Question 12 Marks
The following table shows the number of children per family in a certain area. Find the mean number of children per family.
AnswerHere we have $k = 6$ values for the variable $x.$
The calculations to find mean are shown in the following table :
Mean $\begin{aligned} \bar{x} &=\frac{\sum f x}{n} \\ &=\frac{117}{52} \\ &=2.25 \end{aligned}$
Thus: mean for number of children per family is $2.25$ children. View full question & answer→Question 22 Marks
The number of cold drink bottles sold by a shopkeeper on different days is shown in the following table. Find the mode for the number of cold drink bottles sold.
AnswerThe maximum frequency is $20$ for the class $12-15$.
Hence the modal class is $12-15$.
This is an inclusive type of frequency distribution and hence we will take the boundary points of this class as $11.5-15.5$
$ \begin{aligned} &\text { Taking, } \quad L=11.5, f_{m}=20, f_{1}=16, f_{2}=18, c=4 \\ &\text { Mode } M_{o}=L+\frac{f_{m}-f_{1}}{2 f_{m}-f_{1}-f_{2}} \times c \\ &=11.5+\frac{20-16}{2(20)-16-18} \times 4 \\ &=11.5+\frac{4}{40-16-18} \times 4 \\ &=11.5+\frac{16}{6} \\ &=11.5+2.6666 \\ &=14.1666 \\ &=14.17 \end{aligned} $
Thus, mode for sale of cold drink bottles is $14.17$.
View full question & answer→Question 32 Marks
The following table gives output of workers in a factory. Find the modal output.
AnswerThe maximum frequency is $48$ for the class $190-210$. Hence the modal class is $190-200$.
Taking, $L=190, f_{m}=48, f_{1}=33, f_{2}=22, c=10$,
Mode $\quad \begin{aligned} M_{0} &=L+\frac{f_{m}-f_{1}}{2 f_{m}-f_{1}-f_{2}} \times c \\ &=190+\frac{48-33}{2(48)-33-22} \times 10 \\ &=190+\frac{15}{96-33-22} \times 10 \\ &=190+\frac{150}{41} \\ &=190+3.6585 \\ &=193.6585 \\ &=193.66 \end{aligned}$
Thus, the modal output is $193.66$ items.
View full question & answer→Question 42 Marks
The following table shows the record of absent students of class during a month. Find the median of number of absent days per student.
AnswerWe will find cumulative frequency distribution as shown below :
Here $n=\Sigma f=53$
Median $M=$ value of the $\left(\frac{n+1}{2} \right).$ th observation
$=$ value of the $\left(\frac{53+1}{2} \right).$ th observation
$=$ value of the $27$ th observation
It can be known from the cumulative frequencies that the $21$ st to the $38$ th observations have value $2 .$
Hence, the $27$ th observation has value $2 . $
$\therefore$ median $M=2$ days.
Thus, the median number of absent days is $2$ days. View full question & answer→Question 52 Marks
The rainfall (in mm.) at $10$ different places of a district was recorded as :
$126, 110, 91, 115, 112, 80, 101, 93, 97, 113$
Find mean rainfall.
AnswerAs the observations have large values, we will calculate mean by short cut method. We will select assumed mean $A = 100$. The observations $x$ and deviations $d = x — A$ are shown in the following table :
Here, $n=10$
Mean $\bar{x}=A+\frac{\Sigma d}{n}$
$ \begin{aligned} &=100+\frac{38}{10} \\ &=100+3.8 \\ &=103.8 \end{aligned} $
Thus, the mean rainfall is $103.8 \mathrm{~mm}$. View full question & answer→Question 62 Marks
There are $11$ employees in an office. The monthly salaries (in $3$) of the lowest paid $7$ employees among them are $4500, 2100, 3400, 3600, 2500, 4200, 1500$. What is the median monthly salary of all employees ?
AnswerSome of the observations are missing here. We do not know the salaries of highest paid $4$ employees.
Suppose these values are $a, b, C, d$ in their increasing order. These values are greater than the given observations as these are salaries of highest paid employees.
We will arrange these data in ascending order. $1500, 2100, 2500, 3400, 3600, 4200, 4500, a, b, c, d.$
Here, $n = 11$
Median $M=$ Value of the $\left(\frac{n+1}{2}\right)$ th observation
$=$ Value of the $\left|\frac{11+1}{2}\right|$ th observation
$=$ Value of the $6$ th observation
$=4200$
Thus, the median salary of these employees is $₹ 4200 .$
View full question & answer→Question 72 Marks
The daily profits of a hawker (in f) for last $10$ days are given below. Find the median profit.
$261.5, 257, 258.5, 260, 265, 249, 255.5, 262.5, 264, 267$
AnswerArranging in ascending order, the observations will be as follows $: 249,255.5,257,258.5,260,261.5,262.5,264,265,267$
Median $M=$ value of the $\left(\frac{n-1}{2}\right)$ th observation
$=$ value of the $\left(\frac{10-1}{2}\right)$ th observation
$=$ value of the $5.5$ th observation
$=\frac{\text { value of the } 5 \text { th observation }+\text { value of the } 6 \text { th observation }}{2}$
$=\frac{260+261.5}{2}$
$=260.75$
Thus, median daily profits of the hawker is ₹ $260.75$.
View full question & answer→Question 82 Marks
The numbers of items produced by a factory in different weeks are $80, 85, 90, 92, 68 80 72, 63, 55.$ Find the median production.
AnswerArrangement of observations in ascending order is as follows :
$55,63,68,72,80,80,85,90,92$
Here, $n=9$
Median $M=\left(\frac{n+1}{2}\right)$ th observation
$=$ value of the $\left(\frac{9+1}{2}\right)$ th observation $=$ value of the $5$ th observation $=80$
Thus, the median production of this factory is $80$ items.
View full question & answer→Question 92 Marks
The population of an area increased by $15 \%, 18 \%, 13 \%, 20 \%$ respectively for four years. Find the average rate increase in the population.
AnswerSince the values of increase in population are given in percentages, we will use the geometric mean for the average value.
Considering the percentage increase in the population, we will get the observations
$\begin{aligned} x_{1}=100 &+15=115 \quad x_{2}=100+18=118, \\ x_{3}=100 &+13=113 \quad x_{4}=100+20=120 \\ G &=\sqrt[4]{x_{1} \times x_{2} \times x_{3} \times x_{4}} \\ &=\sqrt[4]{115 \times 118 \times 113 \times 120} \\ &=\sqrt[4]{184009200} \\ &=\sqrt{13564.9991} \\ &=116.4689 \\ &=116.47 \end{aligned}$
Thus, the average rate of growth in the population over these four years is $16.47 \%.$
Note : To find the $4th$ root in this example, the square root of $184009200$ is taken and its square root is taken again. Similarly, to find the 8th root of a number the process of taking square root should be repeated three times.
View full question & answer→Question 102 Marks
A student gets $35$ marks in theory paper, $15$ marks in practical examination and $5$ marks in oral examination of a subject. The school gives weights $4, 2$ and $1$ respectively to these types of examinations. Find the weighted mean of marks for this student.
Answer$ \begin{aligned} \text { Here } x_{1} &=35, & x_{2} &=15, & x_{3} &=5 \text { and } \\ w_{1} &=4, & w_{2} &=2, & w_{3} &=1 \end{aligned} $
$ \text { Weighted mean } \begin{aligned} \bar{x}_{w} &=\frac{w_{1} x_{1}-w_{2} x_{2}-w_{3} x_{3}}{w_{1}-w_{2}-w_{3}} \\ &=\frac{4(35)+2(15)+1(5)}{4+2+1} \\ &=\frac{140+30+5}{7} \\ &-\frac{175}{7} \\ &=25 \end{aligned} $
Thus, the weighted mean of marks of this student is $25 .$
View full question & answer→Question 112 Marks
The ratio of women and men employees in an office is $1:2.$ If the means of ages of women and men are $34$ years and $37$ years respectively, find the mean age of all the employees of the office.
AnswerSuppose the no. of women $=a$. Then the no. of men $=2 a$ as the ratio is $1: 2$.
Taking $n_{1}=a$ and $n_{2}=2 a, \bar{x}_{1}=34$ and $\bar{x}_{2}=37$
$ \begin{aligned} \text { Combined mean } \bar{x}_{c} &=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}} \\ &=\frac{a(34)+2 a(37)}{a+2 a} \\ &=\frac{34 a+74 a}{3 a} \\ &=\frac{108 a}{3 a} \\ &=36 \end{aligned} $
Thus, mean age of all the employees in the office is $36$ years.
View full question & answer→Question 122 Marks
A factory owner knows that the mean monthly production from January to March is $350$ items, from April to August it is $254$ items and from September to December it is $315$ items. Find the mean monthly production for that year.
AnswerHere $n_{1}=3$ months, $n_{2}=5$ months, $n_{3}=4$ months,
$ \bar{x}_{1}=350, \quad \bar{x}_{2}=254, \quad \bar{x}_{3}=315 $
Combined mean $\bar{x}_{c}=\frac{n_{1} \bar{x}_{1}-n_{2} \bar{x}_{2}-n_{3} \bar{x}_{3}}{n_{1}-n_{2}-n_{3}}$
$ \begin{aligned} &=\frac{3(350)-5(254)-4(315)}{3-5-4} \\ &=\frac{1050+1270+1260}{12} \\ &=\frac{3580}{12} \\ &=298.3333 \\ &=298.33 \end{aligned} $
Thus, mean monthly production of the factory over the year is $298.33$ items.
View full question & answer→Question 132 Marks
Mean weekly production $(x)$ of factory is $81$ units. Find the mean production cost if cost is given by $y = 3x + 50.$
AnswerMean of production $( x ) \bar{x}=81$
Production cost $y=3 x+50$
$\therefore$ Mean of production cost $\bar{y}=3 \bar{x}+50$
Putting $\bar{x}=81$,
$\bar{y}=3(81)+50=243+50=293$
View full question & answer→Question 142 Marks
Geometric mean of two numbers is $8$, If one number is $4$, find the other number.
Answer$G=8, x_1=4 \text {. }$
Suppose, the another number is $x_2$.
$G=\sqrt{x 1 \cdot x 2}$
$\therefore 8=\sqrt{4 \cdot X 2}$
$\therefore 64=4 X 2$
$\therefore X 2=\frac{64}{4}=16$
Hence, another number is $16.$
View full question & answer→Question 152 Marks
The mean and mode of a variable are $5.5$ and $6.4$ respectively. Find its median.
Answer$\bar{x}=5.5, M _0=6.4, M =\text { ? }$
Using the empirical formula.
$M_0=3 M-2 \bar{x}$
$\therefore 6.4=3 M-2(5.5)$
$\therefore 6.4+11.0=3 M$
$\therefore M=\frac{17.4}{3}=5.8$
View full question & answer→Question 162 Marks
What are the factors to be considered while choosing an appropriate average ?
AnswerWhile choosing an appropriate average the following factors are to be considered :
$(1)$ The nature of data.
$(2)$The nature of variable Involved,
$(3)$ The purpose of study,
$(4)$ The type of classification used and
$(5)$ The need of average for further stastical analysis.
View full question & answer→Question 172 Marks
Which type of data have median as a better measure of central tendency than mean ?
AnswerWhen the data given is of qualitative nature. median is better measure of central tendency than mean. If the observations of the data are not uniformly distributed around the average, then median is better than mean. When continuous frequency distribution has open ended classes median is a better measure of central tendency than mean.
View full question & answer→Question 182 Marks
Explain the combined mean.
AnswerIf the number of observations and mean of two or more groups are given, then the mean of the combined group is called the combined mean. It is denoted by $\overline{X C}$.
Suppose, k groups having the observations $n _1, n _2 \ldots \ldots . . n _{ K }$ have means respectively $\bar{X} 1, \overline{X 2}, \overline{X 3}$, $\overline{X k}$. then their combined mean is obtained as follows :
$\overline{X c}=\frac{\overline{n 1 X} 1+n 2, \overline{X_2}+n 3, \overline{X 3}, \ldots \ldots}{n 1+n 2+\ldots,+n k}$
View full question & answer→Question 192 Marks
State the advantages of mode.
AnswerThe advantages of mode are as follows :
$(1)$ Mode is easy to understand and calculate.
$(2)$ Its value can be found merely by inspection.
$(3)$ Its value is not affected by too large or too small values.
$(4)$ It can be found graphically.
View full question & answer→Question 202 Marks
Explain what is meant by measure of central tendency.
AnswerIn classified data the values of variable are cancentrated around a certain central value. This characteristic of data is called as Central Tendency
View full question & answer→Question 212 Marks
Find the weighted mean of variable $x$ from the following data :
| Veriable $x$ |
$1500$ |
$800$ |
$200$ |
| Weight $w$ |
$5$ |
$4$ |
$1$ |
Answer
| Veriable $x$ |
Weight $y$ |
$Wx$ |
| $1500$ |
$5$ |
$7500$ |
| $800$ |
$4$ |
$3200$ |
| $200$ |
$1$ |
$200$ |
| Total |
$\sum w=10$ |
$\sum w x=10,900$ |
Weighted mean of $x$ :
$\bar{X} w=\frac{\sum w x}{\sum w}=\frac{10900}{10}=1090$
Hence, weighted mean of $x=1090$. View full question & answer→Question 222 Marks
The number of vehicles per family in families residing in a certain area are given in the following table. Find the mean for the number of vehicles.
| No. of vehicles |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
Total |
| No. of families |
$2$ |
$4$ |
$9$ |
$7$ |
$3$ |
$25$ |
Answer
| No. of vehicles $x$ |
No. of families $f$ |
Cumulative frequency $cf$ |
| $0$ |
$2$ |
$2$ |
| $1$ |
$4$ |
$6$ |
| $2$ |
$9$ |
$15$ |
| $3$ |
$7$ |
$22$ |
| $4$ |
$3$ |
$25$ |
| Total |
$n= 25$ |
|
Median of number of vehicles :
$M =$ Value of $\left(\frac{n+1}{2}\right)$ th observation
$=$ Value of $\left(\frac{25+1}{2}\right)$
$=13$ th observation,
Referring to cf column.
$M=2$ vehicles
Hence, median of number of vehicles $=2$ vehicles. View full question & answer→Question 232 Marks
The mean of marks in Mathematics of $40$ students in class is $76$, whereas the same for the other class of $50$ students is $85$. Find the mean of marks in Mathematics of students in both the classes together.
AnswerHere, $n _1,=40 ; \overline{ x 1}=76: n _2=50: \overline{x 2}=85$
Mean of marks of mathematics of the students of both class is combined mean.
$\therefore \overline{X c}=\frac{n 1 \overline{x 1}+n 2 \overline{x 2}}{n 1+22}$
$\therefore \overline{X c}=\frac{(40 \times 76)+(52 \times 85)}{40+50}=\frac{3040+4250}{90}$
$\therefore \overline{X C}=\frac{7290}{90}=81 \text { marks }$
View full question & answer→Question 242 Marks
The median of observations $a-5, a+1, a+2, a-3$, a is $10.$ Find a.
AnswerArranging the observations in ascending order, $a - 5, a-3, a, a + 1, a + 2$
Here, $n=5$ and $M= 10.$
$M =$ Value of $\left(\frac{n+1}{2}\right)$ th observation
$\therefore 10=$ Value of $\left(\frac{5+1}{2}\right)=3$ rd observation
$\therefore 10=a$
Hence $a=10$
View full question & answer→Question 252 Marks
AnswerWhen the Importance of all the observations of the data is not equal, then by assigning weightage to each observation in proposition to its Importance, the mean obtained called weighted mean. It is denoted by $\overline{x w}$
Suppose the weights assigned to observations $X _1, X _2, X _3$ ........ $X _{ n }$ are respectively $W _1 W_2$ ............ $W _{ n }$ then weighted mean $\overline{x w}=\frac{W 1 X 1+W 2 X 2+\ldots \ldots W n X n}{W 1+W 2+\ldots+W n}=\frac{\sum W x}{\sum w}$
View full question & answer→Question 262 Marks
Comment on the mode for the following data showing the time taken (in seconds) for $8$ competitors in a running race:
$25.2, 26.5, 28.6, 32.1, 29.0, 29.3, 31.3, 27.8$
AnswerTime taken (in seconds) by $8$ competitors in a running competition is as follows :
$25.2, 26.5, 28.6, 32.1, 29.0, 29.3, 31.3, 27.8$
Here, each observation occurs only once. Therefore by definition of mode, it cannot be found for the given data. But it may be obtained by using the empirical formula of mode.
View full question & answer→Question 272 Marks
The following table gives the number of cakes sold each day at a bakery. Find the mode for sale of cakes.
| No. of cakes |
$10$ |
$12$ |
$13$ |
$16$ |
$17$ |
$18$ |
| No. of days |
$5$ |
$9$ |
$25$ |
$16$ |
$10$ |
$7$ |
AnswerFind the mode for sale of cakes :
| No of cakes |
$10$ |
$12$ |
$13$ |
$16$ |
$17$ |
$18$ |
| No of days |
$5$ |
$9$ |
$25$ |
$16$ |
$10$ |
$7$ |
$10$ cakes sold in $5$ days
$12$ cakes sold in $9$ days
$13$ cakes sold in $25$ days
$16$ cakes sold in $16$ days
$17$ cakes sold in $10$ days
$18$ cakes sold in $7$ days
Here the maximum nuber of days sold cakes are $13$ View full question & answer→Question 282 Marks
A taxi travelled $15 \ km$ on Monday and $254 \ km$ on Tuesday. Find the average distance trvel1ed over these two days using geometric mean.
AnswerHere, $x_1$ = Travelling of taxi on Monday $=15 \ km$
$x_2$ = Travelling of taxi on Tuesday $= 254 \ km$
Average distance travelled by taxi:
$G=\sqrt{x_1 \times x_2}$
$=\sqrt{15 \times 254}$
$=\sqrt{3810}$
$=61.725$
$\approx 61.73 \mathrm{~km}$
Hence, the average distance travelled by a taxi $=61.73 \mathrm{~km}$.
View full question & answer→Question 292 Marks
The value of a machine depreciates at the rate of $10 \%, 7 \%, 5 \%$ and $2 \%$ in its first four years respectively. Find the average rate of depreciation using an appropriate method.
AnswerHere, the depreciation values of a machine are given in percentage. Therefore, an appropriate average of depreciation Is geometric mean.
Considering the percentage depreciation of value of a machine, we get the following observation :
$x_1=100-10=90$
$x_2=100-7=93$
$x_3=100-5=95$
$x_4=100-2=98$
Average of depreciation :
Hence, average rate of depreciation is four years $= (100 – 93.95) = 6.05 \%.$ View full question & answer→Question 302 Marks
The following data show the number of books read by $8$ students of a class during last month : $2, 1, 5, 9, 1, 3, 2, 4$ Find the average number of books read using geometric mean.
AnswerHere, $x_1=2 ; x_2=1: x_3=5 ; x_4=9 ; x_5=1 ; x_6=3 ; x_7=2$ and $x_8=8$
Geometric mean of books read :
Hence, mean of books read $= 2.61$ books.
[Note: To find 8th root of a number, find ( the square root again of square root ; found, first, repeat this process till the last square roots. Here, square root process is done thrice.] View full question & answer→Question 312 Marks
The mean marks of a student in $7$ theory papers is $62$. What should be the mean marks in $3$ practical examinations so that his mean marks for the entire examination is $68 ?$ (The marks of each theory paper ai practical examination are same.)
AnswerHere, $\mathrm{n}_1=$ No. of theory papers $=7$
$\bar{x}_1=$ Mean of marks of theory papers $=62$
$\mathrm{n}_2=$ No. of papers for practical exam $=3$
$\overline{\mathrm{x}}_2=$ Mean of marks of practical exam $=$ ?
$\bar{x}_c=$ Mean of marks of theory and practical $=68$
Now, $\bar{x}_c=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}$
Putting $\bar{x}_c=68, n_1=7, \bar{x}_1=62$ and $n_2=3$ in the formula,
$68=\frac{(7 \times 62)+\left(3 \times \bar{x}_2\right)}{7+3} $
$ 68 \times 10=434+3 \overline{\mathrm{x}}_2 $
$ 3 \overline{\mathrm{x}}_2=680-434 $
$ 3 \overline{\mathrm{x}}_2=246 $
$ \overline{\mathrm{x}}_2=\frac{246}{3} $
$ \overline{\mathrm{x}}_2=82 \text { marks }$
Hence, mean of marks of practical exam $=82$ marks.
View full question & answer→Question 322 Marks
The mean daily wage paid to $75$ skilled workers of a factory was 280 Rs. whereas the mean daily wage paid to $125$ unskilled workers was $150$ Rs.. Find the mean wage of all the workers.
AnswerHere, $\mathrm{n}_1=$ No. of skilled workers $=75$
$\bar{x}_1=$ Mean of daily wages of skilled workers $= ₹ 280$
$n_2=$ No. of unskilled workers $=125$
$\bar{x}_2=$ Mean of daily wages of unskilled workers $=₹ 150$
Mean of daily wages of workers:
Combined mean $\bar{x}_c=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}$
Putting $n_1=75, \bar{x}_1=280, n_2=125$ and $\bar{x}_2=150$ in the formula,
$\bar{x}_c=\frac{(75 \times 280)+(125 \times 150)}{75+125}$
$\therefore \bar{x}_c=\frac{21000+18750}{200}$
$\therefore \bar{x}_c=\frac{39750}{200}$
$\therefore \bar{x}_c=₹ 198.75$
Hence, the mean of daily wages of workers $=198.75$.
View full question & answer→Question 332 Marks
The mean age of $4$ participants in a relay race was calculated to be $24$ years. Later it was found that one of the participant’s age was actually $27$ years, which was wrongly recorded as $25$ years. If there is rule wherein it is not possible to participate in this race if the mean age is more than $25$ years, can they participate in this race even after the correction of age ?
AnswerHere, Mean of age of $4$ participants $=24$ years
$\therefore$ Sum of age of $4$ participants $=4 \times 24=96$ years
The age of one participants was taken $25$ years instead of 27 years.
$\therefore$ True sum of age of $4$ participants $=96-25+27=98$ years
$\therefore$ Correct mean of age of $4$ participants $=\frac{98}{4}=24.5$ years
If the age of one participant is more than $25$ years, he cannot participate in the race.
The mean after the correction of age is $24.5$ years. Therfore, a participants can participate in the race.
View full question & answer→Question 342 Marks
The mean of $8$ observations is $52$. If in the calculation, by mistake one observation was taken $64$ instead of $46$, find the correct mean.
View full question & answer→Question 352 Marks
The mean of $20$ observations is $29.6.$ On inquiry, it is found that one observation was taken $21$ instead of $-21.$ Correct this mistake and calculate the correct value of mean.
View full question & answer→Question 362 Marks
The mean of $10$ observations is $- 15.5$. If in the calculation, one observation was taken $-3$ instead of $3,$ find the correct value of mean.
View full question & answer→Question 372 Marks
The average weekly salary of employees of a factory is $410 Rs.$ There are $20$ employees in the office whose average weekly salary is $490$.The average weekly salary of the remaining workers is $370 Rs.,$ find the number of workers.
View full question & answer→Question 382 Marks
The average age of persons of a group is $25.02$ years. If in the group the average age of males is $28.8$ years and that of females it is $22.5$ years, then find proportion of males and females in the group.
View full question & answer→Question 392 Marks
The mean of $20$ observations is $28.4$, that of $18$ observations it is $20.5$ and of $12$ observations it is $24$, find the combined mean.
View full question & answer→Question 402 Marks
Find the geometric mean of observations $9, 8, 3.$
View full question & answer→Question 412 Marks
Find geometric mean of the numbers $1, 9, 81.$
View full question & answer→Question 422 Marks
Find mean and geometric mean of observations $32$ and $128$. Also veri1r that $\bar{x}>G$
View full question & answer→Question 432 Marks
If the geometric mean of numbers $3.6$ and a Is $7.2$, find a.
View full question & answer→Question 442 Marks
Find the mean of the observations $-8,12,0,1,-4,2,-2$, $7,-3,5,-1$. If each observation is multiplied by $\frac{1}{2}$, find the new value of mean.
Answer$\overline{0.82,}$ New value $=0.41$
View full question & answer→Question 452 Marks
Find the mean of observations $1.9,1.99,0.9,0.99$, $0.199,1.09,0.09,1.0,1.19$ and $0.19. $
$(1)$ If each observation is increased by $3$ and then divided by $2$ , find the new mean.
$(2)$ Obtain the deviations of observations from their mean and prove that $\sum(x-$ $\bar{x})=0 .$
Answer$(1)$ $0.9539$
$(2)$ $1.97695$
View full question & answer→Question 462 Marks
The labour paid per hour to $10$ labourers are respectively Rs. $3.50,4.00,4.25,3.00,4.60,3.90,4.50,3.60,5.00,3.75$. Find the average labour by direct method and short cut method.
View full question & answer→Question 472 Marks
The weight of players of a cricket eleven are respectively $60,57,52,60,65,68,60,55,63,52$ and $59 \mathrm{~kg}$, find mean of weight.
Answer$59.18 \mathrm{~kg}$
View full question & answer→Question 482 Marks
Without classifying the following observations of a sample survey, obtain mean, median andmode:
$20,25,20,30,36,28,27,25,30,29,38,40,35,28,39$, $47,40,42,37,35,42,35,37,35,50$.
Answer$\bar{x}=34, M=35, M_{0}=35$
View full question & answer→