Home » » $$\begin{array}{c|c} class& 1 - 3 & 4 - 6 & 7 - 9\\ \hline Frequency & 5 & 8 & 5\end{array}$$Find the standard deviation of the data using the table above...

# $$\begin{array}{c|c} class& 1 - 3 & 4 - 6 & 7 - 9\\ \hline Frequency & 5 & 8 & 5\end{array}$$Find the standard deviation of the data using the table above...

### Question

$$\begin{array}{c|c} class& 1 - 3 & 4 - 6 & 7 - 9\\ \hline Frequency & 5 & 8 & 5\end{array}$$

Find the standard deviation of the data using the table above

### Options

A) 5

B) $$\sqrt{6}\0 C) \(\frac{5}{3}$$

D) $$\sqrt{5}$$

The correct answer is A.

### Explanation:

$$\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 10 & 10 & -3 & 9 & 90\\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \\ \hline & 18 & & 90 & & & 450 \end{array}$$
x = $$\frac{\sum fx}{\sum f}$$
= $$\frac{90}{18}$$
= 5
S.D = $$\frac{\sum f(x - x)^2}{\sum f}$$
= $$\frac{450}{18}$$
= $$\sqrt{25}$$
= 5

## Dicussion (1)

• $$\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 10 & 10 & -3 & 9 & 90\\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \\ \hline & 18 & & 90 & & & 450 \end{array}$$
x = $$\frac{\sum fx}{\sum f}$$
= $$\frac{90}{18}$$
= 5
S.D = $$\frac{\sum f(x - x)^2}{\sum f}$$
= $$\frac{450}{18}$$
= $$\sqrt{25}$$
= 5

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