Chapter 15: Statistics

Ans: We know that,\[X = A + h\left( {\dfrac{1}{N}\sum {f_i}{d_i}} \right)\]

Mean deviation about mean = \[ = \dfrac{1}{N}\left( {\sum {f_i}\left| {{x_i} - X} \right|} \right)\]

Let us assume \[{\text{A  =  10}}\] and \[{\text{h  =  4}}\]

$X = A + h\left( {\dfrac{1}{N}\sum {f_i}{d_i}} \right)$

$= 10 + 4\left( {\dfrac{1}{{25}}\left( { - 5} \right)} \right)$

$= 10 - \dfrac{4}{5} = 10 - 0.8$

$= 9.2$ 

Mean deviation about mean = \[ = \dfrac{1}{N}\left( {\sum {f_i}\left| {{x_i} - X} \right|} \right) = \dfrac{{96}}{{25}} = 3.84\]

Ans: With the help of the basic formula,

Median \[ = l + \dfrac{{\dfrac{N}{2} - C}}{f} \times h\]

$= 12 + \dfrac{{(10 - 9)}}{3} \times 6 = 14$

Since, $\dfrac{N}{2} = 10$ ${\text{median class is }}12 - 18$

${\text{Median}} = l + \dfrac{{\dfrac{N}{2} - C}}{f} \times h$ 

${\text{Median}} = 12 + \dfrac{{10 - 9}}{3} \times 6=14$ 

${\text{MD}} = \dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}} = \dfrac{{140}}{{20}} = 7$. 

So the required mean deviation is 7.

Ans: ${\text{Mean}} = \dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}$

  $\sigma  = \sqrt {\dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - \left( {\dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}} \right)}$ 

${\text{Mean}} = \dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}} = \dfrac{{239}}{{40}} = 5.975 = 6$

$\sigma  = \sqrt {\dfrac{{\sum {f_i}{d_i}^2}}{{\sum {f_i}}} - \left( {\dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)} $

$= \sqrt {\dfrac{{325}}{{40}} - {{\left( {\dfrac{{ - 1}}{{40}}} \right)}^2}} $

$= \sqrt {8.124}  = 2.85$ 

Ans: We know that,

\[{\sigma ^2} = \dfrac{{\sum {f_i}{d_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)^2}\]

The mean $\bar{x}=\dfrac{200.5\times 13+201.5\times 27+202.5\times 18+203.5\times 10+204.5\times 1+205.5\times 1}{70}=\dfrac{14137}{70}=202.5$

${\sigma ^2} = \dfrac{{\sum {f_i}{d_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)^2}$

$= \dfrac{{102}}{{70}} - {\left( {\dfrac{{ - 38}}{{70}}} \right)^2}$

$= 1.4571 - 0.2946 = 1.1655$

$\sigma  = \sqrt {1.1655}  = 1.08gm $ 

Ans: Given: First-term of an A.P is \[a\].

In addition, the common difference is \[d.\]

Mean \[ = \dfrac{{\sum {x_i}}}{n} = \dfrac{1}{n}\left[ {\dfrac{n}{2}[2a + \left( {n - 1} \right)d]} \right] = a + \dfrac{{n - 1}}{2}d\]

Standard deviation \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {{x_i} - a} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {{x_i} - a} \right)}}{n}} \right)}^2}} \]

Mean \[ = \dfrac{{\sum {x_i}}}{n} = \dfrac{1}{n}\left[ {\dfrac{n}{2}[2a + \left( {n - 1} \right)d]} \right] = a + \dfrac{{n - 1}}{2}d\]

$\sum{\left( {{x_i} - a} \right)}=d\left[1+2+3+........+(n-1)\right]=\dfrac{d(n-1)n}{2}$

$\sum{\left( {{x_i} - a} \right)^{2}}={d}^{2}\left[1^2+2^2+3^2+........+(n-1)^{2}\right]= {d^2}\dfrac{n(n-1)(2n-1)}{6}$

Standard deviation \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {{x_i} - a} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {{x_i} - a} \right)}}{n}} \right)}^2}} \]

$= \sqrt {\dfrac{{{d^2}\left( {n - 1} \right)\left( {2n - 1} \right)}}{6} - \dfrac{{{d^2}{{\left( {n - 1} \right)}^2}}}{4}} $

 $= d\sqrt {\dfrac{{n - 1}}{2}\left[ {\dfrac{{\left( {2n - 1} \right)}}{3} - \dfrac{{\left( {n - 1} \right)}}{2}} \right]} $

$= d\sqrt {\dfrac{{n - 1}}{2}\left[ {\dfrac{{2\left( {2n - 1} \right) - 3(n - 1)}}{6}} \right]} $

$= d\sqrt {\dfrac{{n - 1}}{2}\left[ {\dfrac{{\left( {n + 1} \right)}}{6}} \right]}  = d\sqrt {\dfrac{{{n^2} - 1}}{{12}}} $ 

Ans: We know that,

\[\sigma  = \sqrt {\dfrac{{\sum {d_i}^2}}{n} - {{\left( {\dfrac{{\sum {d_i}^2}}{n}} \right)}^2}} \]

For Ravi

Let assumed mean =45

$\sigma  = \sqrt {\dfrac{{\sum {d_i}^2}}{n} - {{\left( {\dfrac{{\sum {d_i}}}{n}} \right)}^2}} $

$= \sqrt {\dfrac{{1699}}{{10}} - {{\left( {\dfrac{{ - 9}}{{10}}} \right)}^2}} $

$= \sqrt {169.9 - 0.81} $

$= \sqrt {169.09}  = 13.003$

$\overline x  = {\text{A}} + \dfrac{{\sum {d_i}}}{{\sum {f_i}}} = 45 - \dfrac{{9}}{{10}} = 44.1$ 

${\text{for Hashina,}}$

$\sigma  = \sqrt {\dfrac{{\sum {d_i}^2}}{n} - {{\left( {\dfrac{{\sum {d_i}}}{n}} \right)}^2}} $

$= \sqrt {\dfrac{{5698}}{{10}} - {{\left( {\dfrac{{20}}{{10}}} \right)}^2}} $

$= \sqrt {596.8 - 2} $

$= \sqrt {594.8}  = 24.38$ 

And, $\overline x  = {\text{A}} + \dfrac{{\sum {d_i}}}{{\sum {f_i}}} = 55 - \dfrac{{20}}{{10}} = 53$ 

For Ravi, \[\dfrac{\sigma }{{\overline x }} \times 100 = \dfrac{{13.003}}{{44.1}} \times 100 = 29.48\]

For hashina, \[\dfrac{\sigma }{{\overline x }} \times 100 = \dfrac{{24.38}}{{53}} \times 100 = 46.01\]

Hence hashina is more intelligent and consistent.

Ans: Given: $n = 100,\overline x  = 40$, $\sigma  = 10$

Two Observations were wrongly recorded as \[30{\text{ and }}70\] in the place of \[3{\text{ and }}27\]

We know that,${\sigma ^2} = \dfrac{{\sum {x_i}^2}}{n} - {\left( {\dfrac{{\sum {x_i}}}{n}} \right)^2}$

$n = 100,\overline x  = 40$,$\sigma  = 10$

$\dfrac{{\sum {x_i}}}{n} = 40 \Rightarrow \sum {x_i} = 40 \times 100$

$\sum {x_i} = 4000$ 

Corrected $\sum {x_i} = 4000 - 30 - 70 + 3 + 27$$ = 3930$

Corrected mean = $\dfrac{{3930}}{{100}} = 39.3$

${\sigma ^2} = \dfrac{{\sum {x_i}^2}}{n} - {\left( {\dfrac{{\sum {x_i}}}{n}} \right)^2}$

$100 = \dfrac{{\sum {x_i}^2}}{{100}} - 1600$

$\sum {x_i}^2 = 170000$ 

Corrected $\sum {x_i}^2 = 170000 - {\left( {30} \right)^2} - {\left( {70} \right)^2} + {3^2} + {\left( {27} \right)^2}$

$= 164938$

${\text{corrected }}\sigma  = \sqrt {\dfrac{{164938}}{{100}} - {{\left( {39.3} \right)}^2}}  = \sqrt {\left( {1649.38 - 1544.49} \right)} $

$= \sqrt {104.9}  = 10.24$ 

Ans: Given: Mistaken mean of the $10$ observations $ = 45$

Mistaken variance of the $10$ observations $ = 16$

The correct reading is $25$ instead of $52$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to find the sum of all the observations. Subtract the incorrect value and add the correct value to find the correct total. Here $\overline x $ is the mean.

Use the formula ${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} }}{n} - {\left( {\overline x } \right)^2}$, where ${\sigma ^2}$ is the variance, to calculate $\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}}$. Now, subtract ${\left( {52} \right)^2}$ from $\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}}$ and add ${\left( {25} \right)^2}$.

The incorrect mean of the given $10$ observations is $45$. Assuming this mean as $\overline x $ and using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$,

$\Rightarrow 45 = \dfrac{{\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)}}{{10}}$

$\Rightarrow \left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right) = 450$ 

That means the sum of all the observations (including the incorrect one) is $450$. Since one reading is to be $25$ instead of $52$, so the correct sum must be,

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)_{co}} = 450 - 52 + 25$

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)_{co}} = 423$ 

Therefore, the correct mean is ${\overline x _{co}} = \dfrac{{{{\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)}_{co}}}}{{10}} = 42.3$.

Now, using the formula ${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} }}{n} - {\left( {\overline x } \right)^2}$ for the variance of the incorrect mean, where ${\sigma ^2} = 16$ is provided,

$\Rightarrow \dfrac{{\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)}}{{10}} - {\left( {45} \right)^2} = 16$

$\Rightarrow \sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}}  = 20410$ 

That means the incorrect sum of squares of all the observations is $20410$. Since $52$ is to be replaced with $25$, so the correct value of $\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}}$ is,

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)_{co}} = 20410 - {\left( {52} \right)^2} + {\left( {25} \right)^2}$

$\Rightarrow {\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)_{co}} = 18331$ 

Hence, the variance for the correct mean is,

$\Rightarrow {\left( {{\sigma ^2}} \right)_{co}} = \dfrac{{{{\left( {\sum\limits_{i = 1}^{10} {{{\left( {{x_i}} \right)}^2}} } \right)}_{co}}}}{{10}} - {\left( {\overline x } \right)^2}_{co}$

$\Rightarrow {\left( {{\sigma ^2}} \right)_{co}} = \dfrac{{18331}}{{10}} - {\left( {42.3} \right)^2}_{co}$

$\Rightarrow {\left( {{\sigma ^2}} \right)_{co}} = 43.81$ 

Ans: The correct option is (B).

Given: The data $3$, $10$, $10$, $4$, $7$, $10$, $5$.

Use the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$ to calculate the mean of the given data. Here $\overline x $ is the mean.

Now, use the formula $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$ to calculate the mean deviation.

The total number of observations is $n = 7$. Using the formula $\overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}}{n}$, where $\overline x $ is the mean,

$\Rightarrow \overline x  = \dfrac{{\left( {\sum\limits_{i = 1}^7 {{x_i}} } \right)}}{7}$

$\Rightarrow \overline x  = \dfrac{{3 + 10 + 10 + 4 + 7 + 10 + 5}}{7}$

$\Rightarrow \overline x  = 7$ 

Now, the mean deviation is given as $M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$. So for the given data,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^7 {\left| {{x_i} - 7} \right|} }}{7}$

$\Rightarrow M.D. = \dfrac{{\left| {3 - 7} \right| + \left| {10 - 7} \right| + \left| {10 - 7} \right| + \left| {4 - 7} \right| + \left| {7 - 7} \right| + \left| {10 - 7} \right| + \left| {5 - 7} \right|}}{7}$

$\Rightarrow M.D. = 2.57$ 

Correct Answer: B

Ans: The correct option is (B).

Given: $n$ observations ${x_1},{x_2},....,{x_n}$.

The mean value of these observations is $\overline x $.

Use the definition of the mean deviation which is ‘the ratio of the summation of absolute deviation of each data from the mean and the total number of observations given’.

According to the definition of the mean deviation it is the ratio of the summation of absolute deviation of each data from the mean and the total number of observations given.

Now, the absolute deviation of $n$ observations ${x_1},{x_2},....,{x_n}$ from their mean $\overline x $ is given as $\left| {{x_1} - \overline x } \right|,\left| {{x_2} - \overline x } \right|,....,\left| {{x_n} - \overline x } \right|$ respectively. Taking the sum of these deviations and dividing by the number of observations gives,

$\Rightarrow M.D. = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \overline x } \right|} }}{n}$

Correct Answer: B