 # S O C Labor force participation rates (percent employed), percent high school graduates, and mean income for males and females in ten states are reported here. Calculate the mean and the standard for both groups for each variable and describe the differences. Are males and females unequal on any of these variables? How great is the gender inequality? Labor Force Participation Rate Percentage HS Graduates Mean Income State Male Female Male Female Male Female A 65.8 54.3 81.0 81.9 55,623 50,012 B 76.7 63.0 88.4 89.7 52,345 51,556 C 71.8 57.2 82.4 84.6 55,789 48,231 D 76.1 66.6 89.5 90.9 48,907 46,289 E 75.1 63.1 86.9 88.7 62,023 58,034 F 69.9 61.1 86.3 86.4 55,000 53,897 G 73.6 59.6 87.1 87.6 49,145 47,148 H 70.5 60.3 87.0 87.6 51,897 50,659 I 66.3 55.2 81.7 84.1 51,238 45,289 J 74.5 67.1 89.1 91.6 60,746 56,489 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

#### Solutions

Chapter
Section
Chapter 4, Problem 4.7P
Textbook Problem
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## S O C Labor force participation rates (percent employed), percent high school graduates, and mean income for males and females in ten states are reported here. Calculate the mean and the standard for both groups for each variable and describe the differences. Are males and females unequal on any of these variables? How great is the gender inequality? Labor Force Participation Rate Percentage HS Graduates Mean Income State Male Female Male Female Male Female A 65.8 54.3 81.0 81.9 55,623 50,012 B 76.7 63.0 88.4 89.7 52,345 51,556 C 71.8 57.2 82.4 84.6 55,789 48,231 D 76.1 66.6 89.5 90.9 48,907 46,289 E 75.1 63.1 86.9 88.7 62,023 58,034 F 69.9 61.1 86.3 86.4 55,000 53,897 G 73.6 59.6 87.1 87.6 49,145 47,148 H 70.5 60.3 87.0 87.6 51,897 50,659 I 66.3 55.2 81.7 84.1 51,238 45,289 J 74.5 67.1 89.1 91.6 60,746 56,489

Expert Solution
To determine

To find:

The mean and standard deviation of male and female for the given variable

### Explanation of Solution

Given:

The following table shows the data of 3 different variables of 10 states.

 Labor Force Participation Rate Percentage HS Graduates Mean Income State Male Female Male Female Male Female A 65.8 54.3 81.0 81.9 55, 623 50, 012 B 76.7 63.0 88.4 89.7 52, 345 51, 556 C 71.8 57.2 82.4 84.6 55, 789 48, 231 D 76.1 66.6 89.5 90.9 48, 907 46, 289 E 75.1 63.1 86.9 88.7 62, 023 58, 034 F 69.9 61.1 86.3 86.4 55, 000 53, 897 G 73.6 59.6 87.1 87.6 49, 145 47, 148 H 70.5 60.3 87.0 87.6 51, 897 50, 659 I 66.3 55.2 81.7 84.1 51, 238 45, 289 J 74.5 67.1 89.1 91.6 60, 746 56, 489

Formula used:

Let the data values be Xi’s.

The formula to calculate mean is given by,

X¯=i=1NXiN

The formula to calculate standard deviation is given by,

s=(XiX¯)2N

Where, N is the population size.

Calculation:

Consider the labor force participation rate for males.

The size of the population is 10.

The mean is given by,

X¯=i=1NXiN

Substitute 10 for N, 65.8 for X1, 76.7 for X2 and so on in the above mentioned formula,

X¯=65.8+76.7+...+74.510=720.310=72.03 ……(1)

Consider the following table of sum of squares,

 (Xi) (Xi−X¯) (Xi−X¯)2 65.8 −6.23 38.8129 76.7 4.67 21.8089 71.8 −0.23 0.0529 76.1 4.07 16.5649 75.1 3.07 9.4249 69.9 −2.13 4.5369 73.6 1.57 2.4649 70.5 −1.53 2.3409 66.3 −5.73 32.8329 74.5 2.47 6.1009 ∑(Xi−X¯)=0 ∑(Xi−X¯)2=134.941

From equation (1), substitute 65.8 for X1 and 72.03 for X¯ in (X1X¯).

(X1X¯)=(65.872.03)(X1X¯)=6.23

Square the both sides of the equation.

(X1X¯)2=(6.23)2(X1X¯)2=38.8129

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=38.8129+21.8089+...+6.1009=134.941 ……(2)

The standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 134.941 for (XiX¯)2 and 10 for N in the above mentioned formula,

s=134.94110=13.49413.67

Thus, standard deviation for the labor force participation rate for males is 3.67.

Consider the labor force participation rate for females.

The mean is given by,

X¯=i=1NXiN

Substitute 10 for N, 54.3 for X1, 63.0 for X2 and so on in the above mentioned formula,

X¯=54.3+63+...+67.110=607.510=60.75 ……(3)

Consider the following table of sum of squares,

 (Xi) (Xi−X¯) (Xi−X¯)2 54.3 −6.45 41.6025 63 2.25 5.0625 57.2 −3.55 12.6025 66.6 5.85 34.2225 63.1 2.35 5.5225 61.1 0.35 0.1225 59.6 −1.15 1.3225 60.3 −0.45 0.2025 55.2 −5.55 30.8025 67.1 6.35 40.3225 ∑(Xi−X¯)=0 ∑(Xi−X¯)2=171.785

From equation (3), substitute 54.3 for X1 and 60.75 for X¯ in (X1X¯).

(X1X¯)=(54.360.75)(X1X¯)=6.45

Square the both sides of the equation.

(X1X¯)2=(6.45)2(X1X¯)2=41.6025

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=41.6025+5.0625+...+40.3225=171.785 ……(4)

The standard deviation is given by,

s=(XiX¯)2N

From equation (4), substitute 171.785 for (XiX¯)2 and 10 for N in the above mentioned formula,

s=171.78510=17.17854.144

Thus, standard deviation for the labor force participation rate for females is 4.144.

Consider the percentage HS graduates of males.

The mean is given by,

X¯=i=1NXiN

Substitute 10 for N, 81 for X1, 88.4 for X2 and so on in the above mentioned formula,

X¯=81+88.4+...+89.110=859.410=85.94 ……(5)

Consider the following table of sum of squares,

 (Xi) (Xi−X¯) (Xi−X¯)2 81 −4.94 24.4036 88.4 2.46 6.0516 82.4 −3.54 12.5316 89.5 3.56 12.6736 86.9 0.96 0.9216 86.3 0.36 0.1296 87.1 1.16 1.3456 87 1.06 1.1236 81.7 −4.24 17.9776 89.1 3.16 9.9856 ∑(Xi−X¯)=0 ∑(Xi−X¯)2=87.144

From equation (5), substitute 81 for X1 and 85.94 for X¯ in (X1X¯).

(X1X¯)=(8185.94)(X1X¯)=4.94

Square the both sides of the equation.

(X1X¯)2=(4.94)2(X1X¯)2=24.4036

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=24.4036+6.0516+...+9.9856=87.144 ……(6)

The standard deviation is given by,

s=(XiX¯)2N

From equation (6), substitute 87.144 for (XiX¯)2 and 10 for N in the above mentioned formula,

s=87

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