Question

course: probability and statistics
consider the data below, calculate the mean and standard for grouped data.
1 3 1 1 1
1 2 2 1 0
1 1 0 0 0
1 5 0 3 6
3 0 3 1 1
3 6 6 1 1
the table is from the data

x	f
0	5
1	13
2	2
3	5
5	1
6	3

Explanation:

Step1: Calculate the mid - points

Let the values be \(x_i\) and frequencies be \(f_i\). The mid - points are already given as \(x = 0,1,2,3,5,6\) with frequencies \(f = 5,13,2,5,1,3\) respectively.

Step2: Calculate the mean \(\bar{x}\)

The formula for the mean of grouped data is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_if_i}{\sum_{i = 1}^{n}f_i}\).
First, calculate \(\sum_{i = 1}^{n}x_if_i\):
\[

$$\begin{align*} \sum_{i = 1}^{n}x_if_i&=(0\times5)+(1\times13)+(2\times2)+(3\times5)+(5\times1)+(6\times3)\\ &=0 + 13+4 + 15+5+18\\ &=55 \end{align*}$$

\]
\(\sum_{i = 1}^{n}f_i=5 + 13+2+5+1+3=29\).
So, \(\bar{x}=\frac{55}{29}\approx1.90\).

Step3: Calculate \((x_i-\bar{x})^2f_i\) for each \(i\)

For \(x_1 = 0\): \((0 - 1.90)^2\times5=( - 1.90)^2\times5 = 3.61\times5=18.05\).
For \(x_2 = 1\): \((1 - 1.90)^2\times13=( - 0.90)^2\times13 = 0.81\times13 = 10.53\).
For \(x_3 = 2\): \((2 - 1.90)^2\times2=(0.10)^2\times2 = 0.01\times2=0.02\).
For \(x_4 = 3\): \((3 - 1.90)^2\times5=(1.10)^2\times5 = 1.21\times5 = 6.05\).
For \(x_5 = 5\): \((5 - 1.90)^2\times1=(3.10)^2\times1 = 9.61\times1=9.61\).
For \(x_6 = 6\): \((6 - 1.90)^2\times3=(4.10)^2\times3 = 16.81\times3 = 50.43\).
\(\sum_{i = 1}^{n}(x_i-\bar{x})^2f_i=18.05+10.53+0.02+6.05+9.61+50.43 = 94.69\).

Step4: Calculate the standard deviation \(s\)

The formula for the standard deviation of grouped data is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2f_i}{\sum_{i = 1}^{n}f_i - 1}}\).
\(s=\sqrt{\frac{94.69}{29 - 1}}=\sqrt{\frac{94.69}{28}}\approx\sqrt{3.38}\approx1.84\).

Answer:

Mean is approximately \(1.90\) and standard deviation is approximately \(1.84\).