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Lesson Topics Discussion Quiz: Class Homework

Example1

Title: Exploring Deviation & Percentage

Grade Lesson s6-l5

Explanation: The best way to understand SAT-4 is by looking at some examples. Take turns and read each example for easy understanding.

Examples

Topics → Definition Example1 Example2 Example3

Consider the following set of data representing the ages of students in a class: {18, 20, 22, 19, 21, 23, and 20}.

Step: 1

To find the standard deviation Calculate the mean (average) age:

\$Mean = (18 + 20 + 22 + 19 + 21 + 23 + 20) / 7\$ = \$(143) / 7\$ ≈ 20.43

Squared deviation: \$(18 − 20.43) ^ 2\$ = 4.1849
\$(20 − 20.43) ^ 2\$ = 0.1849
\$(22 − 20.43) ^ 2\$ = 2.4649
\$(19 − 20.43) ^ 2\$ = 2.0449
\$(21 − 20.43) ^ 2\$ = 0.3364
\$(23 − 20.43) ^ 2\$ = 6.7249
\$(20 − 20.43) ^ 2\$ = 0.1849

Explanation:

Here, we find the squared deviation of each age from the mean. The mean age is 20.43.

Step: 2

\$"Mean of Squared Deviations" = (4.1849 + 0.1849 + 2.4649 + 2.0449 + 0.3364 + 6.7249 + 0.1849) / 7\$ = \$(16.1257) / 7\$ = 2.304

Take the square root of the mean of squared deviations to find the standard deviation:

\$"Standard Deviation" = \sqrt(2.304)\$ = 1.52.

Explanation:

Here, we calculate the mean of these squared deviations. Therefore, the standard deviation of the ages is approximately 1.52.