Statistics - Class 11 Engineering Maths - Extra Questions

We have,

$$1,2,3,4,5,6$$

We know that

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

$$ \sigma =\sqrt{\dfrac{91}{6}-{{\left( \dfrac{21}{6} \right)}^{2}}} $$

$$ \sigma =\sqrt{\dfrac{91}{6}-{{\left( \dfrac{7}{2} \right)}^{2}}} $$

$$ \sigma =\sqrt{\dfrac{91}{6}-\dfrac{49}{4}} $$

$$ \sigma =\sqrt{\dfrac{364-294}{24}} $$

$$ \sigma =\sqrt{\dfrac{70}{24}} $$

$$ \sigma =\sqrt{\dfrac{35}{12}} $$

$$ \sigma =\sqrt{2.916}\approx 1.70 $$

Hence, the $$S.D$$ is $$1.70$$.

(i) Here the number of observations, $$n = 48$$ which is even.

So $$median = \frac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$$

$$= \dfrac{1}{2}(48/2^{th} term + ((48/2)+1)^{th} term)$$

$$= \dfrac{1}{2}(24^{th} term + (24+1)^{th} term)$$

$$= \dfrac{1}{2} (24^{th} term + 25^{th} term)$$

$$= \dfrac{1}{2} (22+22)$$ [Since all observations from $$19^{th}$$ to $$27^{th}$$ are $$22$$]

$$= \dfrac{1}{2} \times 44$$

$$= 22$$

Hence the median is 22.

(ii) Lower quartile, $$Q_1 = (n/4)^{th} term$$

$$= (48)/4$$

$$= 12^{th} term$$

$$= 20$$

Hence the lower quartile is $$20$$.

(iii)Upper quartile, $$Q_3 = (3n/4)^{th} term$$

$$= (3\times48/4)^{th} term$$

$$= (3\times12)^{th} term$$

$$= 36^{th} term$$

$$= 27$$

Hence the upper quartile is $$27$$.

(iv)Interquartile range $$= Q_3-Q_1$$

$$= 27-20$$

$$= 7$$

Hence the Interquartile range is $$7$$.

We have,

Since the marks that occur the most are 3535 and 3737.
So, the mode of the given data is 3535 and 3737.

$$\textbf{Step -1: Find the mean of first 50 even natural numbers.}$$

                $$\text{First }50\text{ even natural nos. are:}$$

                $$2,4,6,8,\ldots,100$$

                $$\text{Sum of observations}=2+4+6+8+\ldots+100$$

                $$=2(1+2+3+4+\ldots+50)$$

                $$=2\left(\dfrac{50\times51}{2}\right)$$          $${\left[\because\text{Sum of n natural nos.}=\dfrac{n(n+1)}{2}\right]}$$

                $$=50\times51$$

                $$=2550$$

                $$\text{Number of observations, }n=50$$

                $$\text{Mean, }\overline{x}=\dfrac{2550}{50}$$

                $$=51$$

$$\textbf{Step -2: Find the required variance.}$$

                $$\sum_{i=1}^{50} x_i^2=2^2+4^2+6^2+\ldots+100^2$$

                $$=\sum_{k=1}^{50} (2k)^2$$

                $$=4\sum_{k=1}^{50}k^2$$

                $$=4\times\dfrac{50(50+1)(2(50)+1)}{6}$$          $${\left[\because\sum_{k=1}^{n} k^2=\dfrac{n(n+1)(2n+1)}{6}\right]}$$

                $$=\dfrac{2\times50\times51\times101}{3}$$

                $$=171700$$

                $$\because\text{Variance, }\sigma^2=\dfrac{\sum x_i^2}{n}-\overline{x}^2$$

                $$\therefore\sigma^2=\dfrac{171700}{50}-51^2$$

                          $$=\dfrac{171700-130050}{50}$$

                          $$=\dfrac{41650}{50}$$

                          $$=833$$

$$\textbf{Hence, the variance of first 50 even natural numbers is 833.}$$

Here the total number of observations are $$47$$ which is odd.

So median $$=((n+1)/2)^{th}term$$

                  $$= ((47+1)/2 )^{th} term$$

                  $$= (48/2)^{th} term$$

                  $$= 24^{th} term$$

                   $$= 48$$ [Since $$23^{rd}$$ to $$29^{th}$$ observation is $$48$$]

Hence the median of the given data is $$48$$.

Arranging the numbers in ascending order :

$$3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81$$

Median is the middle term of the series, when the series is arranged in ascending/ descending order.

Here, $$n = 12$$ which is even. For a sequence with even number of terms, the median is given by the following formula:

$$\text{Median} = \dfrac{1}{2}\left( \frac{n}{2}^{th}\ \text{term} + \left(\frac{n}{2}+1\right)^{th}\ \text{term}\right)$$

               $$=\dfrac{1}{2}\left( \frac{6}{2}^{th}\ \text{term} + \left(\frac{6}{2}+1\right)^{th}\ \text{term}\right)$$

               $$= \dfrac{1}{2}(6^{th} \text{term} + (6+1)^{th} \text{term})$$

               $$= \dfrac{1}{2}(6^{th} \text{term} + 7^{th} \text{term})$$

                $$= \dfrac{1}{2}(15+17)$$

                $$=  \dfrac{1}{2}\times32$$

                $$= 16$$

Hence, the median is $$16$$.

Arranging the data in the ascending order

$$1,2,2,3,4,5,5,6,7,7,8$$

Here number of terms, $$n = 11$$

Here n is odd.

So median $$= [(n+1)/2 ]^{th}$$ observation

$$= (11+1)/2$$

$$= 12/2$$

$$=6^{th}$$ observation

Here $$6^{th}$$ observation is $$5$$.

Hence the median is $$5$$.

Arranging the numbers in ascending order :

$$0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9$$

Here, $$n = 12$$ which is even

$$Median = \frac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$$

$$= \dfrac{1}{2}(12/2)^{th} term + ((12/2)+1)^{th} term)$$

$$=  \dfrac{1}{2} (6^{th} term + (6+1)^{th} term)$$

$$=\dfrac{1}{2} (6^{th} term + 7^{th} term)$$

$$= \dfrac{1}{2} (4+5)$$

$$= 9/2$$

$$= 4.5$$

Hence the median is $$4.5$$.

We write the marks in cumulative frequency table.

Here the number of observations, $$n = 25$$ which is odd.

$$Median = ((n+1)/2)^{th} term$$

$$= (25+1)/2$$

$$= 26/2$$

$$= 13^{th} term$$

$$= 3$$

Hence his median score is $$3$$.

Here total number of observations, $$n = 70$$ which is even.

$$Median =\dfrac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$$

$$= \dfrac{1}{2}(70/2^{th} term + ((70/2)+1)^{th} term)$$

$$= \dfrac{1}{2}(35^{th} term + (35+1)^{th} term)$$

$$= \dfrac{1}{2} (35^{th} term + 36^{th} term)$$

$$= \dfrac{1}{2} (50+50)$$ [Since all observations from $$21^{st}$$ to $$38^{th}$$ are $$50$$]

$$= \dfrac{1}{2} \times 100$$

$$= 50$$

Hence the median is $$50$$.

Given set is $$5, 7, 6, 8, 9, 0, 6, 8, 1, 8$$.

In this set, $$8$$ occurs maximum number of times.

Hence the mode is $$8$$.

Mode is the number that appears most often in a set of numbers.

(i)Given set is $$3, 2, 0, 1, 2, 3, 5, 3$$.

In this set, $$3$$ occurs the maximum number of times.

Hence the mode is $$3$$.

Here total number of observations, $$n = 36$$ which is even.

$$Median = \dfrac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$$

$$= \dfrac{1}{2}(36/2^{th} term + ((36/2)+1)^{th} term)$$

$$=\dfrac{1}{2}(18^{th} term + (18+1)^{th} term)$$

$$=\dfrac{1}{2}(18^{th} term + 19^{th} term)$$

$$=\dfrac{1}{2}(64+64)$$ [Since $$17^{th}$$ to $$26^{th}$$ observation is $$64$$]

$$= \dfrac{1}{2}\times128$$

$$= 64$$

Hence the median is $$ 64$$.

Arranging the observations in ascending order,

$$21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53$$

As there are $$19$$ observations which is odd so, $$ \left(\dfrac{n+1}2\right)^{th}$$ observation will be the median.

$$ \left(\dfrac{n+1}2\right)=\dfrac{19+1}2$$

                   $$=\dfrac{20}2$$

                   $$=10$$

Hence, $$10^{th}$$ observation that is $$31$$ is the median of the data set.

Given: Marks scored by the boy in 11 subjects$$: 15, 17,16,7,10,12,14,16,19,12,16$$

Ascending order of the marks$$:\ 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19$$

Here the number of observations, $$n = 11$$ which is odd.

Median $$ = \bigg( \dfrac{n+1}{2} \bigg)^{th} \mathrm{term}$$

             $$= \dfrac{11+1}{2}$$

             $$= \dfrac{12}{2}$$

             $$= 6^{th}\ \mathrm{term}$$

             $$= 15$$

Hence the median marks of the boy is $$15$$

Given set is $$9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7$$.

In this set, $$5$$ occurs the maximum number of times.

Hence the mode is $$5$$.

The given
observations are already in ascending order.

Adding a column corresponding to cumulative
frequencies of the given data, we obtain the

 Here $$\mathrm{N}=26$$
which is even

Median is
the mean of 13th and 14th observations. Both of these observations lie in the
cumulative frequency 14, for which the corresponding observation is 7

$$\therefore
Median=\dfrac{13^{\text {th }} \text { observation }+14^{\text {th }} \text {
observation }}{2}=\dfrac{7+7}{2}=7$$

The absolute values of the deviations from median,
ie. $$\left|x_{i}-M\right|$$, are

$$\sum
\limits _{i=1}^{6} f_{i}=26$$ and $$\sum \limits _{i=1}^{6} f_{i}\left|x_{i}-M\right|=84$$

M.D. $$(M)=\dfrac{1}{N}
\sum \limits _{i=1}^{6} f_{i}\left|x_{i}-M\right|=\dfrac{1}{26} \times 84=3.23$$