Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4. Compare the results in each case to the actual calculated value of s .

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Expert Solution & Answer

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements y1,y2,...,yn is defined as,

y¯=1n∑i=1nyi.

Variance:

The variance of a sample of measurements y1,y2,...,yn is defined as the sum of the square of the difference between the measurements and the mean, divided by n−1.

That is,

s2=1n−1∑i=1n(yi−y¯)2.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, s=s2.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, n=45.

Consider the following table for necessary calculations.

yi	yi2
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
∑i=145yi=440.6	∑i=145yi2=5,067.38

Hence, (∑i=145yi)2=194,128.4.

Substitute n=45, and ∑i=145yi=440.6 in the formula of mean.

That is,

y¯=440.645≈9.79.

Thus, the mean value is 9.79.

Substitute n=45, (∑i=145yi)2=194,128.4, and ∑i=145yi2=5,067.38 in the s2 formula.

That is,

s2=145−1[5,067.38−145(194,128.4)]=144[5,067.38−4,313.964]≈17.12

Thus, the value of s is calculated below:

s=s2=17.12≈4.14

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

s=35.1−5.74=7.35

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, n=25.

Consider the following table for necessary calculations.

yi	yi2
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
∑i=125yi=80.63	∑i=125yi2=500.746

Hence, (∑i=125yi)2=6,501.197.

Substitute n=25, and ∑i=125yi=80.63 in the formula of mean.

That is,

y¯=80.6325≈3.23.

Thus, the mean value is 3.23.

Substitute n=25, (∑i=125yi)2=6,501.197, and ∑i=125yi2=500.746 in the s2 formula.

That is,

s2=125−1[500.746−125(6,501.197)]=124[500.746−260.0479]≈10.03

Thus, the value of s is calculated below:

s=s2=10.13≈3.17

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

s=12.48−0.384=3.04

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, n=40.

Consider the following table for necessary calculations.

yi	yi2
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
∑i=140yi=175.48	∑i=140yi2=906.41

Hence, (∑i=140yi)2=30,793.23.

Substitute n=40, and ∑i=140yi=175.48 in the formula of mean.

That is,

y¯=175.4840≈4.39.

Thus, the mean value is 4.39.

Substitute n=40, (∑i=140yi)2=30,793.23, and ∑i=140yi2=906.41 in the s2 formula.

That is,

s2=140−1[906.41−140(30,793.23)]=139[906.41−769.83]≈3.5021

Thus, the value of s is calculated below:

s=s2=3.5021≈1.87

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

s=11.88−2.614=2.3175

Thus, it can be said that the estimated value of s is somewhat close to the actual value.

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Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Expert Solution & Answer

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements y1,y2,...,yn is defined as,

y¯=1n∑i=1nyi.

Variance:

The variance of a sample of measurements y1,y2,...,yn is defined as the sum of the square of the difference between the measurements and the mean, divided by n−1.

That is,

s2=1n−1∑i=1n(yi−y¯)2.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, s=s2.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, n=45.

Consider the following table for necessary calculations.

yi	yi2
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
∑i=145yi=440.6	∑i=145yi2=5,067.38

Hence, (∑i=145yi)2=194,128.4.

Substitute n=45, and ∑i=145yi=440.6 in the formula of mean.

That is,

y¯=440.645≈9.79.

Thus, the mean value is 9.79.

Substitute n=45, (∑i=145yi)2=194,128.4, and ∑i=145yi2=5,067.38 in the s2 formula.

That is,

s2=145−1[5,067.38−145(194,128.4)]=144[5,067.38−4,313.964]≈17.12

Thus, the value of s is calculated below:

s=s2=17.12≈4.14

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

s=35.1−5.74=7.35

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, n=25.

Consider the following table for necessary calculations.

yi	yi2
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
∑i=125yi=80.63	∑i=125yi2=500.746

Hence, (∑i=125yi)2=6,501.197.

Substitute n=25, and ∑i=125yi=80.63 in the formula of mean.

That is,

y¯=80.6325≈3.23.

Thus, the mean value is 3.23.

Substitute n=25, (∑i=125yi)2=6,501.197, and ∑i=125yi2=500.746 in the s2 formula.

That is,

s2=125−1[500.746−125(6,501.197)]=124[500.746−260.0479]≈10.03

Thus, the value of s is calculated below:

s=s2=10.13≈3.17

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

s=12.48−0.384=3.04

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, n=40.

Consider the following table for necessary calculations.

yi	yi2
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
∑i=140yi=175.48	∑i=140yi2=906.41

Hence, (∑i=140yi)2=30,793.23.

Substitute n=40, and ∑i=140yi=175.48 in the formula of mean.

That is,

y¯=175.4840≈4.39.

Thus, the mean value is 4.39.

Substitute n=40, (∑i=140yi)2=30,793.23, and ∑i=140yi2=906.41 in the s2 formula.

That is,

s2=140−1[906.41−140(30,793.23)]=139[906.41−769.83]≈3.5021

Thus, the value of s is calculated below:

s=s2=3.5021≈1.87

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

s=11.88−2.614=2.3175

Thus, it can be said that the estimated value of s is somewhat close to the actual value.

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Answer 3

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Expert Solution & Answer

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements y1,y2,...,yn is defined as,

y¯=1n∑i=1nyi.

Variance:

The variance of a sample of measurements y1,y2,...,yn is defined as the sum of the square of the difference between the measurements and the mean, divided by n−1.

That is,

s2=1n−1∑i=1n(yi−y¯)2.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, s=s2.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, n=45.

Consider the following table for necessary calculations.

yi	yi2
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
∑i=145yi=440.6	∑i=145yi2=5,067.38

Hence, (∑i=145yi)2=194,128.4.

Substitute n=45, and ∑i=145yi=440.6 in the formula of mean.

That is,

y¯=440.645≈9.79.

Thus, the mean value is 9.79.

Substitute n=45, (∑i=145yi)2=194,128.4, and ∑i=145yi2=5,067.38 in the s2 formula.

That is,

s2=145−1[5,067.38−145(194,128.4)]=144[5,067.38−4,313.964]≈17.12

Thus, the value of s is calculated below:

s=s2=17.12≈4.14

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

s=35.1−5.74=7.35

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, n=25.

Consider the following table for necessary calculations.

yi	yi2
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
∑i=125yi=80.63	∑i=125yi2=500.746

Hence, (∑i=125yi)2=6,501.197.

Substitute n=25, and ∑i=125yi=80.63 in the formula of mean.

That is,

y¯=80.6325≈3.23.

Thus, the mean value is 3.23.

Substitute n=25, (∑i=125yi)2=6,501.197, and ∑i=125yi2=500.746 in the s2 formula.

That is,

s2=125−1[500.746−125(6,501.197)]=124[500.746−260.0479]≈10.03

Thus, the value of s is calculated below:

s=s2=10.13≈3.17

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

s=12.48−0.384=3.04

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, n=40.

Consider the following table for necessary calculations.

yi	yi2
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
∑i=140yi=175.48	∑i=140yi2=906.41

Hence, (∑i=140yi)2=30,793.23.

Substitute n=40, and ∑i=140yi=175.48 in the formula of mean.

That is,

y¯=175.4840≈4.39.

Thus, the mean value is 4.39.

Substitute n=40, (∑i=140yi)2=30,793.23, and ∑i=140yi2=906.41 in the s2 formula.

That is,

s2=140−1[906.41−140(30,793.23)]=139[906.41−769.83]≈3.5021

Thus, the value of s is calculated below:

s=s2=3.5021≈1.87

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

s=11.88−2.614=2.3175

Thus, it can be said that the estimated value of s is somewhat close to the actual value.

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Answer 4

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Expert Solution & Answer

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements y1,y2,...,yn is defined as,

y¯=1n∑i=1nyi.

Variance:

The variance of a sample of measurements y1,y2,...,yn is defined as the sum of the square of the difference between the measurements and the mean, divided by n−1.

That is,

s2=1n−1∑i=1n(yi−y¯)2.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, s=s2.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, n=45.

Consider the following table for necessary calculations.

yi	yi2
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
∑i=145yi=440.6	∑i=145yi2=5,067.38

Hence, (∑i=145yi)2=194,128.4.

Substitute n=45, and ∑i=145yi=440.6 in the formula of mean.

That is,

y¯=440.645≈9.79.

Thus, the mean value is 9.79.

Substitute n=45, (∑i=145yi)2=194,128.4, and ∑i=145yi2=5,067.38 in the s2 formula.

That is,

s2=145−1[5,067.38−145(194,128.4)]=144[5,067.38−4,313.964]≈17.12

Thus, the value of s is calculated below:

s=s2=17.12≈4.14

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

s=35.1−5.74=7.35

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, n=25.

Consider the following table for necessary calculations.

yi	yi2
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
∑i=125yi=80.63	∑i=125yi2=500.746

Hence, (∑i=125yi)2=6,501.197.

Substitute n=25, and ∑i=125yi=80.63 in the formula of mean.

That is,

y¯=80.6325≈3.23.

Thus, the mean value is 3.23.

Substitute n=25, (∑i=125yi)2=6,501.197, and ∑i=125yi2=500.746 in the s2 formula.

That is,

s2=125−1[500.746−125(6,501.197)]=124[500.746−260.0479]≈10.03

Thus, the value of s is calculated below:

s=s2=10.13≈3.17

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

s=12.48−0.384=3.04

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, n=40.

Consider the following table for necessary calculations.

yi	yi2
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
∑i=140yi=175.48	∑i=140yi2=906.41

Hence, (∑i=140yi)2=30,793.23.

Substitute n=40, and ∑i=140yi=175.48 in the formula of mean.

That is,

y¯=175.4840≈4.39.

Thus, the mean value is 4.39.

Substitute n=40, (∑i=140yi)2=30,793.23, and ∑i=140yi2=906.41 in the s2 formula.

That is,

s2=140−1[906.41−140(30,793.23)]=139[906.41−769.83]≈3.5021

Thus, the value of s is calculated below:

s=s2=3.5021≈1.87

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

s=11.88−2.614=2.3175

Thus, it can be said that the estimated value of s is somewhat close to the actual value.

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Answer 5

Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Expert Solution & Answer

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements y1,y2,...,yn is defined as,

y¯=1n∑i=1nyi.

Variance:

The variance of a sample of measurements y1,y2,...,yn is defined as the sum of the square of the difference between the measurements and the mean, divided by n−1.

That is,

s2=1n−1∑i=1n(yi−y¯)2.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, s=s2.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, n=45.

Consider the following table for necessary calculations.

yi	yi2
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
∑i=145yi=440.6	∑i=145yi2=5,067.38

Hence, (∑i=145yi)2=194,128.4.

Substitute n=45, and ∑i=145yi=440.6 in the formula of mean.

That is,

y¯=440.645≈9.79.

Thus, the mean value is 9.79.

Substitute n=45, (∑i=145yi)2=194,128.4, and ∑i=145yi2=5,067.38 in the s2 formula.

That is,

s2=145−1[5,067.38−145(194,128.4)]=144[5,067.38−4,313.964]≈17.12

Thus, the value of s is calculated below:

s=s2=17.12≈4.14

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

s=35.1−5.74=7.35

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, n=25.

Consider the following table for necessary calculations.

yi	yi2
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
∑i=125yi=80.63	∑i=125yi2=500.746

Hence, (∑i=125yi)2=6,501.197.

Substitute n=25, and ∑i=125yi=80.63 in the formula of mean.

That is,

y¯=80.6325≈3.23.

Thus, the mean value is 3.23.

Substitute n=25, (∑i=125yi)2=6,501.197, and ∑i=125yi2=500.746 in the s2 formula.

That is,

s2=125−1[500.746−125(6,501.197)]=124[500.746−260.0479]≈10.03

Thus, the value of s is calculated below:

s=s2=10.13≈3.17

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

s=12.48−0.384=3.04

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, n=40.

Consider the following table for necessary calculations.

yi	yi2
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
∑i=140yi=175.48	∑i=140yi2=906.41

Hence, (∑i=140yi)2=30,793.23.

Substitute n=40, and ∑i=140yi=175.48 in the formula of mean.

That is,

y¯=175.4840≈4.39.

Thus, the mean value is 4.39.

Substitute n=40, (∑i=140yi)2=30,793.23, and ∑i=140yi2=906.41 in the s2 formula.

That is,

s2=140−1[906.41−140(30,793.23)]=139[906.41−769.83]≈3.5021

Thus, the value of s is calculated below:

s=s2=3.5021≈1.87

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

s=11.88−2.614=2.3175

Thus, it can be said that the estimated value of s is somewhat close to the actual value.

Answer 6

Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Expert Solution & Answer

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements y1,y2,...,yn is defined as,

y¯=1n∑i=1nyi.

Variance:

The variance of a sample of measurements y1,y2,...,yn is defined as the sum of the square of the difference between the measurements and the mean, divided by n−1.

That is,

s2=1n−1∑i=1n(yi−y¯)2.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, s=s2.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, n=45.

Consider the following table for necessary calculations.

yi	yi2
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
∑i=145yi=440.6	∑i=145yi2=5,067.38

Hence, (∑i=145yi)2=194,128.4.

Substitute n=45, and ∑i=145yi=440.6 in the formula of mean.

That is,

y¯=440.645≈9.79.

Thus, the mean value is 9.79.

Substitute n=45, (∑i=145yi)2=194,128.4, and ∑i=145yi2=5,067.38 in the s2 formula.

That is,

s2=145−1[5,067.38−145(194,128.4)]=144[5,067.38−4,313.964]≈17.12

Thus, the value of s is calculated below:

s=s2=17.12≈4.14

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

s=35.1−5.74=7.35

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, n=25.

Consider the following table for necessary calculations.

yi	yi2
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
∑i=125yi=80.63	∑i=125yi2=500.746

Hence, (∑i=125yi)2=6,501.197.

Substitute n=25, and ∑i=125yi=80.63 in the formula of mean.

That is,

y¯=80.6325≈3.23.

Thus, the mean value is 3.23.

Substitute n=25, (∑i=125yi)2=6,501.197, and ∑i=125yi2=500.746 in the s2 formula.

That is,

s2=125−1[500.746−125(6,501.197)]=124[500.746−260.0479]≈10.03

Thus, the value of s is calculated below:

s=s2=10.13≈3.17

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

s=12.48−0.384=3.04

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, n=40.

Consider the following table for necessary calculations.

yi	yi2
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
∑i=140yi=175.48	∑i=140yi2=906.41

Hence, (∑i=140yi)2=30,793.23.

Substitute n=40, and ∑i=140yi=175.48 in the formula of mean.

That is,

y¯=175.4840≈4.39.

Thus, the mean value is 4.39.

Substitute n=40, (∑i=140yi)2=30,793.23, and ∑i=140yi2=906.41 in the s2 formula.

That is,

s2=140−1[906.41−140(30,793.23)]=139[906.41−769.83]≈3.5021

Thus, the value of s is calculated below:

s=s2=3.5021≈1.87

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

s=11.88−2.614=2.3175

Thus, it can be said that the estimated value of s is somewhat close to the actual value.

Answer 7

Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Answer 8

Expert Solution & Answer

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements y1,y2,...,yn is defined as,

y¯=1n∑i=1nyi.

Variance:

The variance of a sample of measurements y1,y2,...,yn is defined as the sum of the square of the difference between the measurements and the mean, divided by n−1.

That is,

s2=1n−1∑i=1n(yi−y¯)2.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, s=s2.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, n=45.

Consider the following table for necessary calculations.

yi	yi2
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
∑i=145yi=440.6	∑i=145yi2=5,067.38

Hence, (∑i=145yi)2=194,128.4.

Substitute n=45, and ∑i=145yi=440.6 in the formula of mean.

That is,

y¯=440.645≈9.79.

Thus, the mean value is 9.79.

Substitute n=45, (∑i=145yi)2=194,128.4, and ∑i=145yi2=5,067.38 in the s2 formula.

That is,

s2=145−1[5,067.38−145(194,128.4)]=144[5,067.38−4,313.964]≈17.12

Thus, the value of s is calculated below:

s=s2=17.12≈4.14

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

s=35.1−5.74=7.35

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, n=25.

Consider the following table for necessary calculations.

yi	yi2
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
∑i=125yi=80.63	∑i=125yi2=500.746

Hence, (∑i=125yi)2=6,501.197.

Substitute n=25, and ∑i=125yi=80.63 in the formula of mean.

That is,

y¯=80.6325≈3.23.

Thus, the mean value is 3.23.

Substitute n=25, (∑i=125yi)2=6,501.197, and ∑i=125yi2=500.746 in the s2 formula.

That is,

s2=125−1[500.746−125(6,501.197)]=124[500.746−260.0479]≈10.03

Thus, the value of s is calculated below:

s=s2=10.13≈3.17

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

s=12.48−0.384=3.04

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, n=40.

Consider the following table for necessary calculations.

yi	yi2
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
∑i=140yi=175.48	∑i=140yi2=906.41

Hence, (∑i=140yi)2=30,793.23.

Substitute n=40, and ∑i=140yi=175.48 in the formula of mean.

That is,

y¯=175.4840≈4.39.

Thus, the mean value is 4.39.

Substitute n=40, (∑i=140yi)2=30,793.23, and ∑i=140yi2=906.41 in the s2 formula.

That is,

s2=140−1[906.41−140(30,793.23)]=139[906.41−769.83]≈3.5021

Thus, the value of s is calculated below:

s=s2=3.5021≈1.87

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

s=11.88−2.614=2.3175

Thus, it can be said that the estimated value of s is somewhat close to the actual value.

Answer 12

Ch. 1.1 - For each of the following situations, identify the...Ch. 1.2 - Are some cities more windy than others? Does...Ch. 1.2 - Of great importance to residents of central...Ch. 1.2 - The top 40 stocks on the over-the-counter (OTC)...Ch. 1.2 - Given here is the relative frequency histogram...Ch. 1.2 - The relative frequency histogram given next was...Ch. 1.2 - The self-reported heights of 105 students in a...Ch. 1.2 - An article in Archaeometry presented an analysis...Ch. 1.3 - Resting breathing rates for college-age students...Ch. 1.3 - It has been projected that the average and...

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4. Compare the results in each case to the actual calculated value of s .

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Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Answer to Problem 17E

Explanation of Solution

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Chapter 1 Solutions

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4. Compare the results in each case to the actual calculated value of s .

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Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4. Compare the results in each case to the actual calculated value of s.

Answer to Problem 17E

Explanation of Solution

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Chapter 1 Solutions

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.