Question

Calculate the mean, variance, and standard deviation for the following distribution:
$\begin{array}{cccccccc}\text{Class}& 30-40& 40-50& 50-60& 60-70& 70-80& 80-90& 90-100\\ \text{Frequency}& 3& 7& 12& 15& 8& 3& 2\end{array}$

Answer 1

Step 1: Finding mean
$\begin{array}{cccc}30-40& 3& \frac{30+40}{2}=35& 35\times 3=105\\ 40-50& 7& \frac{40+50}{2}=45& 45\times 7=315\\ 50-60& 12& \frac{50+60}{2}=55& 55\times 12=660\\ 60-70& 15& \frac{60+70}{2}=65& 65\times 15=975\\ 70-80& 8& \frac{70+80}{2}=75& 75\times 8=600\\ 80-90& 3& \frac{80+90}{2}=85& 85\times 3=255\\ 90-100& 2& \frac{90+100}{2}=95& 95\times 2=190\\ & \sum f_i=50& & \sum f_i{x}_i=3100\end{array}$
Mean $\overline{x}=\frac{\sum x_i{f}_i}{\sum f_i}$
$\Rightarrow \overline{x}=\frac{3100}{50}$
$\Rightarrow \overline{x}=62$

Step 2: Finding variance and standard deviation


$\begin{array}{cccc}\text{Frequency}& \text{Mid - point}& (x_i-\overline{x}{)}^2& f_i(x_i-\overline{x}{)}^2\\ 3& 35& (35-62{)}^2=(27{)}^2=729& 3\times 729=2187\\ 7& 45& (45-62{)}^2=(17{)}^2=289& 7\times 289=2023\\ 12& 55& (55-62{)}^2=(7{)}^2=49& 12\times 49=588\\ 15& 65& (65-62{)}^2=3^2=9& 15\times 9=135\\ 8& 75& (75-62{)}^2=(13{)}^2=169& 8\times 169=1352\\ 3& 85& (85-62{)}^2=(23{)}^2=529& 3\times 529=1587\\ 2& 95& (95-62{)}^2=(33{)}^2=1089& 2\times 1089=2178\\ \sum f_i=50& & \sum f_i(x_i-\overline{x}{)}^2=10050& \end{array}$
Variance $\left({\sigma }^2\right)=\frac{1}{N}\sum f_i(x_i-\overline{x}{)}^2$
$=\frac{1}{50}\times 10050[\because N=\sum f_i=50]$
$=201$
Standard deviation $\left(\sigma \right)=\sqrt{201}=14.17$

hence, the mean is $62$, variance is $201$ and the standard deviation is $14.17$