Question
Determine the mean and standard deviation for the following distribution:
Marks
$2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $13$ $14$ $15$ $16$
Frequency
$1$ $6$ $6$ $8$ $8$ $2$ $2$ $3$ $0$ $2$ $1$ $0$ $0$ $0$ $1$

Answer

$\text{Marks}$
$f_i$
$f_ix_i$
$\text{d}_\text{i}=\text{x}_\text{i}-\bar{\text{x}}$
$f_id_i$
$f_id_i^2$
$2$ $1$ $2$ $-4$ $-4$ $16$
$3$ $6$ $18$ $-3$ $-18$ $54$
$4$ $6$ $24$ $-2$ $-12$ $24$
$5$ $8$ $40$ $-1$ $-8$ $8$
$6$ $8$ $40$ $-1$ $-8$ $8$
$7$ $2$ $14$ $1$ $2$ $2$
$8$ $2$ $16$ $2$ $4$ $8$
$9$ $3$ $27$ $3$ $9$ $27$
$10$ $0$ $0$ $4$ $0$ $0$
$11$ $2$ $22$ $5$ $10$ $50$
$12$ $1$ $12$ $6$ $6$ $36$
$13$ $0$ $0$ $7$ $0$ $0$
$14$ $0$ $0$ $8$ $0$ $0$
$15$ $0$ $0$ $9$ $0$ $0$
$16$ $1$ $16$ $10$ $10$ $100$
$Total$
$\sum\text{f}_\text{i}=40$
$\sum\text{f}_\text{i}\text{x}_\text{i}=239$
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-1$
$\sum\text{f}_\text{i}\text{x}^2_\text{i}=325$
$\therefore$ Mean $\bar{\text{x}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{239}{40}=5.975\approx6$
and $\sigma=\sqrt{\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2}$
$=\sqrt{\frac{325}{40}-\Big(\frac{-1}{40}\Big)^2}$
$=\sqrt{8.125-0.000625}$
$=\sqrt{8.124375}$
$=2.85$

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