Question

Consider the following frequency distribution. Class Frequency 2 up to 4 20 4 up to 6 60 6 up to 8 80 8 up to 10 20

Answer 1

Answer:

a.) Population mean, $\mu$ = 6.11

b.) Population Standard deviation, $\sigma$ = 1.67

    Population Variance, $\sigma^2$ = 2.78

Step-by-step explanation:

Class     Mid point    Frequency   $x_i \times f_i$      $(x_i - \mu)$     $f(x_i - \mu)^2$      

2 - 4                3          20                 60            -3.11          193.58

4 - 6                5          60               300            -1.11            73.93

6 - 8                7          80                560            0.889       63.23

8 - 10              9          20               180              2.89         167.04

total number of elements  = 20 + 60 + 80 + 20 = 180

Population size, N  = 180

a.)   $\sum_{i=1}^{4}$ $x_i \times f_i$ = 60 + 300 + 560 + 180 = 1100

         Population mean, $\mu$ = $\frac{1100}{180} = 6.11$

b.) $\sum_{i=1}^{4}f(x_i - \mu)^2$ = 193.58 + 73.93 + 63.23 + 167.04 = 497.78

Population Standard deviation,

             $\sigma$ = $\sqrt{\frac{\sum_{i=1}^{4}f(x_i - \mu)^2}{N-1} } = \sqrt{\frac{497.78}{(180 - 1)} } = \sqrt{2.781} = 1.667$

Population Variance, $\sigma^2$ = 2.781