Q1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Solution

Given that the mean of 8 observations is 9, the sum of all observations is \(9\times 8=72\). The first six observations are 6, 7, 10, 12, 12 and 13. Let the remaining two observations be \(x\) and \(y\).

\[ \begin{aligned} 72&=6+7+10+12+12+13+x+y\\ 72&=60+x+y\\ x+y&=12 \end{aligned} \]

The variance is given as \(9.25\). Using the formula \(\sigma^{2}=\dfrac{1}{n}\sum (x_i-\overline{x})^{2}\), with \(n=8\) and \(\overline{x}=9\), we write

\[ \begin{aligned} 9.25&=\dfrac{1}{8}\Big[(6-9)^2+(7-9)^2+(10-9)^2+(12-9)^2+(12-9)^2+(13-9)^2+(x-9)^2+(y-9)^2\Big]\\ 74&=3^2+2^2+1^2+3^2+3^2+4^2+(x-9)^2+(y-9)^2\\ 74&=9+4+1+9+9+16+(x-9)^2+(y-9)^2\\ (x-9)^2+(y-9)^2&=74-48\\ (x-9)^2+(y-9)^2&=26 \end{aligned} \]

Expanding and simplifying gives

\[ \begin{aligned} x^2-18x+81+y^2-18y+81&=26\\ x^2+y^2-18(x+y)+162&=26\\ x^2+y^2-18(x+y)&=-136 \end{aligned} \]

Since \(x+y=12\), substituting this value yields

\[ \begin{aligned} x^2+y^2-18\times12&=-136\\ x^2+y^2&=80 \end{aligned} \]

Now using \(y=12-x\) in \(x^2+y^2=80\), we obtain

\[ \begin{aligned} x^2+(12-x)^2&=80\\ x^2+x^2-24x+144&=80\\ 2x^2-24x+64&=0\\ x^2-12x+32&=0\\ x^2-8x-4x+32&=0\\ x(x-8)-4(x-8)&=0\\ (x-4)(x-8)&=0 \end{aligned} \]

Hence, \(x=4\) or \(x=8\). Correspondingly, \(y=8\) or \(y=4\). Therefore, the remaining two observations are 4 and 8 (in any order).

Q2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Solution

The mean of 7 observations is given as 8, so the total sum of all observations is \(8\times7=56\). The first five observations are 2, 4, 10, 12 and 14. Let the remaining two observations be \(x\) and \(y\).

\[ \begin{aligned} 8&=\dfrac{2+4+10+12+14+x+y}{7}\\ 56&=42+x+y\\ x+y&=14 \end{aligned} \]

The variance is 16. Using the formula \(\sigma^{2}=\dfrac{1}{n}\sum (x_i-\overline{x})^{2}\) with \(n=7\) and \(\overline{x}=8\), we have

\[ \begin{aligned} 16&=\dfrac{1}{7}\Big[(2-8)^2+(4-8)^2+(10-8)^2+(12-8)^2+(14-8)^2+(x-8)^2+(y-8)^2\Big]\\ 112&=(-6)^2+(-4)^2+2^2+4^2+6^2+(x-8)^2+(y-8)^2\\ 112&=36+16+4+16+36+(x-8)^2+(y-8)^2\\ (x-8)^2+(y-8)^2&=112-108\\ (x-8)^2+(y-8)^2&=4 \end{aligned} \]

Expanding the squares gives

\[ \begin{aligned} x^2-16x+64+y^2-16y+64&=4\\ x^2+y^2-16(x+y)+128&=4\\ x^2+y^2-16(x+y)&=-124 \end{aligned} \]

Since \(x+y=14\), substituting this value yields

\[ \begin{aligned} x^2+y^2-16\times14&=-124\\ x^2+y^2&=224-124\\ x^2+y^2&=100 \end{aligned} \]

Now using \(y=14-x\) in \(x^2+y^2=100\), we obtain

\[ \begin{aligned} x^2+(14-x)^2&=100\\ x^2+x^2-28x+196-100&=0\\ 2x^2-28x+96&=0\\ x^2-14x+48&=0\\ x^2-8x-6x+48&=0\\ x(x-8)-6(x-8)&=0 \end{aligned} \]

Hence, \(x=6\) or \(x=8\). Correspondingly, \(y=8\) or \(y=6\). Therefore, the remaining two observations are 6 and 8 (in any order).

Q3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Solution

The mean of 6 observations is 8, so the sum of the observations is \(6\times 8=48\). Each observation is multiplied by 3, therefore the new total sum becomes \(48\times 3\).

The new mean is calculated using \(\overline{x}=\dfrac{\sum x_i}{N}\).

\[ \begin{aligned} \overline{x}&=\dfrac{48\times 3}{6}\\ &=24 \end{aligned} \]

Hence, the new mean is 24.

The standard deviation of the original data is 4, so the variance is \(4^2=16\). Using the variance formula \(\sigma^2=\dfrac{1}{N}\sum x_i^2\), we write

\[ \begin{aligned} \sigma^2&=16\\ 16&=\dfrac{1}{6}\sum x_i^2 \end{aligned} \]

If each observation is multiplied by 3, then every squared term is multiplied by \(3^2=9\). Therefore, the new variance becomes 9 times the original variance.

\[ \begin{aligned} \sigma_1^2&=\dfrac{1}{N}\sum (3x_i)^2\\ &=9\cdot\dfrac{1}{N}\sum x_i^2\\ &=9\times16\\ &=144 \end{aligned} \]

The new standard deviation is the square root of the new variance.

\[ \begin{aligned} \sigma_1&=\sqrt{144}\\ &=12 \end{aligned} \]

Therefore, the new mean is 24 and the new standard deviation is 12.

Q4. Given that \(\overline{x}\) is the mean and \(\sigma^2\) is the variance of \(n\) observations \(x_1,\;x_2,\; ...,\;x_n\) . Prove that the mean and variance of the observations \(ax_1\;, ax_2,\; ax_3,\; ....,\;ax_n\) are \(a\overline{x}\) and \(a^2 \sigma^2\), respectively, \((a \ne 0)\).

Solution

Let \(x_1,x_2,\ldots,x_n\) be \(n\) observations with mean \(\overline{x}\) and variance \(\sigma^2\). By definition,

\[ \begin{aligned} \overline{x}=\dfrac{1}{n}\sum_{i=1}^{n}x_i \end{aligned} \]

Consider the new observations \(ax_1, ax_2, \ldots, ax_n\), where \(a\neq0\). Let their mean be \(\overline{x}_1\).

\[ \begin{aligned} \overline{x}_1 &=\dfrac{1}{n}\sum_{i=1}^{n}ax_i\\ &=\dfrac{a}{n}\sum_{i=1}^{n}x_i\\ &=a\,\overline{x} \end{aligned} \]

Hence, the mean of the new observations is \(a\overline{x}\).

Now let \(\sigma_1^2\) be the variance of the new observations. Using the definition of variance,

\[ \begin{aligned} \sigma_1^2 &=\dfrac{1}{n}\sum_{i=1}^{n}(ax_i-\overline{x}_1)^2 \end{aligned} \]

Since \(\overline{x}_1=a\overline{x}\), we substitute and simplify.

\[ \begin{aligned} \sigma_1^2 &=\dfrac{1}{n}\sum_{i=1}^{n}(ax_i-a\overline{x})^2\\ &=\dfrac{1}{n}\sum_{i=1}^{n}a^2(x_i-\overline{x})^2\\ &=a^2\,\dfrac{1}{n}\sum_{i=1}^{n}(x_i-\overline{x})^2\\ &=a^2\sigma^2 \end{aligned} \]

Therefore, the variance of the transformed observations is \(a^2\sigma^2\).

Hence proved that multiplying each observation by a nonzero constant \(a\) multiplies the mean by \(a\) and the variance by \(a^2\).

Q5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.

Solution

Given that the mean of 20 observations is 10, the total sum of observations is

\[ \begin{aligned} \sum x_i=20\times10=200 \end{aligned} \]

The standard deviation is 2, so the variance is \(2^2=4\). Using the relation \(\sigma^2=\dfrac{\sum x_i^2}{n}-\overline{x}^2\), we obtain

\[ \begin{aligned} 4&=\dfrac{\sum x_i^2}{20}-100\\ \dfrac{\sum x_i^2}{20}&=104\\ \sum x_i^2&=2080 \end{aligned} \]

The wrong observation is 8.

(i) If the wrong item is omitted, the new number of observations becomes 19 and the corrected sum is

\[ \begin{aligned} \sum x_i'&=200-8=192 \end{aligned} \]

Hence, the correct mean is

\[ \begin{aligned} \overline{x}_1=\dfrac{192}{19} \end{aligned} \]

Also, the corrected sum of squares is

\[ \begin{aligned} \sum x_i'^2&=2080-8^2\\ &=2080-64\\ &=2016 \end{aligned} \]

Therefore, the new variance is

\[ \begin{aligned} \sigma_1^2 &=\dfrac{2016}{19}-\left(\dfrac{192}{19}\right)^2\\ &=\dfrac{1440}{361} \end{aligned} \]

Hence, the corrected standard deviation is

\[ \begin{aligned} \sigma_1=\sqrt{\dfrac{1440}{361}}=\dfrac{\sqrt{1440}}{19} \end{aligned} \]

(ii) If the wrong observation 8 is replaced by 12, the new sum becomes

\[ \begin{aligned} \sum x_i''&=200-8+12\\ &=204 \end{aligned} \]

Thus, the corrected mean is

\[ \begin{aligned} \overline{x}_2=\dfrac{204}{20}=10.2 \end{aligned} \]

The corrected sum of squares is

\[ \begin{aligned} \sum x_i''^2&=2080-64+144\\ &=2160 \end{aligned} \]

Hence, the new variance is

\[ \begin{aligned} \sigma_2^2 &=\dfrac{2160}{20}-(10.2)^2\\ &=108-104.04\\ &=3.96 \end{aligned} \]

Therefore, the corrected standard deviation is

\[ \begin{aligned} \sigma_2=\sqrt{3.96}\approx1.99 \end{aligned} \]

So, when the wrong item is omitted, the mean is \(\dfrac{192}{19}\) and the standard deviation is \(\dfrac{\sqrt{1440}}{19}\). When it is replaced by 12, the mean becomes 10.2 and the standard deviation is approximately 1.99.

Q6. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Solution

The given mean of 100 observations is 20. Hence, the total sum of observations is

\[ \begin{aligned} \sum x_i=100\times20=2000 \end{aligned} \]

The standard deviation is 3, so the variance is \(3^2=9\). Using the relation \(\sigma^2=\dfrac{\sum x_i^2}{n}-\overline{x}^2\), we obtain

\[ \begin{aligned} 9&=\dfrac{\sum x_i^2}{100}-400\\ \dfrac{\sum x_i^2}{100}&=409\\ \sum x_i^2&=40900 \end{aligned} \]

The three incorrect observations are 21, 21 and 18. These are to be omitted, so the new number of observations becomes \(100-3=97\).

First, we correct the total sum of observations.

\[ \begin{aligned} \sum x_i'&=2000-(21+21+18)\\ &=2000-60\\ &=1940 \end{aligned} \]

Therefore, the corrected mean is

\[ \begin{aligned} \overline{x}_1=\dfrac{1940}{97}=20 \end{aligned} \]

Next, we correct the sum of squares.

\[ \begin{aligned} \sum x_i'^2 &=40900-(21^2+21^2+18^2)\\ &=40900-(441+441+324)\\ &=40900-1206\\ &=39694 \end{aligned} \]

Now, using the variance formula for the corrected data,

\[ \begin{aligned} \sigma_1^2 &=\dfrac{39694}{97}-20^2\\ &=409.2165-400\\ &\approx9.2165 \end{aligned} \]

Hence, the corrected standard deviation is

\[ \begin{aligned} \sigma_1=\sqrt{9.2165}\approx3.04 \end{aligned} \]

Therefore, after omitting the three incorrect observations, the mean remains 20 and the standard deviation is approximately 3.04.