Question
Find the mean and standard deviation for the following data:
Class interval $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$
Frequency $3$ $2$ $4$ $6$ $5$ $5$ $5$ $2$ $8$ $5$

Answer

We make the table from the given data:
Class marks Mid value $\left(x_i\right)$ $d _{ i }= x _{ i }- a = x _{ i }=45$ $f _{ i }$ $f _{ i } d _{ i }$ $d_i$ $f _{ i } d_i$
$0-10$ $5$ $-40$ $3$ $-120$ $1600$ $4800$
$10-20$ $15$ $-30$ $2$ $-60$ $900$ $1800$
$20-30$ $25$ $-20$ $4$ $-80$ $400$ $1600$
$30-40$ $35$ $-10$ $6$ $-60$ $100$ $800$
$40-50$ $45$ $0$ $5$ $0$ $0$ $0$
$50-60$ $55$ $10$ $5$ $50$ $100$ $500$
$60-70$ $65$ $20$ $5$ $100$ $400$ $2000$
$70-80$ $75$ $30$ $2$ $60$ $900$ $1800$
$80-90$ $85$ $40$ $8$ $320$ $1600$ $12800$
$90-100$ $95$ $50$ $5$ $250$ $2500$ $12500$
      $\sum f _{ i }=45$ $\sum f _{ i } d _{ i }=460$   $\sum f _{ i } d_i=38400$
Let $a = 45.$
$\therefore$ Mean $=a+\frac{\sum f_i d_i}{\sum f_i}$
$=45+\frac{460}{45}$
$=45+10.22$
$=55.22$
$\therefore$ Standard deviation $=\sqrt{\frac{\sum f_i d_i^2}{\sum f_i}-\left(\frac{\sum f_i d_i}{\sum f_i}\right)^2}$
$=\sqrt{\frac{38400}{45}-(10.22)^2}$
$=\sqrt{853.33-104.45}$
$=\sqrt{748.88}$
$=27.36$

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