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A survey regarding the heights (in cm) of $50$ girls of class X of a school was conducted and the following data was obtained:
Find the mean and mode of the above data.
| Height (in cm) | Number of girls |
|---|---|
| 120-130 | 2 |
| 130-140 | 8 |
| 140-150 | 12 |
| 150-160 | 20 |
| 160-170 | 8 |
| Total | 50 |
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Height (in cm) | No. of girls | $x_i$ | $u_i$ | $f_i u_i$
120-130 | 2 | 125 | -2 | -4
130-140 | 8 | 135 | -1 | -8
140-150 | 12 | $145 = a$ | 0 | 0
150-160 | 20 | 155 | 1 | 20
160-170 | 8 | 165 | 2 | 16
Total | 50 | | | 24
Correct table ($1\frac{1}{2}$ Marks)
Mean $$\begin{aligned}& = 145 + \frac{24}{50} \times 10 \\ & = 149.8\end{aligned}$$ (1 Mark)
$\therefore$ mean height is $149.8$ cm (1/2 Mark)
Modal class is $150 - 160$ (1/2 Mark)
Mode $$\begin{aligned}& = 150 + \frac{(20-12)}{(2 \times 20 - 12 - 8)} \times 10 \\ & = 154\end{aligned}$$ (1 Mark)
$\therefore$ modal height is $154$ cm (1/2 Mark)
120-130 | 2 | 125 | -2 | -4
130-140 | 8 | 135 | -1 | -8
140-150 | 12 | $145 = a$ | 0 | 0
150-160 | 20 | 155 | 1 | 20
160-170 | 8 | 165 | 2 | 16
Total | 50 | | | 24
Correct table ($1\frac{1}{2}$ Marks)
Mean $$\begin{aligned}& = 145 + \frac{24}{50} \times 10 \\ & = 149.8\end{aligned}$$ (1 Mark)
$\therefore$ mean height is $149.8$ cm (1/2 Mark)
Modal class is $150 - 160$ (1/2 Mark)
Mode $$\begin{aligned}& = 150 + \frac{(20-12)}{(2 \times 20 - 12 - 8)} \times 10 \\ & = 154\end{aligned}$$ (1 Mark)
$\therefore$ modal height is $154$ cm (1/2 Mark)