The mean and standard deviation of the random variable X .

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

(a)

Expert Solution

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1−p)=1200(0.83)(1−0.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

(b)

To determine

To find: The probability.

(b)

Expert Solution

Answer to Problem 78E

Solution: The probability is P(X≥800)=1_.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1−p) if the condition np(1−p)≥10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1−p)=1200×0.83×(1−0.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1−p)=1200(0.83)(1−0.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=X−μσ=800−99613.012=−15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as −15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 1

The area left to the Z-score, −15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

1−0.0000=1

Hence, the probability is 1.

(c)

To determine

To find: The probability.

(c)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000, trials = 1200, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 2

P(X≤1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.6324=0.3676

Hence, the probability is 0.3676.

(d)

To determine

To find: The probability.

(d)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000, trials = 1150, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 3

P(X≤1000)=0.9999

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.9999=0.0001

Hence, the probability is 0.0001.

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

(a)

Expert Solution

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1−p)=1200(0.83)(1−0.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

(b)

To determine

To find: The probability.

(b)

Expert Solution

Answer to Problem 78E

Solution: The probability is P(X≥800)=1_.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1−p) if the condition np(1−p)≥10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1−p)=1200×0.83×(1−0.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1−p)=1200(0.83)(1−0.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=X−μσ=800−99613.012=−15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as −15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 1

The area left to the Z-score, −15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

1−0.0000=1

Hence, the probability is 1.

(c)

To determine

To find: The probability.

(c)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000, trials = 1200, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 2

P(X≤1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.6324=0.3676

Hence, the probability is 0.3676.

(d)

To determine

To find: The probability.

(d)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000, trials = 1150, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 3

P(X≤1000)=0.9999

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.9999=0.0001

Hence, the probability is 0.0001.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

(a)

Expert Solution

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1−p)=1200(0.83)(1−0.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

(b)

To determine

To find: The probability.

(b)

Expert Solution

Answer to Problem 78E

Solution: The probability is P(X≥800)=1_.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1−p) if the condition np(1−p)≥10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1−p)=1200×0.83×(1−0.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1−p)=1200(0.83)(1−0.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=X−μσ=800−99613.012=−15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as −15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 1

The area left to the Z-score, −15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

1−0.0000=1

Hence, the probability is 1.

(c)

To determine

To find: The probability.

(c)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000, trials = 1200, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 2

P(X≤1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.6324=0.3676

Hence, the probability is 0.3676.

(d)

To determine

To find: The probability.

(d)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000, trials = 1150, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 3

P(X≤1000)=0.9999

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.9999=0.0001

Hence, the probability is 0.0001.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

(a)

Expert Solution

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1−p)=1200(0.83)(1−0.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

(b)

To determine

To find: The probability.

(b)

Expert Solution

Answer to Problem 78E

Solution: The probability is P(X≥800)=1_.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1−p) if the condition np(1−p)≥10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1−p)=1200×0.83×(1−0.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1−p)=1200(0.83)(1−0.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=X−μσ=800−99613.012=−15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as −15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 1

The area left to the Z-score, −15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

1−0.0000=1

Hence, the probability is 1.

(c)

To determine

To find: The probability.

(c)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000, trials = 1200, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 2

P(X≤1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.6324=0.3676

Hence, the probability is 0.3676.

(d)

To determine

To find: The probability.

(d)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000, trials = 1150, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 3

P(X≤1000)=0.9999

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.9999=0.0001

Hence, the probability is 0.0001.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

(a)

Expert Solution

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1−p)=1200(0.83)(1−0.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

(b)

To determine

To find: The probability.

(b)

Expert Solution

Answer to Problem 78E

Solution: The probability is P(X≥800)=1_.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1−p) if the condition np(1−p)≥10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1−p)=1200×0.83×(1−0.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1−p)=1200(0.83)(1−0.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=X−μσ=800−99613.012=−15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as −15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 1

The area left to the Z-score, −15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

1−0.0000=1

Hence, the probability is 1.

(c)

To determine

To find: The probability.

(c)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000, trials = 1200, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 2

P(X≤1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.6324=0.3676

Hence, the probability is 0.3676.

(d)

To determine

To find: The probability.

(d)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000, trials = 1150, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 3

P(X≤1000)=0.9999

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.9999=0.0001

Hence, the probability is 0.0001.

Answer 6

Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

(a)

Expert Solution

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1−p)=1200(0.83)(1−0.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

Answer 7

Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

Answer 8

(a)

Expert Solution

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1−p)=1200(0.83)(1−0.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

Answer 13

(b)

To determine

To find: The probability.

(b)

Expert Solution

Answer to Problem 78E

Solution: The probability is P(X≥800)=1_.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1−p) if the condition np(1−p)≥10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1−p)=1200×0.83×(1−0.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1−p)=1200(0.83)(1−0.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=X−μσ=800−99613.012=−15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as −15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 1

The area left to the Z-score, −15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

1−0.0000=1

Hence, the probability is 1.

Answer 14

(b)

To determine

To find: The probability.

Answer 15

(b)

Expert Solution

Answer to Problem 78E

Solution: The probability is P(X≥800)=1_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1−p) if the condition np(1−p)≥10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1−p)=1200×0.83×(1−0.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1−p)=1200(0.83)(1−0.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=X−μσ=800−99613.012=−15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as −15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 1

The area left to the Z-score, −15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

1−0.0000=1

Hence, the probability is 1.

Answer 20

(c)

To determine

To find: The probability.

(c)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000, trials = 1200, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 2

P(X≤1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.6324=0.3676

Hence, the probability is 0.3676.

Answer 21

(c)

To determine

To find: The probability.

Answer 22

(c)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000, trials = 1200, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 2

P(X≤1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.6324=0.3676

Hence, the probability is 0.3676.

Answer 27

(d)

To determine

To find: The probability.

(d)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000, trials = 1150, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip 3

P(X≤1000)=0.9999

Therefore, the required probability is calculated as:

P(X>1000)=1−P(X≤1000)=1−0.9999=0.0001

Hence, the probability is 0.0001.

Answer 28

(d)

To determine

To find: The probability.

Answer 29

(d)

Expert Solution

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000, trials = 1150, probability = 0.83, and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.