Tag: statistics

Questions Related to statistics

To find the concentration of $SO _2$ in the air (in parts, per

million), the data was collected for 30 localities, in a certain city

and is presented below:

Concentration of $SO _2$ (in ppm) Frequency
0.00-0.04 4
0.04-0.08 9
0.08-0.12 9
0.12-0.16 2
0.16-0.20 4
0.20-0.24 2

Find the mean concentrations of $SO _2$ in the air.

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. $0.099$ ppm

  2. $0.09$ ppm

  3. $0.99$ ppm

  4. $0.0909$ ppm

Show answer
Correct Option: A
Explanation:

Consider the following table, to calculate mean:

$ci$ $f _i$  $x _i$  $f _ix _i$ 
$0.00-0.04$ $4$  $0.02$  $0.08$ 
$0.04-0.08$  $9$  $0.06$  $0.54$ 
$0.08-0.12$  $9$  $0.10$  $0.90$ 
$0.12-0.16$  $2$  $0.14$  $0.28$ 
$0.16-020$  $4$  $0.18$  $0.72$ 
$0.20-0.24$  $2$  $0.22$  $0.44$ 
$N=\Sigma f _i=30$          
 $\Sigma f _ix _i=2.96$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{2.96}{30}=0.0986667 \approx 0.099$
mean concentration of $SO _2$ in air is $0.099ppm$
Hence, option $A$ is correct.

The following table gives the per day income of 50 pupils. Find the arithmetic mean of their per day income.

Income/day (Rs) 70-74 74-78 78-82 82-86 86-90
No. of people    8    10      11    17    4


statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. $75.92$

  2. $79.92$

  3. $80.92$

  4. None of these

Show answer
Correct Option: B
Explanation:

Consider the following table, to calculate mean:

 $ci$  $f _i$  $x _i$  $f _ix _i$
70-74  8  72  576
 74-78  10  76  760
 78-82  11  80  880
 82-86  17  84  1428
 86-90  4  88  352
 $N=\Sigma f _i=50$          
 $\Sigma f _ix _i=3996$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{3996}{50}=79.92$

Hence, option $B$ is correct.

Compute the missing frequencies $'f _1'$ and $'f _2'$ in the following data, if the mean is $166\frac {9}{26}$ and the sum of the observation is 52.

Classes Frequency
140-150 5
150-160 $f _1$
160-170 20
170-180 $f _2$
180-190 6
190-200 2
Total 52
statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. $f _1=7, f _2=3$

  2. $f _1=10, f _2=6$

  3. $f _1=9, f _2=8$

  4. None of these

Show answer
Correct Option: D
Explanation:

Given:- $\Sigma f = 52 $

 classes frequency$ (f _i) $  $ x _i = \cfrac{\text{lower limit + upper limit}}{2} $  $ x _i f _i $ 
140-150  5   145  725
150-160  $ f _1 $  155  $ 155f _1 $ 
 160-170 20  165  3300 
170-180  $ f _2 $  175  $175f _2$ 
180-190  6  185  1110 
190-200  2 195  390 
  $ \Sigma f _i = 52 $    $ \Sigma x _i f _i = 5525 + 155f _1 + 175f _2 $ 

Also $ \Sigma f _i = 33  f _1 + f _2 = 52 $

$ \Rightarrow f _1 + f _2 = 19\longrightarrow eq.(i) $
Now Mean = $ \cfrac{\Sigma x _i f _i}{\Sigma f _i} = \cfrac{5525 + 155f _1 + 175f _2}{52} $ 
Given:- Mean = $ \cfrac{4325}{26} $
$ \Rightarrow \cfrac { 5525+155f _{ 1 }+175f _{ 2 } }{ 52 } =\cfrac { 4325 }{ 26 } $
$ \Rightarrow 5525 + 155f _1 + 175f _2 = 8650 $
$ \Rightarrow 155f _1 + 175f _2 = 8650 - 5525 = 3125$
$ 31f _1 + 35f _2 = 625 \longrightarrow eq.(ii) $
from eq. (i) $ & $ (ii), we get
$ f _2 = 9 \Rightarrow f _1 = 10 $
D) None of these

In a frequency dist. if $\displaystyle d _{i}$ is deviation of variates from a number e and mean = $\displaystyle e+\frac{\Sigma f _{i}d _{i}}{\Sigma f _{i}}$, then e is

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. Lower limit

  2. Assumed mean

  3. Number of observation

  4. Class interval

Show answer
Correct Option: B
Explanation:

Formula of finding Mean using step deviation method is

mean = $\displaystyle e+\frac{\Sigma f _{i}d _{i}}{\Sigma f _{i}}$
where,
$e=$Assumed Mean
$\Sigma f _id _i=$Sum of all $frequency(f _i)\times deviation(d _i)$
$\Sigma f _i=$ Sum of all frequencies
Hence the correct answer is assumed mean.

If the mean of four observations is $20$ and when a constant  is added to each observation the mean becomes $22$ The value of $c$ is?

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. $-2$

  2. $2$

  3. $4$

  4. $6$

Show answer
Correct Option: B
Explanation:

$\displaystyle\text{Let } x _1,x _2,x _3,x _4\text{ be four observations.}$
$\displaystyle\text{According to question}$
$\displaystyle \frac{ x _1+x _2+x _3+x _4}{4}=20$
$\Rightarrow \displaystyle { x _1+x _2+x _3+x _4}=80$
$\displaystyle\text{After adding 'c' to each observation the new A.M becomes 22.}$
$\Rightarrow \displaystyle \frac{ (x _1+c)+(x _2+c)+(x _3+c)+(x _4+c)}{4}=22$
$\Rightarrow \displaystyle  (x _1+x _2+x _3+x _4)+4c=88$
$\Rightarrow \displaystyle  80+4c=88$
$\Rightarrow \displaystyle  4c=8$
$\Rightarrow \displaystyle  c=2$
Options B is correct.

HM of 3 and 5 is

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. $\displaystyle\dfrac{15}{4}$

  2. $\displaystyle\dfrac{15}{8}$

  3. $\displaystyle\dfrac{3}{4}$

  4. $\displaystyle\dfrac{5}{8}$

Show answer
Correct Option: A
Explanation:

Harmonic mean of two numbers $ a $ and $ b = \dfrac {2ab}{a+b} $

So, harmonic mean of $ 3\ and\  5 $ is $ \dfrac {2 \times 3 \times 5}{3+5} = \dfrac {15}{4} $

GM of 4 and 64 is

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. 32

  2. 8

  3. 16

  4. 24

Show answer
Correct Option: C
Explanation:

Geometric mean or GM of two numbers $ a $ and $ b $ is $ \sqrt {ab} $
So, GM of $ 4\  and\  64 $ is $ \sqrt {4 \times 64} = 2 \times 8 = 16 $

The harmonic mean of 20 and 30 is

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. 25

  2. 28

  3. 26

  4. 24

Show answer
Correct Option: D
Explanation:

Harmonic mean of two numbers $ a $ and $ b = \dfrac {2ab}{a+b} $

So, harmonic mean of $ 20\  and\  30 $ is $ \dfrac {2 \times 20 \times 30}{20+30} = 24 $

Find the sum of 5 geometric means between $\displaystyle\frac{1}{3}$ and 243, by taking common ratio positive.

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. 121

  2. 126

  3. 81

  4. 111

Show answer
Correct Option: A
Explanation:

Given that, there are $5$geometric means between the two numbers $\dfrac{1}{3}$and $243$ , we have to find $7=\left( 5+2 \right)$

terms in G.P. of which $\dfrac{1}{3}$ is the first, and$243$ the seventh. Let r be the common ratio;

then $243$  = the seventh term =$\left( \dfrac{1}{3} \right){{r}^{\left( 7-1 \right)}}=\dfrac{1}{3}.{{r}^{6}}$.

 

Therefore,${{r}^{6}}=3.x.243={{3.3.3}^{4}}={{3}^{6}}$;

whence $r=6$

and the series is$\dfrac{1}{3},1,3,9,27,81,243$

(using the standard form a, ar, ar², ar³ …… of a G.P. ).

 

Now, the geometric mean between two given quantities$a,b=\sqrt{ab}$

 

Therefore, the required geometric means are,

$ \sqrt{\dfrac{1}{3}.x.3},\sqrt{1.x.9},\sqrt{3.x.27},\sqrt{9.x.82},\sqrt{27.x.243} $$

$ =1,3,9,27,81 $$

 

Therefore, the sum of the $5$  geometric means is

$1+3+9+27+81=121$

 

Hence, this is the answer.

 

The geometric mean of $10$ observations on a certain variable was calculated as $16.2$. It was later discovered that one of the observations was wrongly recorded as $12.9$; infact it was $21.9$. The correct geometric mean is:

statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. $\left (\dfrac {(16.2)^{9}\times 21.9}{21.9}\right )^{1/10}$

  2. $\left (\dfrac {(16.2)^{10}\times 21.9}{21.9}\right )^{1/10}$

  3. $\left (\dfrac {(16.2)^{10}\times 21.9}{12.9}\right )^{1/10}$

  4. $\left (\dfrac {(16.2)^{11}\times 21.9}{21.9}\right )^{1/11}$

Show answer
Correct Option: A
Explanation:

Geometric mean of $n$ numbers $=(\prod _{i=1}^{n} x _{i})^{1/n}$

Here, $(\prod _{i=1}^{10} x _{i})^{1/10}=16.2$
$\Rightarrow (\prod _{i=1}^{10} x _{i})=(16.2)^{10}$
Now suppose $x _{10}$ was wrongly recorded, so we rewrite above relation as $(\prod _{i=1}^{9} x _{i})\times x _{10}=(16.2)^{10}$
$\Rightarrow (\prod _{i=1}^{9} x _{i})=\dfrac{(16.2)^{10}}{x _{10}}$
Now, the correct value is $21.9$, so multiply both sides by $21.9$ and also put value of $x _{10}=12.9$ in above equation
$(\prod _{i=1}^{9} x _{i})\times 21.9=\dfrac{(16.2)^{10}}{12.9}\times 21.9$
$=$ Correct Geometric mean=$((\prod _{i=1}^{9} x _{i})\times 21.9)^{1/10}$
$=\left(\dfrac{(16.2)^{10}}{12.9}\times 21.9\right)^{1/10}$
$=\left(\dfrac{(16.2)^{10}\times 21.9}{12.9}\right)^{1/10}$
Hence, $(C)$ is correct.