JAMB Mathematics 2011 Paper

Concept:Standard deviation measures how spread out the class marks are from the mean.Explanation:Use the midpoints of each class interval as the values of $x$.For $3-5$, the midpoint is $x = 4$.For $6-8$, the midpoint is $x = 7$.For $9-11$, the midpoint is $x = 10$.Each interval has frequency $f = 2$.First, find the mean $\bar{x}$ using $\bar{x} = \frac{\sum fx}{\sum f}$.$\sum fx = 2(4) + 2(7) + 2(10) = 8 + 14 + 20 = 42$$\sum f = 2 + 2 + 2 = 6$$\bar{x} = \frac{42}{6} = 7$Next, calculate the squared deviations from the mean.$(4 - 7)^2 = 9,\quad (7 - 7)^2 = 0,\quad (10 - 7)^2 = 9$Multiply each squared deviation by its frequency.$\sum f(x - \bar{x})^2 = 2(9) + 2(0) + 2(9) = 18 + 0 + 18 = 36$Now find the standard deviation.$\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}} = \sqrt{\frac{36}{6}} = \sqrt{6}$Answer:$\sqrt{6}$