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Mean - class-X

Description: mean
Number of Questions: 34
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Tags: mean and median statistics measure of central tendency measures of central tendency maths
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What is the value of mean for the following data:

Marks No. of Student
$5-14$ $10$
$15-24$ $18$
$25-34$ $32$
$35-44$ $26$
$45-54$ $14$
$55-64$ $10$
  1. $30$

  2. $29$

  3. $33.68$

  4. $34.21$


Correct Option: C

Mr. R purchases $20\ kgs$. of Wheat, $10\ kgs.$ of rice and $2\ kgs.$ of ghee every month. If the price of wheat is $Rs. 10$ per kg., price of rice is $Rs. 14$ per kg. and price of ghee is $Rs. 120$ per kg. Find the average price per kg. per month. Arithmetic mean $=$?

  1. $48$

  2. $18.125$

  3. $48.125$

  4. $18$


Correct Option: B

If there are two groups containing $30$ and $20$ observations and having $50$ and $60$ as arithmetic means, then the combined arithmetic mean is ________.

  1. $55$

  2. $56$

  3. $54$

  4. $52$


Correct Option: C

What is the value of mean for the following data.

Class interval Frequency
$350-369$ $15$
$370-389$ $27$
$390-409$ $31$
$410-429$ $19$
$430-449$ $13$
$450-469$ $6$
  1. $400$

  2. $400.58$

  3. $394$

  4. $394.50$


Correct Option: B
Explanation:
Class Intervals Mid-values(x) Frequency(f) fx
$350-369$ $359.5$ $15$ $5,392.5$
$370.389$ $379.5$ $27$ $10,246.5$
$390-409$ $399.5$ $31$ $12,384.5$
$410-429$ $419.5$ $19$ $7,970.5$
$430-449$ $439.5$ $13$ $5,713.5$
$450-469$ $459.5$ $6$ $2,757$
$\displaystyle\sum f=111$ $\displaystyle\sum fx=44,464.5$

Arithmetic mean $=44,464.5/111=400.58$.

The following is the distribution of weekly wages of workers in a factory. Calculate the arithmetic mean of the distribution.

Weekly Wages (Rs.) No. of Workers
$240-269$ $7$
$270-299$ $19$
$300-329$ $27$
$330-359$ $15$
$360-389$ $12$
$390-419$ $12$
$420-449$ $8$
  1. $352.3$

  2. $344.5$

  3. $226.7$

  4. $336.7$


Correct Option: D
Explanation:
Class Intervals Mid-values (x) Frequency(f) fx
$240-269$ $254.5$ $7$ $1,781.5$
$270-299$ $284.5$ $19$ $5,405.5$
$300-329$ $314.5$ $27$ $8,491.5$
$330-359$ $344.5$ $15$ $5,167.5$
$360-389$ $374.5$ $12$ $4,494$
$390-419$ $404.5$ $12$ $4,854$
$420-449$ $434.5$ $8$ $3,476$
$\displaystyle\sum f=100$ $\displaystyle\sum fx=33,670$

Arithmetic mean $=33,670/100=336.7$.

The mean weight of $98$ students is found to be $50$ lbs. It is later discovered that the frequency of the class interval $(30-40)$ was wrongly taken as $8$ instead of $10$. Calculate the correct mean.

  1. $49.00$

  2. $49.50$

  3. $49.25$

  4. $49.70$


Correct Option: D
Explanation:

Incorrect mean,
$X=50$kg.
$\displaystyle\sum f _i=98$
Incorrect X$=\displaystyle\frac{Incorrect \displaystyle\sum f _iX _i}{\displaystyle\sum f _i}$
$50=\displaystyle\frac{Incorrect \displaystyle\sum f _iX _i}{98}$
$\therefore$ Incorrect $\displaystyle\sum f _iX _i=98\times 50=4900$
Now, Correct $\displaystyle\sum f _iX _i=$Incorrect $\displaystyle\sum f _iX _i-(8\times 35)+(10\times 35)$
Note, the class-mark of class interval $(30-40)$ is $35$ and for the calculation of the mean, we consider class marks.
Correct $\displaystyle\sum f _iX _i=4900-280+350$
$=4,970$
Also, Correct $\displaystyle\sum f _i=98+2=100$
$\therefore$ Correct Mean$=\displaystyle\frac{Correct \displaystyle\sum f _iX _i}{Correct \displaystyle\sum f _i}$
$=\displaystyle\frac{4970}{100}$
$X=49.70$lbs.

Consider the following frequency distribution.

Class Intervals $0-10$ $10-20$ $20-30$ $30-40$
Frequency $8$ $10$ $12$ $15$

Arithmetic mean $=$?

  1. $39.65$

  2. $22.55$

  3. $32.55$

  4. $23.56$


Correct Option: B
Explanation:
Class Intervals Mid-values(x) Frequency(f) fx
$01-0$ $5$ $8$ $40$
$10-20$ $15$ $10$ $150$
$20-30$ $25$ $12$ $300$
$30-40$ $35$ $15$ $525$
$\displaystyle\sum f=45$ $\displaystyle\sum fx =1,015$

Arithmetic mean $=1,015/45=22.55$.

Coefficient of variation of a distribution is $60$ and its standard deviation is $21$, then its arithmetic mean is?

  1. $36$

  2. $37$

  3. $35$

  4. $38$


Correct Option: A

Arithmetic mean for grouped data can be calculated by _________.

  1. direct method

  2. assumed mean method

  3. step deviation method

  4. all of the above


Correct Option: D
Explanation:

Arithmetic mean refers to the average amount in a given group of data. There are many ways to calculate arithmetic mean like direct method where all the data are added up and then divided by the number of figures in the data in order to ascertain the mean class or assumed mean method and step deviation method, the data of the given class is reduced into smaller units which makes it easy to do calculation and ascertain the mean of the class. 

What needs to be done for calculating mean for a continuous series?

  1. Mid-points of various class intervals are taken

  2. Lower class limits are taken

  3. Upper class limits are taken

  4. A or B or C


Correct Option: A
Explanation:

To calculate the mean of a continuous series, mid points of the various class intervals is taken. For example, if the class is like 10-20 then before calculating the mean mid point that is 15 is calculated for the whole series which is added and divided by the number of terms in order to ascertain the mean. 

For grouped data, Arithmetic mean by Direct Method =

  1. sfX / sf

  2. sd / N

  3. sX / N

  4. None of the above


Correct Option: A
Explanation:

Arithmetic mean refers to the average amount in a given group of data. There are many ways to calculate arithmetic mean for grouped data like direct method where all the data are multiplied with their respective frequencies and then added up which are then divided by the summation of the frequencies or number of figures in the data in order to ascertain the mean. The formula is sfX/ sf where sfd is the summation of frequency multiplied by X for all figures and sf is the frequency or the number of element in the given data. 

For grouped data, Arithmetic mean by Assumed Mean Method =

  1. A + sd/N

  2. A + sfd/sf

  3. sfX/sf

  4. None of these


Correct Option: B
Explanation:

In assumed mean method, any value can be taken as assumed mean whether it is there in the data or not but it should be centrally located in the data so that to simply the big figures in the data in order to ascertain mean of the given data through easy calculations. For grouped data, the formula for assumed mean is A+ sfd/sf where A is the assumed mean, sfd is the summation of frequency multiplied with X-A for all figures and sf is the summation of frequency or the number of element in the given data. 

If 150 is divided in the ratio of 2:3:5, the distribution will be ____.

  1. 45:55:50

  2. 50:80:20

  3. 30:45:75

  4. 50:40:60


Correct Option: C
Explanation:

If 150 is divided in the ratio of 2:3:5, then 

= 150 * 2/10 : 150 * 3/10 : 150 * 5/10 
= 15 * 2 : 15 * 3 : 15 * 5 
= 30 : 45 : 75 

9 times of a number is equal to its square find the number_________.

  1. 8

  2. 11

  3. 9

  4. 10


Correct Option: C
Explanation:

Let the number in the given problem be x. Then, according to the problem 

9x = x2

=> x2 - 9x = 0

=>x ( x-9 ) =0

=> x= 0 or x= 9  

The arithmetic mean of the first 100 natural numbers is _____.

  1. 50

  2. 52

  3. 51

  4. 50.5


Correct Option: D
Explanation:

Arithmetic mean refers to the average amount in a given group of data. So arithmetic mean can be calculated by adding the first term and the last term of the series and then dividing the sum by 2. In the given series the first term is 1 and  the last term is 100, so the 

Arithmetic mean = ( 1+100 ) /2 
                             = 101 /2 
                             = 50.5 

The mean of a sample of size 10 is 15. If the value of each item is doubled, the mean of the sample will be _______.

  1. 15

  2. 30

  3. 11

  4. 22


Correct Option: B
Explanation:

Mean refers to the average amount in a given group of data. So arithmetic mean can be calculated by adding the first term and the last term of the series and then dividing the sum by 2. In the given series the first term 'a' is doubled and  the last term 'b' is also doubled , so the 

Mean = {(a+a)+ (b+b)}  /2 

          = (2a+2b) /2 

          = 2 (a+b) /2 

          = 2 [ (a+b)/2} 

Therefore, the mean is also doubled. So,if the mean was 15 then now it will be 30.  

The sum of first 10 natural numbers is____.

  1. 100

  2. 55

  3. 50

  4. 90


Correct Option: B
Explanation:

The series of first 10 natural numbers is an arithmetic progressions with first tern as 1 and common difference 1. So the sum of the series will be Sn = n/2 { 2a+ ( n-1 ) d } where n is the number of terms in the series, a is the first term and d is the common difference.

S10= 10/2 { 2(1) + ( 10-1 ) 1 }

     = 5 ( 2+9)

    = 5 ( 11 )

    = 55

If the frequency of observations $X _{i}$ is $f _{i} (i = 1, 2, ..... n)$.

  1. $\overline {X} = \dfrac {X _{1}\times X _{2} \times X _{3}\times ......X _{n}}{N}$

  2. $\overline {X} = \dfrac {X _{1} + X _{2} + X _{3} + ......X _{n}}{N}$

  3. $\overline {X} = \dfrac {\sum X _{i}}{N}$

  4. $\overline {X} = \dfrac {\sum f _{i}X _{i}}{\sum f}$


Correct Option: D

Consider following frequency distribution.

Class Intervals $0-10$ $10-20$ $20-30$ $30-40$
Frequency $8$ $10$ $12$ $15$

Arithmetic mean $=$?

  1. $39.65$

  2. $22.55$

  3. $32.55$

  4. $23.56$


Correct Option: B
Explanation:

0−10 | 10−20 | 20−30 | 30−40 | | --- | --- | --- | --- | --- | | Frequency(f) | 8 | 10 | 12 | 15 | | Class mark ( mid points= x) | 5 | 15 | 25 | 35 | | Fx | 40 | 150 | 300 | 525 |

Mean = summation of fx / summation of f

          = 1015/ 45

          = 22.55

The mean salary for a group of $40$ females workers is $Rs. 5,200$ per month and that for a group of $60$ male workers is $Rs. 6,800$ per month. What is the combined salary?

  1. $6,280$

  2. $6,160$

  3. $6,890$

  4. $6,920$


Correct Option: B
Explanation:

The mean salary of 40 females and 60 males are 5,200 and 6,800 rupees respectively. Therefore the total salary of 40 females and 60 males will be 5,200x40 and 6,800x60 rupees respectively. So the average salary of 40 females and 60 males will be

= ( 2.08,000 + 2,08,000 )/ ( 40+60)
= 616000 / 100
= 6,160

The average marks of three batches of students having $70, 50$ and $30$ students respectively are $50, 55$ and $45$. Find the average marks of all the $150$ students, taken together.

  1. $50.67$

  2. $40.67$

  3. $50.60$

  4. $40.60$


Correct Option: A
Explanation:

The average marks of 70, 50 and 30 students respectively is 50, 55 and 45 marks. Therefore the total marks of 70, 50 and 30 students will be 70x50, 50x55, and 30x45 respectively. So the average marks of all the 150 students is 

= ( 3500 + 2750 + 1350 )/ ( 70+50+30)
= 7,600 / 150
= 50.67

Only __________ can be used for all algebraic calculations.

  1. mean

  2. mode

  3. median

  4. All of the above


Correct Option: A
Explanation:

Arithmetic mean refers to the average amount in a given group of data. It is defined as the summation of all the observation in the data which is divided by the number of observations in the data. Since mean includes all the values of the series of observation therefore only this average value is included in the algebraic calculations. 

Which of the following statistics would give you the estimate of the typical examination score of a class of 35 students?

  1. A correlation coefficient

  2. The variance

  3. The standard deviation

  4. The mean


Correct Option: D
Explanation:

Arithmetic mean refers to the average amount in a given group of data. There are many ways to calculate arithmetic mean like direct method where all the data are added up and then divided by the number of figures in the data in order to ascertain the mean class or assumed mean method and step deviation method, the data of the given class is reduced into smaller units which makes it easy to do calculation and ascertain the mean of the class. 

Measures of central tendency are _______________.

  1. inferential statistics that identify the best single value for representing a set of data

  2. descriptive statistics that identify the best single value for representing a set of data

  3. inferential statistics that identify the spread of the scores in a data set

  4. descriptive statistics that identify the spread of the scores in a data set


Correct Option: B
Explanation:

Measures of central tendency is a mathematical concept that measures the central most value of a given series which can represent the whole set of the series. In a given series, the central location of the observation is measured through different tools in order to ascertain the most efficient value that can represent the series.

Three geometric means between 5 and 3125 are___.

  1. 15,75,375

  2. 25,125,625

  3. 11,44,176

  4. 10,40,160


Correct Option: B
Explanation:

The geometric mean is a geometric series which is in the form of arn where a is the first term and r is the common ratio.

T1= 5 and T5 = 3125

Now, T1/T5 = 5/ 3125

=>ar1//ar5 =1/ 625

=> 1/ r4 = 1/625

=> r4 = 625

=> r = 5

So if r=5, then a= 1

T2 = ar2

      = 1(5) 2 = 25

T3 = ar3

      = 1(5) 3 = 125

T4 = ar4

      = 1(5)4 = 625

Therefore, three geometric means between 5 and 3125 are 25, 125 and 625. 

The most appropriate average to be used to compute the average rate of growth in population is:

  1. Arithmetic mean

  2. Median

  3. Geometric mean

  4. Harmonic mean


Correct Option: C
Explanation:

In geometric mean method, the average percentage increase in the population is assumed to be constant decade to decade through which future population is worked out. Therefore, it is the most appropriate average to be used to compute the average rate of growth in population. 

The mean of a sample of size $10$ is $15$. If the value of each item is reduced by $2$, the mean of the sample will be_____.

  1. $15$

  2. $13$

  3. $11$

  4. $22$


Correct Option: B
Explanation:

Mean refers to the average amount in a given group of data. So arithmetic mean can be calculated by adding the first term and the last term of the series and then dividing the sum by 2. In the given series the first term 'a' is decreased by 2 and  the last term 'b' is also decreased by 2 , so the 

Mean = {(a-2)+ (b-2)}  /2 

          = (a+b-4) /2 

          = {(a+b)/2} - 4/2

          = {(a+b)/2} - 2 

Therefore, the mean is also decreased by 2. So,if the mean was 15 then now it will be 13. 

If the average weight of $150$ students is $60$ kg., what will be the number of boys and girls in the school if the average weight of boys and girls is $70$kg and $55$kg. respectively?

  1. $(50,100)$

  2. $(100,50)$

  3. $(40,110)$

  4. $(90,60)$


Correct Option: A

Harmonic mean is a part of _______________.

  1. Positional average

  2. Mathematical average

  3. Both a & b

  4. None of the above


Correct Option: B
Explanation:

Mathematical average refers to all such average where a figure is taken out through mathematical methods from the a given series that represents the whole series. Harmonic mean is a mathematical tool which is used to calculate average of a certain series. Therefore, it is a part of mathematical average. 

If the mean of x and $\displaystyle \frac{1}{x}$ is M, then the mean of x$^3$ and $\displaystyle \frac{1}{x^3}$ is

  1. $\displaystyle \frac{M (M^2 - 3)}{2}$

  2. $M (4M^2 - 3)$

  3. $M^3$

  4. $M^3 + 3$


Correct Option: B
Explanation:

Given,

$M=\dfrac{x+\dfrac{1}{x}}{2}$
$2M=x+\dfrac{1}{x}$                  $ ....... (1)$

On cubing both sides, we get
$(2M)^3=(x+\dfrac{1}{x})^3$
$8M^3=x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})$
$8M^3=x^3+\dfrac{1}{x^3}+3(2M)$
$8M^3=x^3+\dfrac{1}{x^3}+6M$
$x^3+\dfrac{1}{x^3}=8M^3-6M$
$\dfrac{x^3+\dfrac{1}{x^3}}{2}=M(4M^2-3)$

Hence, this is the answer.

Sum of squares of deviation of $10$ observations measured from $5$ is $17$ and sum of squares of observations is $170$ then mean of observation is

  1. $40.3$

  2. $4.5$

  3. $4$

  4. $4.03$

  5. $4.3$


Correct Option: D
Explanation:

$x _1,x _2,x _3,x _4...x _10\x _1^2+x _2^2+x _3^2+x _4^2+....x _10^2=17 (Given) \rightarrow (i)$

$(x _1-5)^2+(x _2-5)^2+(x _3-5)^2+....+(x _10-5)^2=17(Given)\rightarrow (i)$
$Mean (M)=\cfrac{x _1+x-2+x _3+....x _10}{10}\ \Rightarrow x _1+x _2+x _3+....+x _10=10M\rightarrow(iii)$
From equation $(ii)$ we get
$x _1^2+25-10x _1+x _2^2+25-10x _2+x _3^2+25-10x _3+...+x _10^2+25-10x _10=17\ \Rightarrow (x _1^2+x _2^2+....x _10^2)-10(x _1+x _2+x _3+....+x _10)+25\times10=17\ \Rightarrow 170-10(10M)+250=17\ \Rightarrow420-100M=17\ \Rightarrow100M=403\ \Rightarrow M=\cfrac{403}{100}=4.03$

The sum of the deviations of a set of values $x 1, x _2$, ...... $x _n$ measured from $50$ is $-10$ and the sum of deviations of the values from $46$ is $70$. The mean is __________.

  1. $49$

  2. $49.5$

  3. $49.75$

  4. $50$


Correct Option: B
Explanation:

Sum of deviations from $50$ is $-10$


$\Rightarrow \sum(xi - 50) = -10$
     $\sum x _i - 50\sum1 = -10$
     $\sum x _i - 50n = -10$

$\therefore y-50n=-10.....(1)$

Sum of deviations from $46$ is $70$

$\Rightarrow \sum(x _i - 46) = 70$
     $\sum x _i - 46\sum1 = 70$
     $\sum x _i - 46n = 70$

$\therefore y-46n=70.....(2)$


Solving $(1)$ and $(2)$, we get
$4n = 80$ i.e. $n=20$

Putting value of $n$ in $(1)$, we get
$y=990$

Mean $= \dfrac{\sum x _i}{n} = \dfrac{y}{n} = \dfrac{990}{20} = 49.5$

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?

  1. $109$

  2. $106$

  3. $150$

  4. $160$


Correct Option: B
Explanation:

Let m be the mean and s be the standard deviation and find the z score.
$z = (x - m) /s = (0.8 s + m - m) / s = 0.8$
The percentage of student who scored above Jane is (from table of normal distribution).
1 - 0.7881 = 0.2119 = 21.19%
The number of student who scored above Jane is (from table of normal distribution).
21.19% 0f 500 = 106

Following table gives frequency distribution of milk (in litres) given per week by 50 cows.
Find average (mean) amount of milk given by a cow by 'shift of origin method'.

Milk (in litres) 24 - 30 30 - 36 36 - 42 42 - 48 48 - 54 54 - 60 60 - 66 66 - 72  72 - 78 78 - 84 84 - 90
No. of cows 1 3 8 5 5 5 8 4 6 2 3
  1. $51.12$ litres

  2. $54.12$ litres

  3. $57.12$ litres

  4. $60.12$ litres


Correct Option: C
Explanation:

Consider the following table, to calculate mean by "shift of origin method":

$x _i$=mid value of class interval
Assumed mean $a=57$

 $ci$  $f _i$  $x _i$ $d _i=x _i-a$  $f _id _i$
24-30  1 27  27-57= -30  -30
30-36  3 33  33-57= -24  -72
36-42  8 39  39-57= -18  -144
42-48   5 45   45-57= -12  -60
48-54   5 51  51-57= -6  -30
54-60   5 57  57-57=0  0
60-66   8 63   63-57=6   48
66-72   4 69  69-57=12  48
72-78   6 75   75-57=18  108
78-84  2 81   81-57=24  48
84-90 87   87-57=30  90
 $N=\Sigma f _i=50$          
 $\Sigma f _id _i=6$

Mean $\overline x=a +\dfrac {\Sigma f _id _i}{N}$
$\therefore \overline x=57 + \dfrac{6}{50}=57.12$

Average amount of milk given by cow is $ 57.12$ litres
Hence, option $C$ is correct.

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