Ch 3 2 measures of variation

The salaries for the staff of the XYZ Manufacturing Co. are shown here. Find the range.

- Find the ranges for the paints.

Solution W.P(1) : The range is: R = highest value - lowest value = $100,000 - $15,000 = $85,000

The range is: R = highest value - lowest value = 10 – 5 = 5

The range is: R = upper limit of the last class – lower limit of the first class = 14 – 5 = 9

Or The range is: R = 14.5 - 4.5 = 10

Range

W.P(1) : 3-2 Example (3-20/p-125) : (Employee Salaries) Crude Data

Staff SalaryOwner $100,000

Manager 40,000Sales representative 30,000

Workers 25,00015,00018,000

W.P(2) : 3-2 Example Grouped Data (Without Class Interval)

Data (x) Frequency (f) f.x

5 1 56 3 18

7 4 288 8 649 3 2710 5 50

Total = 24 192

W.P(3) : 3-2 Example Grouped Data (With Class Interval)

Class x Frequency (f) f.x5 - 6 5.5 2 117 - 8 7.5 5 37.59 - 10 9.5 8 7611 - 12 11.5 4 4613 - 14 13.5 1 13.5

Total = 20 184

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ROYAL COMMISSION YANBU COLLEGES AND INSTITUTES YANBU INDUSTRIAL COLLEGE

DEPARTMENT OF GENERAL STUDIES SEMESTER I, 1437-38 (H) / 2016-17 (G)

Exercises # Ch3-2 Measures of Variation

Student’s ID: _______________ Student’s Name: ______________________________________________

Course: STAT-201 Section: Date: 24 – 10 – 2016

Instructor’s Name: Mr. Osama Alwusaidi Instructor’s Signature:

The formula for the population variance is:

�

The formula for the variance & standard deviation for data obtained from samples are as follows:

�

Shortcut or Computational Formulas for the S & S 2:

�

Find the sample variance and standard deviation for the amount of European auto sales for a sample of 6 years shown. The data are in millions of dollars.

Solution W.P(4) :

Step 1: !

Step 2:

Step 3: !

- The variance is 1.28 rounded.

- Hence, the sample standard deviation is 1.13.

VarianceandStandardDeviationforUngroupedData

Formula and Shortcut or Computational Formulas Ungrouped Data

σ 2 =X −µ( )2∑N

⇒ σ = σ 2 =X −µ( )2∑N

s2 =X − X( )2∑n−1

⇒ s = s2 =X − X( )2∑n−1

s2 =n X 2∑( )− X∑( )

2

n n−1( ) ⇒ s = s2 =

n X 2∑( )− X∑( )2

n n−1( )

W.P(4) : 3-2 Example (3-23/p-129) : (European Auto Sales)

11.2 11.9 12.0 12.8 13.4 14.3* Source: USA TODAY.

Xii=1

n

∑ = 75.6 ⇒ X =Xi

i=1

n

∑n

=75.6

6=12.6

(xi − x ) (xi − x )2

xi − x( )2i=1

n

∑ =

Data (xi)

11.2 -1.4 1.96

11.9 -0.7 0.49

12.0 -0.6 0.36

12.8 0.2 0.04

13.4 0.8 0.64

14.3 1.7 2.89

75.6 6.4

S2 =(Xi − X)2

i=1

n

∑n−1

=6.46−1

=1.28 ⇒ S = S2 = 1.28 =1.13

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Solution W.P(4) : (use the shortcut formula)Step 1:

Step 2: !

- The variance is: 1.28 rounded.

- Hence, the sample standard deviation is: 1.13.

Using the frequency distribution for Example 2–7, find the mean. The data represent the number of miles run during one week for a sample of 20 runners.

Solution W.P(5) : (use the shortcut formula)

Step 2:

!



Xi2Data (Xi)

11.2 125.411.9 141.612.0 144.012.8 163.813.4 179.614.3 204.575.6 958.9

s2 =n X 2∑( )− X∑( )

2

n n−1( )=

6 958.9( )− 75.6( )2

6 6−1( )=1.28 ⇒ s = s2 = 1.28 =1.13

VarianceandStandardDeviationforGroupedData

Formula and Shortcut or Computational Formulas Grouped Data (With Class Interval)

W.P(5) : 3-2 Example (3-24/p-129) : (Miles Run per Week)

Class limitsClass

boundaries Frequency (f) Midpoints (Xm) f . Xm f . X2m

6 - 10 5.5 - 10.5 1 8 8 6411 - 15 10.5 - 15.5 2 13 26 338

16 - 20 15.5 - 20.5 3 18 54 972

21 - 25 20.5 - 25.5 5 23 115 2645

26 - 30 25.5 - 30.5 4 28 112 3136

31 - 35 30.5 - 35.5 3 33 99 3267

36 - 40 35.5 - 40.5 2 38 76 2888Total = 20 161 490 13310

s2 =n f ⋅Xm

2∑( )− f ⋅Xm∑( )2n n −1( ) =

20 13,310( )− 490( )220 20 −1( ) ≈ 68.7 ⇒ s = s2 = 68.7 ≈ 8.3

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The coefficient of variation, denoted by CVar, The formula for CVar is:

� �

The mean of the number of sales of cars over a 3-month period is 87, and the standard deviation is 5. The mean of the commissions is $5225, and the standard deviation is $773. Compare the variations of the two.

Solution W.P(6) :

The coefficients of variation are:

!

Since the coefficient of variation is larger for commissions, the commissions are more variable than the sales.

CoefficientofVariation(C.V)

For samples,

CVar = sX⋅100

For populations,

CVar = σµ⋅100

W.P(6) : 3-2 Example (3-25/p-132) : (Sales of Automobiles)

s1 =5 s2 =773X1 = 87 X2 = 5225

sales commissions

CVar = 587

⋅ 100 = 5.7% sales

CVar = 7735225

⋅ 100 =14.8% commissions

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The number of yards gained in NFL playoff games by rookie quarterbacks is shown. a

- find the range, variance, and standard deviation and use the shortcut formula for the unbiased estimator to compute the variance and standard deviation.


!


- Hence, the sample standard deviation is: 44.55 rounded.

Exercises3-2

W.P(7) : 3-2 Exercises (15/p-129) : (Football Playoff Statistics)

193 66 136 140 157

163 181 226 135 199

xi2Data (xi)

193 37249

66 4356

136 18496

140 19600

157 24649

163 26569

181 32761

226 51076

135 18225

199 39601

1596 272582

s2 =n X 2∑( )− X∑( )

2

n n−1( )=

10 272582( )− 1596( )2

10 9( )≈ 1984.49 ⇒ s = s2 = 1984.49 ≈ 44.55

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In a study of reaction times to a specific stimulus, a psychologist recorded these data (in seconds).


Step 2:

!



W.P(8) : 3-2 Exercises (22/p-139) : (Reaction Times)

Class limits Class boundaries

Frequency (f) Midpoints (Xm) f . Xm f . X2m

2.1 - 2.7 2.05 - 2.75 12 2.4 28.8 69.12

2.8 - 3.4 2.75 - 3.45 13 3.1 40.3 124.93

3.5 - 4.1 3.45 - 4.15 7 3.8 26.6 101.08

4.2 - 4.8 4.15 - 4.85 5 4.5 22.5 101.25

4.9 - 5.5 4.85 - 5.55 2 5.2 10.4 54.08

5.6 - 6.2 5.55 - 6.25 1 5.9 5.9 34.81

Total = 40 134.5 485.27

s2 =n f ⋅ Xm

2∑( )− f ⋅ Xm∑( )2

n n−1( )=

40 485.27( )− 134.5( )2

40 40−1( )= 0.847 ⇒ s = s2 = 0.847 = 0.92

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The following sample data reflect electricity bills for ten households in San Diego in March.

Compute the range, variance, and standard deviation for these sample data. Discuss which of these three measures you would prefer to use as a measure of variation.

Solution W.P(9) :

Step 1: !

Step 2:

Step 3: !

Solution W.P(9) : (use the shortcut formula) Step 1:

Step 2:

!


$118.20 $67.88 $133.40 $88.42 $110.34 $76.90 $144.56 $127.89 $89.34 $129.10

Xii=1

n

∑ = 1086.03 ⇒ X =X∑n

= 1086.0310

= 108.603

(xi − x ) (xi − x )2

xi − x( )2i=1

n

∑ =

Data (xi)

$118.20 $9.60 $92.10$67.88 -$40.72 $1658.36

$133.40 $24.80 $614.89$88.42 -$20.18 $407.35$110.34 $1.74 $3.02$76.90 -$31.70 $1005.08$144.56 $35.96 $1292.91$127.89 $19.29 $371.99$89.34 -$19.26 $371.06$129.10 $20.50 $420.13

$1086.03 $6236.89

S2 =(xi − x )2

i=1

n

∑n −1

= 6236.8910 −1

= 692.988 ⇒ S = S2 = 3.623 = 1.903

xi2Data (xi)

$118.20 $13971.24$67.88 $4607.69$133.40 $17795.56$88.42 $7818.10$110.34 $12174.92$76.90 $5913.61

$144.56 $20897.59$127.89 $16355.85$89.34 $7981.64

$129.10 $16666.81

$1086.03 $124183.01

s2 =n X 2∑( )− X∑( )2

n n −1( ) =10 124183.01( )− 1086.03( )2

10 9( ) ≈ 692.988 ⇒ s = s2 = 692.988 ≈1.903

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Ages of Accountants The average age of the accountants at Three Rivers Corp. is 26 years, with a standard deviation of 6 years; the average salary of the accountants is $31,000, with a standard deviation of $4000. Compare the variations of age and income. In a study of reaction times to a specific stimulus, a psychologist recorded these data (in seconds).

Solution W.P(10) :

The coefficients of variation are:

!

Since the coefficient of variation is larger for Age, the Age are more variable than the salary of the accountants.


s1 =6 s2 = $4000X1 = 26 X 2 = $31,000

Age Salary

CVar = 626

⋅ 100 = 23.08 % years

CVar = 400031,000

⋅ 100 = 12.9 % $

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Mr. Osama A. Alwusaidi | Yanbu University College | GS Dept. | Saudi Arabia | instructor(B.Ed.), Mathematics. (Ed.M.), Testing, Measurement and Statistics (Psychology)

Mobile: +966-544115001 | Phone: +966-43932961 | Cisco:1574 | Fax: +966-43925394Email: [email protected] | Site: www.rcyci.edu.sa | Address: P.O.Box 31387

Yanbu Industrial City 41912 Saudi Arabia

Education

Ch 3 2 measures of variation