Find the standard deviation of the x distribution.

Question

Want to see more full solutions like this?

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 6.3, Problem 34P

(a)

To determine

Find the standard deviation of the x distribution.

(a)

Expert Solution

Answer to Problem 34P

The standard deviation of the x distribution is 5.25 ounces.

Explanation of Solution

Calculation:

Rule of Thumb:

The formula for standard deviation using range is,

σ=Range4

In the formula, range is the obtained by subtracting the low value from the high value.

The variable x is a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten.

The 95% of data range from 14 to 35 ounces.

The standard deviation is,

σ=35−144=214=5.25

Hence, the standard deviation of the x distribution is 5.25 ounces.

(b)

To determine

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

(b)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=x−μσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z=14−24.55.25=−10.55.25=−2

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

The probability is,

P(x<14)=P(z<−2)=0.0228

Hence, the probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

(c)

To determine

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

(c)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Explanation of Solution

Calculation:

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(x>33)=P(z>1.62)=1−P(z<1.62)=1−0.9474=0.0526

Hence, the probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

(d)

To determine

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

(d)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Explanation of Solution

Calculation:

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z1=14−24.55.25=−10.55.25=−2

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z2=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(14<x<33)=P(−2<z<1.62)=P(z<1.62)−P(z<−2)=0.9474−0.0228=0.9246

Hence, the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

(e)

To determine

Find the cutoff point for the weight of an undernourished kitten.

(e)

Expert Solution

Answer to Problem 34P

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Explanation of Solution

Calculation:

The weight is in the bottom 10% of the probability distribution of weights is called undernourished.

Step by step procedure to obtain probability plot using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 24.5, and Standard deviation as 5.25.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability and Left Tail, for the region of the curve to shade.
Enter the Probability as 0.10.
Click OK.

Output using MINITAB software is given below:

Understandable Statistics: Concepts and Methods, Chapter 6.3, Problem 34P

From Minitab output, the cutoff point for the weight of an undernourished kitten is 17.77.

Hence, the cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Want to see more full solutions like this?

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

Students have asked these similar questions

Calculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.

Microsoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADO

Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?

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 6.3, Problem 34P

(a)

To determine

Find the standard deviation of the x distribution.

(a)

Expert Solution

Answer to Problem 34P

The standard deviation of the x distribution is 5.25 ounces.

Explanation of Solution

Calculation:

Rule of Thumb:

The formula for standard deviation using range is,

σ=Range4

In the formula, range is the obtained by subtracting the low value from the high value.

The variable x is a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten.

The 95% of data range from 14 to 35 ounces.

The standard deviation is,

σ=35−144=214=5.25

Hence, the standard deviation of the x distribution is 5.25 ounces.

(b)

To determine

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

(b)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=x−μσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z=14−24.55.25=−10.55.25=−2

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

The probability is,

P(x<14)=P(z<−2)=0.0228

Hence, the probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

(c)

To determine

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

(c)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Explanation of Solution

Calculation:

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(x>33)=P(z>1.62)=1−P(z<1.62)=1−0.9474=0.0526

Hence, the probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

(d)

To determine

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

(d)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Explanation of Solution

Calculation:

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z1=14−24.55.25=−10.55.25=−2

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z2=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(14<x<33)=P(−2<z<1.62)=P(z<1.62)−P(z<−2)=0.9474−0.0228=0.9246

Hence, the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

(e)

To determine

Find the cutoff point for the weight of an undernourished kitten.

(e)

Expert Solution

Answer to Problem 34P

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Explanation of Solution

Calculation:

The weight is in the bottom 10% of the probability distribution of weights is called undernourished.

Step by step procedure to obtain probability plot using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 24.5, and Standard deviation as 5.25.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability and Left Tail, for the region of the curve to shade.
Enter the Probability as 0.10.
Click OK.

Output using MINITAB software is given below:

Understandable Statistics: Concepts and Methods, Chapter 6.3, Problem 34P

From Minitab output, the cutoff point for the weight of an undernourished kitten is 17.77.

Hence, the cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Want to see more full solutions like this?

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 6.3, Problem 34P

(a)

To determine

Find the standard deviation of the x distribution.

(a)

Expert Solution

Answer to Problem 34P

The standard deviation of the x distribution is 5.25 ounces.

Explanation of Solution

Calculation:

Rule of Thumb:

The formula for standard deviation using range is,

σ=Range4

In the formula, range is the obtained by subtracting the low value from the high value.

The variable x is a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten.

The 95% of data range from 14 to 35 ounces.

The standard deviation is,

σ=35−144=214=5.25

Hence, the standard deviation of the x distribution is 5.25 ounces.

(b)

To determine

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

(b)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=x−μσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z=14−24.55.25=−10.55.25=−2

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

The probability is,

P(x<14)=P(z<−2)=0.0228

Hence, the probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

(c)

To determine

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

(c)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Explanation of Solution

Calculation:

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(x>33)=P(z>1.62)=1−P(z<1.62)=1−0.9474=0.0526

Hence, the probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

(d)

To determine

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

(d)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Explanation of Solution

Calculation:

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z1=14−24.55.25=−10.55.25=−2

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z2=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(14<x<33)=P(−2<z<1.62)=P(z<1.62)−P(z<−2)=0.9474−0.0228=0.9246

Hence, the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

(e)

To determine

Find the cutoff point for the weight of an undernourished kitten.

(e)

Expert Solution

Answer to Problem 34P

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Explanation of Solution

Calculation:

The weight is in the bottom 10% of the probability distribution of weights is called undernourished.

Step by step procedure to obtain probability plot using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 24.5, and Standard deviation as 5.25.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability and Left Tail, for the region of the curve to shade.
Enter the Probability as 0.10.
Click OK.

Output using MINITAB software is given below:

Understandable Statistics: Concepts and Methods, Chapter 6.3, Problem 34P

From Minitab output, the cutoff point for the weight of an undernourished kitten is 17.77.

Hence, the cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Want to see more full solutions like this?

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 6.3, Problem 34P

(a)

To determine

Find the standard deviation of the x distribution.

(a)

Expert Solution

Answer to Problem 34P

The standard deviation of the x distribution is 5.25 ounces.

Explanation of Solution

Calculation:

Rule of Thumb:

The formula for standard deviation using range is,

σ=Range4

In the formula, range is the obtained by subtracting the low value from the high value.

The variable x is a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten.

The 95% of data range from 14 to 35 ounces.

The standard deviation is,

σ=35−144=214=5.25

Hence, the standard deviation of the x distribution is 5.25 ounces.

(b)

To determine

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

(b)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=x−μσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z=14−24.55.25=−10.55.25=−2

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

The probability is,

P(x<14)=P(z<−2)=0.0228

Hence, the probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

(c)

To determine

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

(c)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Explanation of Solution

Calculation:

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(x>33)=P(z>1.62)=1−P(z<1.62)=1−0.9474=0.0526

Hence, the probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

(d)

To determine

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

(d)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Explanation of Solution

Calculation:

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z1=14−24.55.25=−10.55.25=−2

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z2=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(14<x<33)=P(−2<z<1.62)=P(z<1.62)−P(z<−2)=0.9474−0.0228=0.9246

Hence, the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

(e)

To determine

Find the cutoff point for the weight of an undernourished kitten.

(e)

Expert Solution

Answer to Problem 34P

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Explanation of Solution

Calculation:

The weight is in the bottom 10% of the probability distribution of weights is called undernourished.

Step by step procedure to obtain probability plot using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 24.5, and Standard deviation as 5.25.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability and Left Tail, for the region of the curve to shade.
Enter the Probability as 0.10.
Click OK.

Output using MINITAB software is given below:

Understandable Statistics: Concepts and Methods, Chapter 6.3, Problem 34P

From Minitab output, the cutoff point for the weight of an undernourished kitten is 17.77.

Hence, the cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Want to see more full solutions like this?

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

Answer 5

Chapter 6.3, Problem 34P

(a)

To determine

Find the standard deviation of the x distribution.

(a)

Expert Solution

Answer to Problem 34P

The standard deviation of the x distribution is 5.25 ounces.

Explanation of Solution

Calculation:

Rule of Thumb:

The formula for standard deviation using range is,

σ=Range4

In the formula, range is the obtained by subtracting the low value from the high value.

The variable x is a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten.

The 95% of data range from 14 to 35 ounces.

The standard deviation is,

σ=35−144=214=5.25

Hence, the standard deviation of the x distribution is 5.25 ounces.

(b)

To determine

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

(b)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=x−μσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z=14−24.55.25=−10.55.25=−2

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

The probability is,

P(x<14)=P(z<−2)=0.0228

Hence, the probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

(c)

To determine

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

(c)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Explanation of Solution

Calculation:

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(x>33)=P(z>1.62)=1−P(z<1.62)=1−0.9474=0.0526

Hence, the probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

(d)

To determine

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

(d)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Explanation of Solution

Calculation:

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z1=14−24.55.25=−10.55.25=−2

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z2=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(14<x<33)=P(−2<z<1.62)=P(z<1.62)−P(z<−2)=0.9474−0.0228=0.9246

Hence, the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

(e)

To determine

Find the cutoff point for the weight of an undernourished kitten.

(e)

Expert Solution

Answer to Problem 34P

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Explanation of Solution

Calculation:

The weight is in the bottom 10% of the probability distribution of weights is called undernourished.

Step by step procedure to obtain probability plot using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 24.5, and Standard deviation as 5.25.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability and Left Tail, for the region of the curve to shade.
Enter the Probability as 0.10.
Click OK.

Output using MINITAB software is given below:

Understandable Statistics: Concepts and Methods, Chapter 6.3, Problem 34P

From Minitab output, the cutoff point for the weight of an undernourished kitten is 17.77.

Hence, the cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Answer 6

Chapter 6.3, Problem 34P

(a)

To determine

Find the standard deviation of the x distribution.

(a)

Expert Solution

Answer to Problem 34P

The standard deviation of the x distribution is 5.25 ounces.

Explanation of Solution

Calculation:

Rule of Thumb:

The formula for standard deviation using range is,

σ=Range4

In the formula, range is the obtained by subtracting the low value from the high value.

The variable x is a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten.

The 95% of data range from 14 to 35 ounces.

The standard deviation is,

σ=35−144=214=5.25

Hence, the standard deviation of the x distribution is 5.25 ounces.

Answer 7

Chapter 6.3, Problem 34P

(a)

To determine

Find the standard deviation of the x distribution.

Answer 8

(a)

Expert Solution

Answer to Problem 34P

The standard deviation of the x distribution is 5.25 ounces.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

Rule of Thumb:

The formula for standard deviation using range is,

σ=Range4

In the formula, range is the obtained by subtracting the low value from the high value.

The variable x is a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten.

The 95% of data range from 14 to 35 ounces.

The standard deviation is,

σ=35−144=214=5.25

Hence, the standard deviation of the x distribution is 5.25 ounces.

Answer 13

(b)

To determine

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

(b)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=x−μσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z=14−24.55.25=−10.55.25=−2

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

The probability is,

P(x<14)=P(z<−2)=0.0228

Hence, the probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Answer 14

(b)

To determine

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

Answer 15

(b)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=x−μσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z=14−24.55.25=−10.55.25=−2

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

The probability is,

P(x<14)=P(z<−2)=0.0228

Hence, the probability that a healthy 10-week-old kitten would weigh less than 14 ounces is 0.0228.

Answer 20

(c)

To determine

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

(c)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Explanation of Solution

Calculation:

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(x>33)=P(z>1.62)=1−P(z<1.62)=1−0.9474=0.0526

Hence, the probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Answer 21

(c)

To determine

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

Answer 22

(c)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(x>33)=P(z>1.62)=1−P(z<1.62)=1−0.9474=0.0526

Hence, the probability that a healthy 10-week-old kitten would weigh more than 33 ounces is 0.0526.

Answer 27

(d)

To determine

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

(d)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Explanation of Solution

Calculation:

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z1=14−24.55.25=−10.55.25=−2

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z2=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(14<x<33)=P(−2<z<1.62)=P(z<1.62)−P(z<−2)=0.9474−0.0228=0.9246

Hence, the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Answer 28

(d)

To determine

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

Answer 29

(d)

Expert Solution

Answer to Problem 34P

The probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

Substitute x as 14, μ as 24.5 and σ as 5.25 in the formula of z score.

z1=14−24.55.25=−10.55.25=−2

Substitute x as 33, μ as 24.5 and σ as 5.25 in the formula of z score.

z2=33−24.55.25=8.55.25=1.62

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.

Locate the value –2.0 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0228.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.62.

Locate the value 1.6 in column z.
Locate the value 0.02 in top row.
The intersecting value of row and column is 0.9474.

The probability is,

P(14<x<33)=P(−2<z<1.62)=P(z<1.62)−P(z<−2)=0.9474−0.0228=0.9246

Hence, the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces is 0.9246.

Answer 34

(e)

To determine

Find the cutoff point for the weight of an undernourished kitten.

(e)

Expert Solution

Answer to Problem 34P

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Explanation of Solution

Calculation:

The weight is in the bottom 10% of the probability distribution of weights is called undernourished.

Step by step procedure to obtain probability plot using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 24.5, and Standard deviation as 5.25.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability and Left Tail, for the region of the curve to shade.
Enter the Probability as 0.10.
Click OK.

Output using MINITAB software is given below:

Understandable Statistics: Concepts and Methods, Chapter 6.3, Problem 34P

From Minitab output, the cutoff point for the weight of an undernourished kitten is 17.77.

Hence, the cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Answer 35

(e)

To determine

Find the cutoff point for the weight of an undernourished kitten.

Answer 36

(e)

Expert Solution

Answer to Problem 34P

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation:

The weight is in the bottom 10% of the probability distribution of weights is called undernourished.

Step by step procedure to obtain probability plot using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
Enter the Mean as 24.5, and Standard deviation as 5.25.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose Probability and Left Tail, for the region of the curve to shade.
Enter the Probability as 0.10.
Click OK.

Output using MINITAB software is given below:

Understandable Statistics: Concepts and Methods, Chapter 6.3, Problem 34P

From Minitab output, the cutoff point for the weight of an undernourished kitten is 17.77.

Hence, the cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Answer 41

Ch. 6.1 - Statistical Literacy Which, if any, of the curves...Ch. 6.1 - Statistical Literacy Look at the normal curve in...Ch. 6.1 - Prob. 3P Ch. 6.1 - Critical Thinking Sketch a normal curve (a) with...Ch. 6.1 - Basic Computation: Empirical Rule What percentage...Ch. 6.1 - Basic Computation: Empirical Rule What percentage...Ch. 6.1 - Distribution: Heights of Coeds Assuming that the...Ch. 6.1 - Distribution: Rhode Island Red Chicks The...Ch. 6.1 - Archaeology: Tree Rings At Burnt Mesa Pueblo,...Ch. 6.1 - Vending Machine: Soft Drinks A vending machine...

Find the standard deviation of the x distribution.

Concept explainers

Videos

Find the standard deviation of the x distribution.

Answer to Problem 34P

Explanation of Solution

Find the probability that a healthy 10-week-old kitten would weigh less than 14 ounces.

Answer to Problem 34P

Explanation of Solution

Find the probability that a healthy 10-week-old kitten would weigh more than 33 ounces.

Answer to Problem 34P

Explanation of Solution

Find the probability that a healthy 10-week-old kitten would weigh between 14 and 33 ounces.

Answer to Problem 34P

Explanation of Solution

Find the cutoff point for the weight of an undernourished kitten.

Answer to Problem 34P

Explanation of Solution

Want to see more full solutions like this?

Chapter 6 Solutions