P(X=7)=770.371−0.37−7P(X=7)=1×0.370.77−7=0.0002

Hence, the required probabilities are 0.0036 and 0.0002 respectively.

Answer 34

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

e.

Expert Solution

Answer to Problem 1PP

The probability is 0.0288.

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

P(X≥5)=P(X=5)+P(X=6)+P(X=7)

It is known that,

P(X=5)=0.025P(X=6)=0.0036P(X=7)=0.0002

Now,

P(X≥5)=P(X=5)+P(X=6)+P(X=7)=0.025+0.0036+0.0002=0.0288

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.

Answer 35

e.

To determine

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

Answer 36

e.

Expert Solution

Answer to Problem 1PP

The probability is 0.0288.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation:The probability that five or more pathogenic isolates are resistant is calculated as shown below:

P(X≥5)=P(X=5)+P(X=6)+P(X=7)

It is known that,

P(X=5)=0.025P(X=6)=0.0036P(X=7)=0.0002

Now,

P(X≥5)=P(X=5)+P(X=6)+P(X=7)=0.025+0.0036+0.0002=0.0288

The probability that there are five or more pathogenic isolates that are resistant is 0.0288.

Answer 41

Ch. 7 - Prob. 1PP Ch. 7 - Prob. 20AP Ch. 7 - Prob. 21AP Ch. 7 - Prob. 25AP

To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

Solution Summary: The author explains that the binomial distribution assumes that there are only two outcomes for each trial, which are treated as success and failure.

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To state: The assumptions of the binomial distribution and check if the given case satisfies the required assumptions.

Answer to Problem 1PP

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To find: The probability of success for the given example. Also, compute the value of n .

Answer to Problem 1PP

Explanation of Solution

To calculate: The probability that exactly five pathogenic isolates are resistant.

Answer to Problem 1PP

Explanation of Solution

To calculate: The probability that exactly six pathogenic isolates are resistant and the probability that there are exactly seven resistant pathogenic isolates.

Answer to Problem 1PP

Explanation of Solution

To calculate: The probability of five or more pathogenic isolates are resistant using the principle of addition.

Answer to Problem 1PP

Explanation of Solution

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