Question 1: Find the standard deviation for the following distribution:  

\displaystyle \begin{array}  {|c | c| c| c| c | c | c | c |} \hline    x: & 4.5 & 14.5 & 24.5 & 34.5 & 44.5 & 54.5 & 64.5 \\ \hline   f: & 1 & 5 & 12 & 22 & 17 & 9 & 4 \\ \hline \end{array}

Answer:

Median value of x is 34.5

Calculation of Variance and Standard Deviation

\displaystyle \begin{array}  {|c | c| c| c| c | c | c | } \hline x_i   &  f_i  &  d_i   &  u_i    & f_iu_i   & {u_i}^2   & f_i {u_i}^2 \\ & & = x_i - 34.5 & = \frac{x_i - 34.5}{10} &  &  &  \\ \hline 4.5 & 1 & -30 & -3 & -3 & 9 & 9 \\ \hline 14.5 & 5 & -20 & -2 & -10 & 4 & 20 \\ \hline 24.5 & 12 & -10 & -1& -12 & 1 & 12 \\ \hline 34.5 & 22 & 0  & 0 & 0 & 0 & 0 \\ \hline 44.5 & 17 & 10 & 1 & 17 & 1 & 17 \\ \hline 54.5 & 9 & 20 & 2 & 18 & 4 & 36 \\ \hline 64.5 & 4 & 30 & 3 & 12  & 9   & 36  \\ \hline & N = \sum f_i = 70 & & & \sum f_i u_i = 22 & & \sum f_i {u_i}^2 = 130 \\ \hline \end{array}

\displaystyle \text{Here } N = 70, \hspace{1.0cm} \sum f_iu_i = 22, \hspace{1.0cm} \sum f_i {u_i}^2 = 130, \hspace{1.0cm} h = 10

\displaystyle Var(X) = h^2 \Bigg[  \Bigg(  \frac{1}{N}  \sum f_i {u_i}^2 \Bigg) -  \Bigg(  \frac{1}{N} \sum f_iu_i  \Bigg)^2  \Bigg]

\displaystyle \Rightarrow Var(X) = 10^2 \Bigg[  \Bigg(  \frac{1}{70}  \times 130 \Bigg) -  \Bigg(  \frac{1}{70} \times 22  \Bigg)^2  \Bigg]  \\ \\ \\ { \hspace{2.0cm} = 100 \Bigg[ \frac{13}{7} - \frac{121}{1225}   \Bigg] = 100( 1.857 -  0.098) = 175.822}

\displaystyle \text{Hence } S.D. = \sqrt{Var(X)} = \sqrt{175.822} = 13.259

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Question 2: Table below shows the frequency f with which x alpha particles were radiated from a diskette.

\displaystyle \begin{array}  {|c | c| c| c| c | c | c | c |c | c| c| c| c | c | } \hline    x: & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 &  8 & 9 &  10 & 11 & 12 \\ \hline   f: & 51 & 203 & 383 & 525 & 532 & 408 & 273  & 139 & 43 & 27 &  10 & 4 &  2 \\ \hline \end{array}

Calculate the mean and variance.

Answer:

\displaystyle \text{Mean } \overline{X} = \frac{\sum f_ix_i}{\sum f_i}  = \frac{10078}{2600} = 3.88

Calculation of Variance and Standard Deviation

\displaystyle \begin{array}  {|c | c| c| c| c | c | c| } \hline x_i   &  f_i  & f_i x_i  & d_i = x_i - \overline{X}   &  {d_i}^2    & f_id_i   &  f_i {d_i}^2 \\  \hline 0 & 51 & 0 & -3.88 & 15.05 & -197.88 & 767.55 \\ \hline 1 & 203 & 203  & -2.88 &  8.29 & -584.64 & 1682.87 \\ \hline 2 & 383 & 766  &-1.88 & 3.53 & -720.04 & 1351.99 \\ \hline 3 & 525 & 1575  & -0.88 & 0.77 & -462 & 404.25 \\ \hline 4 & 532 & 2128 & 0.12 & 0.014 & 63.84 & 7.448 \\ \hline 5 & 408 & 2040 & 1.12  & 1.25 & 456.96 & 510 \\ \hline 6 & 273 & 1638 & 2.12  & 4.49 & 578.76 & 1225.77 \\ \hline 7 & 139 & 973  & 3.12 & 9.73 &  433.68 & 1352.47 \\ \hline 8 & 43 & 344  & 4.12  & 16.97 & 177.16 & 729.71 \\ \hline 9 & 27 & 243  & 5.12  & 26.21 & 138.24 & 707.67 \\ \hline 10 & 10 & 100  & 6.12 & 37.45 & 61.2 & 374.5 \\ \hline 11 & 4 & 44 & 7.12  &  50.69 &  28.48 & 202.76 \\ \hline 12 &  2 & 24 &  8.12 & 65.93 &  16.24 & 131.86 \\ \hline & N = \sum f_i &  \sum f_ix_i  & &   & \sum f_i d_i & \sum f_i {d_i}^2  \\ &  = 2600  & = 10078 &   &   & = -10 & = 9448.848 \\ \hline \end{array}

\displaystyle \text{Here } N = 2600, \hspace{1.0cm} \sum f_id_i = -10, \hspace{1.0cm} \sum f_i {d_i}^2 = 9448.848

\displaystyle Var(X) = \Bigg(  \frac{1}{N}  \sum f_i {d_i}^2 \Bigg) -  \Bigg(  \frac{1}{N} \sum f_id_i  \Bigg)^2   

\displaystyle \Rightarrow Var(X) = \frac{9448.848}{2600} - \Bigg( \frac{-10}{2600} \Bigg)^2 = 3.6341 - 0.000015  = 3.63415

\displaystyle \text{Hence } S.D. = \sqrt{Var(X)} = \sqrt{3.63415} = 1.9063

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Question 3: Find the mean, and standard deviation for the following data:

\displaystyle \text{(i) } \begin{array}  {|l | c| c| c| c | c | c| } \hline    \text{Year render}: & 10 & 20 & 30 & 40 & 50 & 60  \\ \hline   \text{Number of persons ( cumulative)}: & 15 & 32 & 51 & 78 & 97 & 109 \\ \hline \end{array}

\displaystyle \text{(ii) } \begin{array}  {|c | 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}

Answer:

(i)

Computation of mean and standard deviation

\displaystyle  \begin{array}  {|l | c| c| c| c | c |  } \hline x_i & \text{Cumulative Frequency} & f_i & x_i f_i & {x_i}^2 f_i  \\ \hline 10 & 15 & 15 & 150 &  1500 \\ \hline 20 & 32 & 17 & 340 &  6800 \\ \hline 30 & 51 & 19 & 570 & 17100 \\ \hline 40 & 78 & 27 & 1080 & 43200  \\ \hline 50 & 97 & 19 & 950 & 47500  \\ \hline 60 & 109 & 12 & 720 & 43200 \\ \hline & & \sum f_i = 109  & \sum f_i x_i = 3810 & \sum f_i {x_i}^2 = 159300 \\ \hline \end{array}

\displaystyle  \text{Mean } = \frac{\sum f_i x_i }{N} = \frac{3810}{109} = 34.95

\displaystyle Var(X) = \Bigg(  \frac{1}{N}  \sum f_i {x_i}^2 \Bigg) -  \Bigg(  \frac{1}{N} \sum f_ix_i  \Bigg)^2  \\ \\ \\ { \hspace{2.0cm} = \frac{159300}{109} -  \Bigg( \frac{3810}{109} \Bigg)^2 = 1461.46789 - 1221.7910 = 239.67} 

\displaystyle \text{Hence } S.D. = \sqrt{Var(X)} = \sqrt{239.67} = 15.4815

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(ii)

Computation of mean and standard deviation

\displaystyle  \begin{array}  {|l | c| c| c| c | c |  } \hline x_i &   f_i &   x_i f_i &  {x_i}^2 f_i  \\ \hline 2 &  1 & 2  & 4   \\ \hline 3 &  6 & 18  & 54   \\ \hline 4 & 6 & 24  &  96  \\ \hline 5 &  8  & 40  & 200   \\ \hline 6 &  8  & 48  &  288  \\ \hline 7 &  2  & 14  &  98  \\ \hline 8 &  2  & 16  &  128  \\ \hline 9 &  3  &  27 & 243   \\ \hline 10 & 0  & 0  &  0  \\ \hline 11 & 2  & 22  &  242  \\ \hline 12 & 1  & 12  &  144  \\ \hline 13 &  0 &  0 &  0  \\ \hline 14 & 0  & 0  & 0   \\ \hline 15 & 0  &  0 &  0  \\ \hline 16 &  1 & 16  &  256  \\ \hline & N = \sum f_i = 40  & \sum f_i x_i = 239 & \sum f_i {x_i}^2 = 1753 \\ \hline \end{array}

\displaystyle  \text{Mean } = \frac{\sum f_i x_i }{N} = \frac{239}{40} = 5.975

\displaystyle Var(X) = \Bigg(  \frac{1}{N}  \sum f_i {x_i}^2 \Bigg) -  \Bigg(  \frac{1}{N} \sum f_ix_i  \Bigg)^2  \\ \\ \\ { \hspace{2.0cm} = \frac{17530}{409} -  \Bigg( \frac{239}{40} \Bigg)^2 = 43.825 -  5.975 = 8.12 } 

\displaystyle \text{Hence } S.D. = \sqrt{Var(X)} = \sqrt{8.12} = 2.85

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Question 4: Find the standard deviation for the following data:

\displaystyle \text{(i) } \begin{array}  {|c | c| c| c| c | c | } \hline    x: & 3 & 8 & 13 & 18 & 23  \\ \hline   f: & 7 & 10 & 15 & 10 & 6 \\ \hline \end{array}

\displaystyle \text{(ii) } \begin{array}  {|c | c| c| c| c | c | c| } \hline    x: & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline   f: & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}

Answer:

(i)

\displaystyle \text{Mean } \overline{X} = \frac{\sum f_ix_i}{\sum f_i}  = \frac{614}{48} = 12.79

Calculation of Variance and Standard Deviation

\displaystyle \begin{array}  {|c | c| c| c| c | c | c| } \hline x_i   &  f_i  & f_i x_i  & d_i = x_i - \overline{X}   &  {d_i}^2    & f_id_i   &  f_i {d_i}^2 \\  \hline 3 & 7 & 21 & -9.79 & 95.84 & - 68.53 & 670.88 \\ \hline 8 & 10 & 80  & -4.79 & 22.94 & -47.9  & 229.4 \\ \hline 13 & 15  &195 & 0.21 & 0.04 & 0.6  &  0.6 \\ \hline 18 & 10 & 180 & 5.21 & 27.14 & 52.1 &  271.4 \\ \hline 23 & 6 & 138 & 10.21 & 104.24 & 61.26 &  625.44 \\ \hline & N = \sum f_i &  \sum f_ix_i  & &   & \sum f_i d_i & \sum f_i {d_i}^2  \\ &  = 48  & = 614 &   &   & = -2.47 & = 2797.32 \\ \hline \end{array}

\displaystyle \text{Here } N = 48, \hspace{1.0cm} \sum f_id_i = -2.47, \hspace{1.0cm} \sum f_i {d_i}^2 = 1797.32

\displaystyle Var(X) = \Bigg(  \frac{1}{N}  \sum f_i {d_i}^2 \Bigg) -  \Bigg(  \frac{1}{N} \sum f_id_i  \Bigg)^2   

\displaystyle \Rightarrow Var(X) = \frac{1797.32}{48} - \Bigg( \frac{-2.47}{48} \Bigg)^2 =37.44417 - 0.00265  = 37.4415 = 37.44

\displaystyle \text{Hence } S.D. = \sqrt{Var(X)} = \sqrt{37.4415} = 6.1189 = 6.12

(ii)

\displaystyle \text{Mean } \overline{X} = \frac{\sum f_ix_i}{\sum f_i}  = \frac{277}{60} = 4.62

Calculation of Variance and Standard Deviation

\displaystyle \begin{array}  {|c | c| c| c| c | c | c| } \hline x_i   &  f_i  & f_i x_i  & d_i = x_i - \overline{X}   &  {d_i}^2    & f_id_i   &  f_i {d_i}^2 \\  \hline 2 & 4  & 8 & -2.62 & 6.86 & -10.48 &  27.44 \\ \hline 3 &  9 &  27 & -1.62 & 2.62 & -14.58  &  23.58 \\ \hline 4 & 16  &  64 & -0.62 & 0.38 & -9.92  &  6.08 \\ \hline 5 &  14 &  70 & 0.38 &  0.14 &  5.32 &  1.96 \\ \hline 6 &  11 & 66  & 1.38 &  1.90 &  15.18 &  20.90 \\ \hline 7 &  6 & 42  & 2.38 &  5.66 &  14.28 &  33.96 \\ \hline & N = \sum f_i &  \sum f_ix_i  & &   & \sum f_i d_i & \sum f_i {d_i}^2  \\ &  = 60  & = 277 &   &   & = -0.2 & = 113.92 \\ \hline \end{array}

\displaystyle \text{Here } N = 60, \hspace{1.0cm} \sum f_id_i = -0.2, \hspace{1.0cm} \sum f_i {d_i}^2 = 113.92

\displaystyle Var(X) = \Bigg(  \frac{1}{N}  \sum f_i {d_i}^2 \Bigg) -  \Bigg(  \frac{1}{N} \sum f_id_i  \Bigg)^2   

\displaystyle \Rightarrow Var(X) = \frac{113.92}{60} - \Bigg( \frac{-0.2}{60} \Bigg)^2 =1.8987 - 0.0000011  = 1.8987 

\displaystyle \text{Hence } S.D. = \sqrt{Var(X)} = \sqrt{1.8987} = 1.3779 = 1.38