Respuesta :
Answer:
[tex]Mean = 68.9[/tex]
[tex]s^2 =18.1[/tex] --- Variance
Step-by-step explanation:
Given
[tex]\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}[/tex]
Solving (a): Calculate the mean.
The given data is a grouped data. So, first we calculate the class midpoint (x)
For 51 - 58.
[tex]x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5[/tex]
For 59 - 66
[tex]x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5[/tex]
For 67 - 74
[tex]x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5[/tex]
For 75 - 82
[tex]x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5[/tex]
For 83 - 90
[tex]x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5[/tex]
So, the table becomes:
[tex]\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}[/tex]
The mean is then calculated as:
[tex]Mean = \frac{\sum fx}{\sum f}[/tex]
[tex]Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}[/tex]
[tex]Mean = \frac{2547.5}{37}[/tex]
[tex]Mean = 68.9[/tex] -- approximated
Solving (b): The sample variance:
This is calculated as:
[tex]s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}[/tex]
So, we have:
[tex]s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}[/tex]
[tex]s^2 =\frac{652.8}{36}[/tex]
[tex]s^2 =18.1[/tex] -- approximated