Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

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 6, Problem 49SE

a.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

a.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is about 5.113%.

Explanation of Solution

Calculation:

The examination contains 50 multiple choice questions. Each question has four options. The students have 75% probability to answer any question correctly if they attend lecture and do homework.

Here, the student must answer correctly to 43 questions to obtain grade A.

The formula for mean in binomial distribution is,

μ=np

The value of mean is,

μ=50(0.75)=37.5

The formula for standard deviation is,

σ=np(1−p)

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 43.

That is, c=43

P(x≥43)=P(x≥43−0.5)=P(x≥42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥42.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 42.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 1

From the graph, the probability value is 0.05113.

Thus, the percentage of the students is 5.113%.

b.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

b.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 57.99%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 35 to 39 questions to obtain grade C.

If the binomial probability represents “between than c1 and c2” then the normal probability is P(c1−0.5≤x≤c2+0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 35 and the value 0.5 added to 39.

That is, c1=35 and c2=39

P(x<43)=P(x≤43−0.5)=P(x≤42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(34.5≤x≤39.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Middle Tail for the region of the curve to shade.
Enter the X value 1 as 34.5 and X value 2 as 39.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 2

From the graph, the probability value is 0.5799

Thus, the percentage of the students is 57.99%.

c.

To determine

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

c.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will pass the examination is about 99.55%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 30 or more questions to pass the examination.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x>30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 3

From the graph, the probability value is 0.9955.

Thus, the percentage of the students is 99.55%.

d.

To determine

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

d.

Expert Solution

Answer to Problem 49SE

The students who did not attend the classes and not done the homework will not pass the examination by guess.

Explanation of Solution

Calculation:

Here, the student did not attend the class and has not done the homework.

Approximately 25% students of did not attend the classes and not done the homework.

The value of mean is,

μ=50(0.25)=12.5

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x≥30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 12.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 4

From the graph, the probability value is 0.000. Thus, the percentage of the students is 0.

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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 6, Problem 49SE

a.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

a.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is about 5.113%.

Explanation of Solution

Calculation:

The examination contains 50 multiple choice questions. Each question has four options. The students have 75% probability to answer any question correctly if they attend lecture and do homework.

Here, the student must answer correctly to 43 questions to obtain grade A.

The formula for mean in binomial distribution is,

μ=np

The value of mean is,

μ=50(0.75)=37.5

The formula for standard deviation is,

σ=np(1−p)

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 43.

That is, c=43

P(x≥43)=P(x≥43−0.5)=P(x≥42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥42.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 42.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 1

From the graph, the probability value is 0.05113.

Thus, the percentage of the students is 5.113%.

b.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

b.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 57.99%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 35 to 39 questions to obtain grade C.

If the binomial probability represents “between than c1 and c2” then the normal probability is P(c1−0.5≤x≤c2+0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 35 and the value 0.5 added to 39.

That is, c1=35 and c2=39

P(x<43)=P(x≤43−0.5)=P(x≤42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(34.5≤x≤39.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Middle Tail for the region of the curve to shade.
Enter the X value 1 as 34.5 and X value 2 as 39.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 2

From the graph, the probability value is 0.5799

Thus, the percentage of the students is 57.99%.

c.

To determine

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

c.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will pass the examination is about 99.55%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 30 or more questions to pass the examination.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x>30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 3

From the graph, the probability value is 0.9955.

Thus, the percentage of the students is 99.55%.

d.

To determine

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

d.

Expert Solution

Answer to Problem 49SE

The students who did not attend the classes and not done the homework will not pass the examination by guess.

Explanation of Solution

Calculation:

Here, the student did not attend the class and has not done the homework.

Approximately 25% students of did not attend the classes and not done the homework.

The value of mean is,

μ=50(0.25)=12.5

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x≥30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 12.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 4

From the graph, the probability value is 0.000. Thus, the percentage of the students is 0.

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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 6, Problem 49SE

a.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

a.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is about 5.113%.

Explanation of Solution

Calculation:

The examination contains 50 multiple choice questions. Each question has four options. The students have 75% probability to answer any question correctly if they attend lecture and do homework.

Here, the student must answer correctly to 43 questions to obtain grade A.

The formula for mean in binomial distribution is,

μ=np

The value of mean is,

μ=50(0.75)=37.5

The formula for standard deviation is,

σ=np(1−p)

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 43.

That is, c=43

P(x≥43)=P(x≥43−0.5)=P(x≥42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥42.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 42.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 1

From the graph, the probability value is 0.05113.

Thus, the percentage of the students is 5.113%.

b.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

b.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 57.99%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 35 to 39 questions to obtain grade C.

If the binomial probability represents “between than c1 and c2” then the normal probability is P(c1−0.5≤x≤c2+0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 35 and the value 0.5 added to 39.

That is, c1=35 and c2=39

P(x<43)=P(x≤43−0.5)=P(x≤42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(34.5≤x≤39.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Middle Tail for the region of the curve to shade.
Enter the X value 1 as 34.5 and X value 2 as 39.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 2

From the graph, the probability value is 0.5799

Thus, the percentage of the students is 57.99%.

c.

To determine

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

c.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will pass the examination is about 99.55%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 30 or more questions to pass the examination.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x>30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 3

From the graph, the probability value is 0.9955.

Thus, the percentage of the students is 99.55%.

d.

To determine

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

d.

Expert Solution

Answer to Problem 49SE

The students who did not attend the classes and not done the homework will not pass the examination by guess.

Explanation of Solution

Calculation:

Here, the student did not attend the class and has not done the homework.

Approximately 25% students of did not attend the classes and not done the homework.

The value of mean is,

μ=50(0.25)=12.5

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x≥30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 12.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 4

From the graph, the probability value is 0.000. Thus, the percentage of the students is 0.

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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 6, Problem 49SE

a.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

a.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is about 5.113%.

Explanation of Solution

Calculation:

The examination contains 50 multiple choice questions. Each question has four options. The students have 75% probability to answer any question correctly if they attend lecture and do homework.

Here, the student must answer correctly to 43 questions to obtain grade A.

The formula for mean in binomial distribution is,

μ=np

The value of mean is,

μ=50(0.75)=37.5

The formula for standard deviation is,

σ=np(1−p)

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 43.

That is, c=43

P(x≥43)=P(x≥43−0.5)=P(x≥42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥42.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 42.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 1

From the graph, the probability value is 0.05113.

Thus, the percentage of the students is 5.113%.

b.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

b.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 57.99%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 35 to 39 questions to obtain grade C.

If the binomial probability represents “between than c1 and c2” then the normal probability is P(c1−0.5≤x≤c2+0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 35 and the value 0.5 added to 39.

That is, c1=35 and c2=39

P(x<43)=P(x≤43−0.5)=P(x≤42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(34.5≤x≤39.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Middle Tail for the region of the curve to shade.
Enter the X value 1 as 34.5 and X value 2 as 39.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 2

From the graph, the probability value is 0.5799

Thus, the percentage of the students is 57.99%.

c.

To determine

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

c.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will pass the examination is about 99.55%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 30 or more questions to pass the examination.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x>30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 3

From the graph, the probability value is 0.9955.

Thus, the percentage of the students is 99.55%.

d.

To determine

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

d.

Expert Solution

Answer to Problem 49SE

The students who did not attend the classes and not done the homework will not pass the examination by guess.

Explanation of Solution

Calculation:

Here, the student did not attend the class and has not done the homework.

Approximately 25% students of did not attend the classes and not done the homework.

The value of mean is,

μ=50(0.25)=12.5

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x≥30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 12.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 4

From the graph, the probability value is 0.000. Thus, the percentage of the students is 0.

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Answer 5

Chapter 6, Problem 49SE

a.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

a.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is about 5.113%.

Explanation of Solution

Calculation:

The examination contains 50 multiple choice questions. Each question has four options. The students have 75% probability to answer any question correctly if they attend lecture and do homework.

Here, the student must answer correctly to 43 questions to obtain grade A.

The formula for mean in binomial distribution is,

μ=np

The value of mean is,

μ=50(0.75)=37.5

The formula for standard deviation is,

σ=np(1−p)

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 43.

That is, c=43

P(x≥43)=P(x≥43−0.5)=P(x≥42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥42.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 42.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 1

From the graph, the probability value is 0.05113.

Thus, the percentage of the students is 5.113%.

b.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

b.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 57.99%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 35 to 39 questions to obtain grade C.

If the binomial probability represents “between than c1 and c2” then the normal probability is P(c1−0.5≤x≤c2+0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 35 and the value 0.5 added to 39.

That is, c1=35 and c2=39

P(x<43)=P(x≤43−0.5)=P(x≤42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(34.5≤x≤39.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Middle Tail for the region of the curve to shade.
Enter the X value 1 as 34.5 and X value 2 as 39.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 2

From the graph, the probability value is 0.5799

Thus, the percentage of the students is 57.99%.

c.

To determine

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

c.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will pass the examination is about 99.55%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 30 or more questions to pass the examination.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x>30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 3

From the graph, the probability value is 0.9955.

Thus, the percentage of the students is 99.55%.

d.

To determine

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

d.

Expert Solution

Answer to Problem 49SE

The students who did not attend the classes and not done the homework will not pass the examination by guess.

Explanation of Solution

Calculation:

Here, the student did not attend the class and has not done the homework.

Approximately 25% students of did not attend the classes and not done the homework.

The value of mean is,

μ=50(0.25)=12.5

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x≥30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 12.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 4

From the graph, the probability value is 0.000. Thus, the percentage of the students is 0.

Answer 6

Chapter 6, Problem 49SE

a.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

a.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is about 5.113%.

Explanation of Solution

Calculation:

The examination contains 50 multiple choice questions. Each question has four options. The students have 75% probability to answer any question correctly if they attend lecture and do homework.

Here, the student must answer correctly to 43 questions to obtain grade A.

The formula for mean in binomial distribution is,

μ=np

The value of mean is,

μ=50(0.75)=37.5

The formula for standard deviation is,

σ=np(1−p)

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 43.

That is, c=43

P(x≥43)=P(x≥43−0.5)=P(x≥42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥42.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 42.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 1

From the graph, the probability value is 0.05113.

Thus, the percentage of the students is 5.113%.

Answer 7

Chapter 6, Problem 49SE

a.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

Answer 8

a.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is about 5.113%.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The examination contains 50 multiple choice questions. Each question has four options. The students have 75% probability to answer any question correctly if they attend lecture and do homework.

Here, the student must answer correctly to 43 questions to obtain grade A.

The formula for mean in binomial distribution is,

μ=np

The value of mean is,

μ=50(0.75)=37.5

The formula for standard deviation is,

σ=np(1−p)

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

Continuity correction:

The binomial probability is converted to a normal distribution probability by using the continuity correction.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 43.

That is, c=43

P(x≥43)=P(x≥43−0.5)=P(x≥42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥42.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 42.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 1

From the graph, the probability value is 0.05113.

Thus, the percentage of the students is 5.113%.

Answer 13

b.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

b.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 57.99%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 35 to 39 questions to obtain grade C.

If the binomial probability represents “between than c1 and c2” then the normal probability is P(c1−0.5≤x≤c2+0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 35 and the value 0.5 added to 39.

That is, c1=35 and c2=39

P(x<43)=P(x≤43−0.5)=P(x≤42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(34.5≤x≤39.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Middle Tail for the region of the curve to shade.
Enter the X value 1 as 34.5 and X value 2 as 39.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 2

From the graph, the probability value is 0.5799

Thus, the percentage of the students is 57.99%.

Answer 14

b.

To determine

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

Answer 15

b.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 57.99%.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Here, the student must answer correctly to 35 to 39 questions to obtain grade C.

If the binomial probability represents “between than c1 and c2” then the normal probability is P(c1−0.5≤x≤c2+0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 35 and the value 0.5 added to 39.

That is, c1=35 and c2=39

P(x<43)=P(x≤43−0.5)=P(x≤42.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(34.5≤x≤39.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Middle Tail for the region of the curve to shade.
Enter the X value 1 as 34.5 and X value 2 as 39.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 2

From the graph, the probability value is 0.5799

Thus, the percentage of the students is 57.99%.

Answer 20

c.

To determine

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

c.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will pass the examination is about 99.55%.

Explanation of Solution

Calculation:

Here, the student must answer correctly to 30 or more questions to pass the examination.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x>30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 3

From the graph, the probability value is 0.9955.

Thus, the percentage of the students is 99.55%.

Answer 21

c.

To determine

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

Answer 22

c.

Expert Solution

Answer to Problem 49SE

The percentage of the students who have done their homework and attended lectures will pass the examination is about 99.55%.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

Here, the student must answer correctly to 30 or more questions to pass the examination.

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x>30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 37.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 3

From the graph, the probability value is 0.9955.

Thus, the percentage of the students is 99.55%.

Answer 27

d.

To determine

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

d.

Expert Solution

Answer to Problem 49SE

The students who did not attend the classes and not done the homework will not pass the examination by guess.

Explanation of Solution

Calculation:

Here, the student did not attend the class and has not done the homework.

Approximately 25% students of did not attend the classes and not done the homework.

The value of mean is,

μ=50(0.25)=12.5

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x≥30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 12.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 4

From the graph, the probability value is 0.000. Thus, the percentage of the students is 0.

Answer 28

d.

To determine

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

Answer 29

d.

Expert Solution

Answer to Problem 49SE

The students who did not attend the classes and not done the homework will not pass the examination by guess.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

Here, the student did not attend the class and has not done the homework.

Approximately 25% students of did not attend the classes and not done the homework.

The value of mean is,

μ=50(0.25)=12.5

The value of standard deviation is,

σ=50(0.75)(0.25)=9.375=3.06

If the binomial probability represents “more than c” then the normal probability is P(x≤c−0.5) for any number c.

By using continuity correction, the value 0.5 is subtracted from 30.

That is, c=30

P(x≥30)=P(x≥30−0.5)=P(x≥29.5)

Thus, the binomial probability to a normal probability by using continuity correction is P(x≥29.5).

Software Procedure:

Step-by-step procedure to obtain the probability value using the MINITAB software:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 12.5 and Standard deviation as 3.06.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the X value as 29.5.
Click OK.

Output using the MINITAB software is given below:

EBK ESSENTIALS OF STATISTICS FOR BUSINE, Chapter 6, Problem 49SE , additional homework tip 4

From the graph, the probability value is 0.000. Thus, the percentage of the students is 0.

Answer 34

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Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

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Find the percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination.

Answer to Problem 49SE

Explanation of Solution

Find the percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination.

Answer to Problem 49SE

Explanation of Solution

Find the percentage of the students who have done their homework and attended lectures will pass the examination.

Answer to Problem 49SE

Explanation of Solution

Find the probability that the students who did not attend the classes and not done the homework will answer 30 or more questions correctly on the multiple choice examinations.

Answer to Problem 49SE

Explanation of Solution

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Chapter 6 Solutions