Question

1. The following are the number of hours that 10 police officers have spent being trained in how to handle encounters with people who are mentally ill:

4 17 12 9 6 10 1 5 9 3

Calculate the (a) range, (b) inter-quartile range, (c) variance, and (d) standard deviation.
(Use N)

Answer 1

Answer:

$Range = 16$

$Inter\ Quartile\ Range = 6.75$

$Variance = 20.44$

$Standard\ Deviation = 4.52$

Step-by-step explanation:

Given

4, 17, 12, 9, 6, 10, 1, 5, 9, 3

Calculating the range;

$Range = Highest - Lowest$

From the given data;

Highest = 17 and Lowest = 1

Hence;

$Range = 17 - 1$

$Range = 16$

Calculating the Inter-quartile Range

Inter quartile range (IQR) is calculates as thus

$IQR = Q_3 - Q_1$

Where

Q3 = Upper Quartile and Q1 = Lower Quartile

Start by arranging the data in ascending order

1, 3, 4, 5, 6, 9, 9, 10, 12, 17

N = Number of data; N = 10

---------------------------------------------------------------------------------

Calculating Q3

$Q_3 = \frac{3}{4}(N+1) th\ item$

Substitute 10 for N

$Q_3 = \frac{3}{4}(10+1) th\ item$

$Q_3 = \frac{3}{4}(11) th\ item$

$Q_3 = \frac{33}{4} th\ item$

$Q_3 = 8.25 th\ item$

Express 8.25 as 8 + 0.25

$Q_3 = (8 + 0.25) th\ item$

$Q_3 = 8th\ item + 0.25 th\ item$

Express 0.25 as fraction

$Q_3 = 8th\ item +\frac{1}{4} th\ item$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

From the arranged data;

$8th\ item = 10$ and $9th\ item = 12$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

$Q_3 = 10 +\frac{1}{4} (12 - 10)$

$Q_3 = 10 +\frac{1}{4} (2)$

$Q_3 = 10 +0.5$

$Q_3 = 10.5$

Calculating Q1

$Q_1 = \frac{1}{4}(N+1) th\ item$

Substitute 10 for N

$Q_1 = \frac{1}{4}(10+1) th\ item$

$Q_1 = \frac{1}{4}(11) th\ item$

$Q_1 = \frac{11}{4} th\ item$

$Q_1 = 2.75 th\ item$

Express 2.75 as 2 + 0.75

$Q_1 = (2 + 0.75) th\ item$

$Q_1 = 2nd\ item + 0.75 th\ item$

Express 0.75 as fraction

$Q_1 = 2nd\ item +\frac{3}{4} th\ item$

$Q_1 = 2nd\ item +\frac{3}{4} (3rd\ item - 2nd\ item)$

From the arranged data;

$2nd\ item = 3$ and $3rd\ item = 4$

$Q_1 = 3 +\frac{3}{4} (4 - 3)$

$Q_1 = 3 +\frac{3}{4} (1)$

$Q_1 = 3 +0.75$

$Q_1 = 3 .75$

---------------------------------------------------------------------------------

Recall that

$IQR = Q_3 - Q_1$

$IQR = 10.5 - 3.75$

$IQR = 6.75$

Calculating Variance

Start by calculating the mean

$Mean = \frac{1+3+4+5+6+9+9+10+12+17}{10}$

$Mean = \frac{76}{10}$

$Mean = 7.6$

Subtract the mean from each data, then square the result

$(1 - 7.6)^2 = (-6.6)^2 = 43.56$

$(3 - 7.6)^2 = (-4.6)^2 = 21.16$

$(4 - 7.6)^2 = (-3.6)^2 = 12.96$

$(5 - 7.6)^2 = (-2.6)^2 = 6.76$

$(6 - 7.6)^2 = (-1.6)^2 = 2.56$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(10 - 7.6)^2 = (2.4)^2 = 5.76$

$(12 - 7.6)^2 = (4.4)^2 = 19.36$

$(17 - 7.6)^2 = (9.4)^2 = 88.36$

Sum the result

$43.56 + 21.16 + 12.96 + 6.76 + 2.56 + 1.96 + 1.96 + 5.76 + 19.36 + 88.36 = 204.4$

Divide by number of observation;

$Variance = \frac{204.4}{10}$

$Variance = 20.44$

Calculating Standard Deviation (SD)

$SD = \sqrt{Variance}$

$SD = \sqrt{20.44}$

$SD = 4.52$ (Approximated)