Find the mean and variance for each of the data in Exercies 1 to 5.
Q1. 6, 7, 10, 12, 13, 4, 8, 12
Solution
The given observations are treated as individual data. To find the mean, we first add all the values of \(x_i\). Using this mean, the deviations \(|x_i-\overline{x}|\) and their squares \(|x_i-\overline{x}|^2\) are then calculated for each observation, as shown in the table.
| \(x_i\) | \(|x_i-\overline{x}|\) | \(|x_i-\overline{x}|^2\) |
|---|---|---|
| 6 | 3 | 9 |
| 7 | 2 | 4 |
| 10 | 1 | 1 |
| 12 | 3 | 9 |
| 13 | 4 | 16 |
| 4 | 5 | 25 |
| 8 | 1 | 1 |
| 12 | 3 | 9 |
| \(\sum x_i=72\) | \(\sum (x_i-\overline{x})^2=74\) |
First, the arithmetic mean is obtained by dividing the sum of observations by their number.
$$ \begin{aligned} \overline{x}&=\dfrac{72}{8}\\ &=9 \end{aligned} $$Using this mean, the squared deviations are summed as \(74\). The variance is then calculated by dividing this total by the number of observations.
$$ \begin{aligned} \sigma^2&=\dfrac{1}{N}\sum (x_i-\overline{x})^2\\ &=\dfrac{1}{8}\times 74\\ &=9.25 \end{aligned} $$Hence, the mean of the data is \(9\) and the variance is \(9.25\).
Q2. First n natural numbers
Solution
Let the first \(n\) natural numbers be \(1,2,3,\dots ,n\). The total number of observations is \(N=n\).
First, we find the arithmetic mean. The sum of the first \(n\) natural numbers is \(\dfrac{n(n+1)}{2}\). Hence,
\[ \begin{aligned} \overline{x}&=\dfrac{1}{n}\sum_{i=1}^{n} i\\ &=\dfrac{1}{n}\times \dfrac{n(n+1)}{2}\\ &=\dfrac{n+1}{2} \end{aligned} \]Next, to find the variance, we use the formula \(\sigma^2=\dfrac{1}{n}\sum x_i^2-\overline{x}^2\). For the first \(n\) natural numbers, the sum of squares is \(\dfrac{n(n+1)(2n+1)}{6}\).
\[ \begin{aligned} \sigma^2&=\dfrac{1}{n}\sum_{i=1}^{n} i^2-\overline{x}^2\\ &=\dfrac{1}{n}\times \dfrac{n(n+1)(2n+1)}{6}-\left(\dfrac{n+1}{2}\right)^2\\ &=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^2}{4} \end{aligned} \]Taking \((n+1)\) as common factor and simplifying, we obtain
\[ \begin{aligned} \sigma^2&=(n+1)\left(\dfrac{2n+1}{6}-\dfrac{n+1}{4}\right)\\ &=(n+1)\left(\dfrac{4n+2-3n-3}{12}\right)\\ &=(n+1)\left(\dfrac{n-1}{12}\right)\\ &=\dfrac{n^2-1}{12} \end{aligned} \]Hence, for the first \(n\) natural numbers, the mean is \(\dfrac{n+1}{2}\) and the variance is \(\dfrac{n^2-1}{12}\).
Q3. First 10 multiples of 3
Solution
The first 10 multiples of 3 are \(3,6,9,12,15,18,21,24,27,30\). Hence, the total number of observations is \(N=10\).
First, we calculate the arithmetic mean by dividing the sum of all observations by their number.
\[ \begin{aligned} \sum x_i&=3+6+9+12+15+18+21+24+27+30\\ &=165 \end{aligned} \] \[ \begin{aligned} \overline{x}&=\dfrac{1}{10}\times 165\\ &=16.5 \end{aligned} \]Next, to find the variance, we use the formula \(\sigma^2=\dfrac{1}{N}\sum (x_i-\overline{x})^2\).
\[ \begin{aligned} \sigma^2&=\dfrac{1}{10}\Big[(3-16.5)^2+(6-16.5)^2+(9-16.5)^2+(12-16.5)^2\\ &\quad +(15-16.5)^2+(18-16.5)^2+(21-16.5)^2+(24-16.5)^2+(27-16.5)^2+(30-16.5)^2\Big] \end{aligned} \]Evaluating each squared deviation and adding, we get
\[ \begin{aligned} \sum (x_i-\overline{x})^2&=182.25+110.25+56.25+20.25+2.25+2.25+20.25+56.25+110.25+182.25\\ &=742.5 \end{aligned} \] \[ \begin{aligned} \sigma^2&=\dfrac{742.5}{10}\\ &=74.25 \end{aligned} \]Hence, the mean of the first 10 multiples of 3 is \(16.5\) and the variance is \(74.25\).
Q4.
| \(x_i\) | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
| \(f_i\) | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Solution
The data are given in discrete form with corresponding frequencies. To find the mean and variance, we first compute the products \(f_ix_i\) and then determine the mean \(\overline{x}\). Using this mean, the deviations \(|x_i-\overline{x}|\), their squares, and the products \(f_i|x_i-\overline{x}|^2\) are calculated as shown in the table.
| \(x_i\) | \(f_i\) | \(f_ix_i\) | \(|x_i-\overline{x}|\) | \(|x_i-\overline{x}|^2\) | \(f_i|x_i-\overline{x}|^2\) |
|---|---|---|---|---|---|
| 6 | 2 | 12 | 13 | 169 | 338 |
| 10 | 4 | 40 | 9 | 81 | 324 |
| 14 | 7 | 98 | 5 | 25 | 175 |
| 18 | 12 | 216 | 1 | 1 | 12 |
| 24 | 8 | 192 | 5 | 25 | 200 |
| 28 | 4 | 112 | 9 | 81 | 324 |
| 30 | 3 | 90 | 11 | 121 | 363 |
| \(N=\sum f_i=40\) | \(\sum f_ix_i=760\) | \(\sum f_i|x_i-\overline{x}|^2=1736\) | |||
First, the arithmetic mean is obtained by dividing the total of \(f_ix_i\) by the total frequency.
$$ \begin{aligned} \overline{x}&=\dfrac{\sum f_ix_i}{N}\\ &=\dfrac{760}{40}\\ &=19 \end{aligned} $$Using this mean, the squared deviations are multiplied by their respective frequencies and summed. The variance is then calculated by dividing this total by \(N\).
$$ \begin{aligned} \sigma^2&=\dfrac{1}{N}\sum f_i(x_i-\overline{x})^2\\ &=\dfrac{1}{40}\times 1736\\ &=43.4 \end{aligned} $$Hence, the mean of the distribution is \(19\) and the variance is \(43.4\).
Q5.
| \(x_i\) | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
| \(f_i\) | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Solution
The data are given in discrete form with corresponding frequencies. To determine the mean and variance, we first calculate the products \(f_ix_i\). Using the mean \(\overline{x}\), the deviations \(|x_i-\overline{x}|\), their squares, and the products \(f_i|x_i-\overline{x}|^2\) are then obtained as shown in the table.
| \(x_i\) | \(f_i\) | \(f_ix_i\) | \(|x_i-\overline{x}|\) | \(|x_i-\overline{x}|^2\) | \(f_i|x_i-\overline{x}|^2\) |
|---|---|---|---|---|---|
| 92 | 3 | 276 | 8 | 64 | 192 |
| 93 | 2 | 186 | 7 | 49 | 98 |
| 97 | 3 | 291 | 3 | 9 | 27 |
| 98 | 2 | 196 | 2 | 4 | 8 |
| 102 | 6 | 612 | 2 | 4 | 24 |
| 104 | 3 | 312 | 4 | 16 | 48 |
| 109 | 3 | 327 | 9 | 81 | 243 |
| \(N=\sum f_i=22\) | \(\sum f_ix_i=2200\) | \(\sum f_i|x_i-\overline{x}|^2=640\) | |||
First, the arithmetic mean is obtained by dividing the total of \(f_ix_i\) by the total frequency.
$$ \begin{aligned} \overline{x}&=\dfrac{\sum f_ix_i}{N}\\ &=\dfrac{2200}{22}\\ &=100 \end{aligned} $$Using this mean, the squared deviations are multiplied by their respective frequencies and added. The variance is then calculated by dividing this total by \(N\).
$$ \begin{aligned} \sigma^2&=\dfrac{1}{N}\sum f_i(x_i-\overline{x})^2\\ &=\dfrac{1}{22}\times 640\\ &=29.09 \end{aligned} $$Hence, the mean of the distribution is \(100\) and the variance is \(29.09\).
Q6. Find the mean and standard deviation using short-cut method.
| \(x_i\) | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
| \(f_i\) | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |
Solution
An assumed mean \(a=64\) is taken to simplify calculations. Let \(y_i=x_i-a\). Using this transformation, we compute \(y_i\), \(y_i^2\), \(f_iy_i\), and \(f_iy_i^2\) as shown in the table.
| \(x_i\) | \(f_i\) | \(y_i=x_i-64\) | \(y_i^2\) | \(f_iy_i\) | \(f_iy_i^2\) |
|---|---|---|---|---|---|
| 60 | 2 | -4 | 16 | -8 | 32 |
| 61 | 1 | -3 | 9 | -3 | 9 |
| 62 | 12 | -2 | 4 | -24 | 48 |
| 63 | 29 | -1 | 1 | -29 | 29 |
| 64 | 25 | 0 | 0 | 0 | 0 |
| 65 | 12 | 1 | 1 | 12 | 12 |
| 66 | 10 | 2 | 4 | 20 | 40 |
| 67 | 4 | 3 | 9 | 12 | 36 |
| 68 | 5 | 4 | 16 | 20 | 80 |
| \(\sum f_i=100\) | \(\sum f_iy_i=0\) | \(\sum f_iy_i^2=286\) |
Since \(\sum f_iy_i=0\), the mean is obtained directly from the assumed mean.
$$ \begin{aligned} \overline{x}&=a+\dfrac{\sum f_iy_i}{N}\\ &=64+\dfrac{0}{100}\\ &=64 \end{aligned} $$Here the common difference \(h=1\). Using the short-cut formula for variance,
$$ \begin{aligned} \sigma^2&=\dfrac{h^2}{N}\sum f_iy_i^2\\ &=\dfrac{1}{100}\times 286\\ &=2.86 \end{aligned} $$The standard deviation is the positive square root of the variance.
$$ \begin{aligned} \sigma&=\sqrt{2.86}\\ &\approx1.69 \end{aligned} $$Hence, the mean of the distribution is \(64\) and the standard deviation is approximately \(1.69\).
Find the mean and variance for the following frequency distributions in Exercises 7 and 8.
Q7.
| classes | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |
| Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Solution
The data are grouped into equal class intervals of width \(h=30\). The class marks \(x_i\) are calculated and an assumed mean \(a=105\) is chosen to simplify the calculations. Using this assumed mean, we compute \(y_i=\dfrac{x_i-a}{h}\), along with \(f_iy_i\) and \(f_iy_i^2\), as shown in the table.
| \(Class\) | \(f_i\) | \(x_i\) | \(y_i=\dfrac{x_i-105}{30}\) | \(f_iy_i\) | \(y_i^2\) | \(f_iy_i^2\) |
|---|---|---|---|---|---|---|
| 0–30 | 2 | 15 | -3 | -6 | 9 | 18 |
| 30–60 | 3 | 45 | -2 | -6 | 4 | 12 |
| 60–90 | 5 | 75 | -1 | -5 | 1 | 5 |
| 90–120 | 10 | 105 | 0 | 0 | 0 | 0 |
| 120–150 | 3 | 135 | 1 | 3 | 1 | 3 |
| 150–180 | 5 | 165 | 2 | 10 | 4 | 20 |
| 180–210 | 2 | 195 | 3 | 6 | 9 | 18 |
| \(\sum f_i=30\) | \(\sum f_iy_i=2\) | \(\sum f_iy_i^2=76\) |
The mean is calculated using the step-deviation formula \(\overline{x}=a+\dfrac{\sum f_iy_i}{N}\times h\).
$$ \begin{aligned} \overline{x}&=105+\dfrac{2}{30}\times 30\\ &=105+2\\ &=107 \end{aligned} $$The variance is obtained using \(\sigma^2=\dfrac{h^2}{N}\left[\sum f_iy_i^2-\dfrac{(\sum f_iy_i)^2}{N}\right]\).
$$ \begin{aligned} \sigma^2&=\dfrac{30^2}{30}\left[76-\dfrac{4}{30}\right]\\ &=30\times \dfrac{2276}{30}\\ &=2276 \end{aligned} $$Hence, the mean of the distribution is \(107\) and the variance is \(2276\).
Q8.
| Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequencies | 5 | 8 | 15 | 16 | 6 |
Solution
The data are grouped into equal class intervals of width \(h=10\). The class marks \(x_i\) are obtained for each class, and an assumed mean \(a=25\) is chosen from the central class to simplify the calculations. Using this assumed mean, we compute \(y_i=\dfrac{x_i-a}{h}\), along with \(y_i^2\), \(f_iy_i\), and \(f_iy_i^2\), as shown in the table.
| \(Class\) | \(f_i\) | \(x_i\) | \(y_i=\dfrac{x_i-25}{10}\) | \(y_i^2\) | \(f_iy_i\) | \(f_iy_i^2\) |
|---|---|---|---|---|---|---|
| 0–10 | 5 | 5 | -2 | 4 | -10 | 20 |
| 10–20 | 8 | 15 | -1 | 1 | -8 | 8 |
| 20–30 | 15 | 25 | 0 | 0 | 0 | 0 |
| 30–40 | 16 | 35 | 1 | 1 | 16 | 16 |
| 40–50 | 6 | 45 | 2 | 4 | 12 | 24 |
| \(N=\sum f_i=50\) | \(\sum f_iy_i=10\) | \(\sum f_iy_i^2=68\) |
The mean is obtained using the step-deviation formula \(\overline{x}=a+\dfrac{\sum f_iy_i}{N}\times h\).
$$ \begin{aligned} \overline{x}&=25+\dfrac{10}{50}\times 10\\ &=25+2\\ &=27 \end{aligned} $$The variance is calculated using \(\sigma^2=\dfrac{h^2}{N^2}\left[N\sum f_iy_i^2-(\sum f_iy_i)^2\right]\).
$$ \begin{aligned} \sigma^2&=\dfrac{10^2}{50^2}\left[50\times 68-100\right]\\ &=\dfrac{100}{2500}\left[3400-100\right]\\ &=\dfrac{1}{25}\times 3300\\ &=132 \end{aligned} $$Hence, the mean of the distribution is \(27\) and the variance is \(132\).
Q9. Find the mean, variance and standard deviation using short-cut method
| Height in cms | 70-75 | 75-80 | 80-85/td> | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |
| No. of Children | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
Solution
The data are grouped into equal class intervals of width \(h=5\). The class marks \(x_i\) are calculated and the assumed mean is taken as \(a=92.5\), corresponding to the central class. Using this assumed mean, we compute \(y_i=\dfrac{x_i-a}{h}\), along with \(y_i^2\), \(f_iy_i\), and \(f_iy_i^2\), as shown in the table.
| Class | \(f_i\) | \(x_i\) | \(y_i=\dfrac{x_i-92.5}{5}\) | \(y_i^2\) | \(f_iy_i\) | \(f_iy_i^2\) |
|---|---|---|---|---|---|---|
| 70–75 | 3 | 72.5 | -4 | 16 | -12 | 48 |
| 75–80 | 4 | 77.5 | -3 | 9 | -12 | 36 |
| 80–85 | 7 | 82.5 | -2 | 4 | -14 | 28 |
| 85–90 | 7 | 87.5 | -1 | 1 | -7 | 7 |
| 90–95 | 15 | 92.5 | 0 | 0 | 0 | 0 |
| 95–100 | 9 | 97.5 | 1 | 1 | 9 | 9 |
| 100–105 | 6 | 102.5 | 2 | 4 | 12 | 24 |
| 105–110 | 6 | 107.5 | 3 | 9 | 18 | 54 |
| 110–115 | 3 | 112.5 | 4 | 16 | 12 | 48 |
| \(N=\sum f_i=60\) | \(\sum f_iy_i=6\) | \(\sum f_iy_i^2=254\) |
The mean is obtained using the short-cut (step-deviation) formula \(\overline{x}=a+\dfrac{\sum f_iy_i}{N}\times h\).
$$ \begin{aligned} \overline{x}&=92.5+\dfrac{6}{60}\times 5\\ &=92.5+0.5\\ &=93 \end{aligned} $$The variance is calculated using \(\sigma^2=\dfrac{h^2}{N^2}\left[N\sum f_iy_i^2-(\sum f_iy_i)^2\right]\).
$$ \begin{aligned} \sigma^2&=\dfrac{5^2}{60^2}\left[60\times 254-6^2\right]\\ &=\dfrac{25}{3600}\left[15240-36\right]\\ &=\dfrac{25}{3600}\times 15204\\ &=105.58 \end{aligned} $$The standard deviation is the positive square root of the variance.
$$ \begin{aligned} \sigma&=\sqrt{105.58}\\ &=10.27 \end{aligned} $$Hence, the mean height is \(93\) cm, the variance is \(105.58\), and the standard deviation is \(10.27\) (approximately).
Q10. The diameters of circles (in mm) drawn in a design are given below:
| Diameters | 33-36 | 37-40 | 41-44 | 45-48 | 49-52 |
| No. of Circles | 15 | 17 | 21 | 22 | 25 |
Solution
The given diameters are first converted into continuous class intervals by applying the correction factor. The class width is \(h=4\), and the class marks \(x_i\) are obtained for each class. Taking the central value \(a=42.5\) as the assumed mean, we compute \(y_i=\dfrac{x_i-a}{h}\), along with \(y_i^2\), \(f_iy_i\), and \(f_iy_i^2\), as shown in the table.
| Class | \(f_i\) | \(x_i\) | \(y_i=\dfrac{x_i-42.5}{4}\) | \(y_i^2\) | \(f_iy_i\) | \(f_iy_i^2\) |
|---|---|---|---|---|---|---|
| 32.5–36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |
| 36.5–40.5 | 17 | 38.5 | -1 | 1 | -17 | 17 |
| 40.5–44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |
| 44.5–48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |
| 48.5–52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |
| \(N=\sum f_i=100\) | \(\sum f_iy_i=25\) | \(\sum f_iy_i^2=199\) |
The mean diameter is calculated using the short-cut (step-deviation) formula \(\overline{x}=a+\dfrac{\sum f_iy_i}{N}\times h\).
$$ \begin{aligned} \overline{x}&=42.5+\dfrac{25}{100}\times 4\\ &=42.5+1\\ &=43.5 \end{aligned} $$The variance is obtained using \(\sigma^2=\dfrac{h^2}{N^2}\left[N\sum f_iy_i^2-(\sum f_iy_i)^2\right]\).
$$ \begin{aligned} \sigma^2&=\dfrac{4^2}{100^2}\left[100\times 199-25^2\right]\\ &=\dfrac{16}{10000}\left[19900-625\right]\\ &=\dfrac{1}{625}\times 19275\\ &=30.84 \end{aligned} $$The standard deviation is the positive square root of the variance.
$$ \begin{aligned} \sigma&=\sqrt{30.84}\\ &=5.55 \end{aligned} $$Hence, the mean diameter of the circles is \(43.5\) mm and the standard deviation is approximately \(5.55\) mm.