Determine the mean and standard deviation for the following distribution:
$\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Marks} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text{Frequency} & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$
$\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Marks} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text{Frequency} & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$
(N/A) First,calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
$\sum f_i = 1+6+6+8+8+2+2+3+0+2+1+0+0+0+1 = 40$.
$\sum f_i x_i = (2\times1) + (3\times6) + (4\times6) + (5\times8) + (6\times8) + (7\times2) + (8\times2) + (9\times3) + (10\times0) + (11\times2) + (12\times1) + (13\times0) + (14\times0) + (15\times0) + (16\times1) = 2+18+24+40+48+14+16+27+0+22+12+0+0+0+16 = 239$.
Mean $\bar{x} = \frac{239}{40} = 5.975$.
To find the standard deviation $\sigma = \sqrt{\frac{\sum f_i x_i^2}{N} - (\bar{x})^2}$,calculate $\sum f_i x_i^2 = (4\times1) + (9\times6) + (16\times6) + (25\times8) + (36\times8) + (49\times2) + (64\times2) + (81\times3) + (100\times0) + (121\times2) + (144\times1) + (169\times0) + (196\times0) + (225\times0) + (256\times1) = 4+54+96+200+288+98+128+243+0+242+144+0+0+0+256 = 1753$.
$\sigma = \sqrt{\frac{1753}{40} - (5.975)^2} = \sqrt{43.825 - 35.700625} = \sqrt{8.124375} \approx 2.85$.
$\sum f_i = 1+6+6+8+8+2+2+3+0+2+1+0+0+0+1 = 40$.
$\sum f_i x_i = (2\times1) + (3\times6) + (4\times6) + (5\times8) + (6\times8) + (7\times2) + (8\times2) + (9\times3) + (10\times0) + (11\times2) + (12\times1) + (13\times0) + (14\times0) + (15\times0) + (16\times1) = 2+18+24+40+48+14+16+27+0+22+12+0+0+0+16 = 239$.
Mean $\bar{x} = \frac{239}{40} = 5.975$.
To find the standard deviation $\sigma = \sqrt{\frac{\sum f_i x_i^2}{N} - (\bar{x})^2}$,calculate $\sum f_i x_i^2 = (4\times1) + (9\times6) + (16\times6) + (25\times8) + (36\times8) + (49\times2) + (64\times2) + (81\times3) + (100\times0) + (121\times2) + (144\times1) + (169\times0) + (196\times0) + (225\times0) + (256\times1) = 4+54+96+200+288+98+128+243+0+242+144+0+0+0+256 = 1753$.
$\sigma = \sqrt{\frac{1753}{40} - (5.975)^2} = \sqrt{43.825 - 35.700625} = \sqrt{8.124375} \approx 2.85$.
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The following values are calculated in respect of heights and weights of the students of a section of Class $XI$:
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| Mean | $162.6 \ cm$ | $52.36 \ kg$ |
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