# The range and standard deviation of the given scores

### 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 4, Problem 4.1P
To determine

## To find:The range and standard deviation of the given scores

Expert Solution

Solution:

The range and standard deviation of the given scores is 40 and 12.28 respectively.

### Explanation of Solution

The given scores are,

10,12,15,20,25,30,32,35,40,50

Formula used:

The formula to calculate the range is,

Range=High ScoreLow score

Let the data values be Xi’s.

The formula to calculate the mean is,

X¯=i=1NXiN

The formula to calculate the standard deviation is,

s=(XiX¯)2N

Where, N is the population size.

Calculation:

Arrange the given data in increasing order.

The data in increasing order is given by,

10,12,15,20,25,30,32,35,40,50

The highest score is 50 and the lowest score is 10.

The range is given by,

Range=High ScoreLow score

Substitute 50 for high score and 10 for low score in the above mentioned formula,

Range=5010=40

The range of population is 40.

The size of the population is 10.

The mean is given by,

X¯=i=1NXiN

Substitute 10 for N, 10 for X1, 12 for X2 and so on in the above mentioned formula,

X¯=10+12+15+20+25+30+32+35+40+5010=26910=26.9 ……(1)

Consider the following table of sum of squares,

 Scores (Xi) (Xi−X¯) (Xi−X¯)2 10 −16.9 285.61 12 −14.9 222.01 15 −11.9 141.61 20 −6.9 47.61 25 −1.9 3.61 30 3.1 9.61 32 5.1 26.01 35 8.1 65.61 40 13.1 171.61 50 23.1 533.61 ∑(Xi−X¯)=0 ∑(Xi−X¯)2=1506.9

From equation (1), substitute 10 for X1 and 26.9 for X¯ in (X1X¯).

(X1X¯)=(1026.9)(X1X¯)=16.9

Square both sides of the equation.

(X1X¯)2=(16.9)2(X1X¯)2=285.61

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

(XiX¯)2=285.61+222.01+..................+171.61+533.61=1506.9 ……(2)

The formula to calculate the standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 1506.9 for (XiX¯)2 and 10 for N in the above mentioned formula,

s=1506.910=150.6912.28

Thus, range is 40 and standard deviation is 12.28.

Conclusion:

Therefore, the range and standard deviation of the given scores is 40 and 12.28 respectively.

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