Question 13 Marks
Find the median of the following data: $133, 73, 89, 108, 94,104, 94, 85, 100, 120$
AnswerArranging the data in ascending order, we have: $73, 85, 89, 94, 100, 104, 108, 120, 133$
Here, the number of observations, $n = 10$ (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{94+100}{2}=97$
Hence, the median of the given data is $49.5$
View full question & answer→Question 23 Marks
Find the median of the following data: $41, 43, 127, 99, 61, 92, 71, 58, 57,$ If $58$ is replaced by $85$, what will be the new median?
AnswerArranging the given data in ascending order, we have:
$41,43,57,58,61,71,92,99,127$
Here, the number of observations $n$ is $9$ (odd).
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $5^{\text {th }}$ observation $=61$
Hence, the median $=61$
If 58 is replaced by 85 , and then the new observations arranged in ascending order are:
$41,43,57,61,71,85,92,99,12$
$\therefore$ New median $=$ value of the $5^{\text {th }}$ observation $=71$
View full question & answer→Question 33 Marks
The following observations have been arranged in ascending order. If the median of the data is $63$, find the value of $x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95$
AnswerArranging the given data in ascending order,
we have: $41, 43, 57, 58, 61, 71, 92, 99, 127$
Here, the number of observations $n$ is $10$.
Since $n$ is even, $\Rightarrow \text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of}\ 5^\text{th}\ \text{observation}+\text{Value of}\ 6^\text{th}\ \text{observation}}{2}$
$\Rightarrow63=\frac{\text{x}+(\text{x+2})}{2}$
$\Rightarrow63=\frac{2\text{x}+2}{2}$
$\Rightarrow63=\frac{2(\text{x}+1)}{2}$
$\Rightarrow63 = \text{x} + 1$
$\Rightarrow\text{x} = 63 - 1$
$\Rightarrow\text{x} = 62$
View full question & answer→Question 43 Marks
The mean weight of $8$ numbers is $15\ kg$. If each number is multiplied by $2$, what will be the new mean?
AnswerLet $x_1, x_2, x_3 \ldots x_8$ be the eight numbers whose mean is $15\ kg$ .
Then,
$15=\frac{x_1+x_2+x_3+\ldots+x_8}{8}$
$x_1+x_2+x_3+\ldots+x_8=15 \times 8$
$x_1+x_2+x_3+\ldots+x_8=120$
Let the new numbers be $2 x_1, 2 x_2, 2 x_3 \ldots 2 x_8$.
Let $M$ be the arithmetic mean of the new numbers.
Then,
$\mathrm{M}=\frac{2 \mathrm{x}_1+2 \mathrm{x}_2+2 \mathrm{x}_3+\ldots+2 \mathrm{x}_8}{8}$
$\Rightarrow \mathrm{M}=\frac{2\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots+\mathrm{x}_8\right)}{8}$
$\Rightarrow \mathrm{M}=\frac{2 \times 120}{8}$
$\Rightarrow \mathrm{M}=30$
View full question & answer→Question 53 Marks
Find the median of the following observations: $46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33$. If $92$ is replaced by $99$ and $41$ by $43$ in the above data, find the new median?
AnswerArranging the given data in ascending order, we have:
$33,35,41,46,55,58,64,77,87,90,92$
Here, the number of observations $n$ is $11$ (odd).
Since the number of observations is odd, therefore.
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $6^{\text {th }}$ observation $=58$
Hence, median $=58$
If $92$ is replaced by $99$ and $41$ by $43$ , then the new observations arranged in ascending order are:
$33,35,43,46,55,58,64,77,87,90,99$
$\therefore$ New median $=$ value of the $6^{\text {th }}$ observation $=58$
View full question & answer→Question 63 Marks
Find the median of the following data: $31, 38, 27, 28, 36, 25, 35, 40$
AnswerArranging the data in ascending order, we have: $25, 27, 28, 31, 35, 36, 38, 40$
Here, the number of observations, $n = 8$ (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 4}^\text{th}\ \text{observation}+\text{Value of 5}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{31+35}{2}=33$
Hence, the median of the given data is $33$
View full question & answer→Question 73 Marks
The numbers of children in $10$ families of a locality are: $2, 4, 3, 4, 2, 3, 5, 1, 1, 5$. Find the mean number of children per family.
AnswerWe have, The mean number of children per family
$=\frac{\text{Sum of the total number of children}}{\text{Total Number of families}}$$\text{Mean}= \frac{2+4+3+4+2+3+5+1+1+5}{10}$
$=\frac{30}{10}$
View full question & answer→Question 83 Marks
Calculate the mean for the following distribution:
|
x:
|
$5$ |
$6$
|
$7$
|
$8$
|
$9$
|
|
y:
|
$4$
|
$8$
|
$14$
|
$11$
|
$3$
|
AnswerCalculation of mean:
|
$x_i$
|
$f_i$
|
$x_if_i$
|
| $5$ |
$4$
|
$20$
|
|
$6$
|
$8$ |
$48$
|
| $7$ |
$14$ |
$98$
|
| $8$ |
$11$
|
$88$
|
| $9$ |
$3$ |
$27$
|
|
Total
|
$\sum\text{f}_\text{i}=40$
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=281$
|
$\therefore\text{Mean weight}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{281}{40}=7.025.$ View full question & answer→Question 93 Marks
The following table shows the weights (in kg) of $15$ workers in a factory:
|
Weight (in kg):
|
$60$
|
$63$
|
$66$
|
$72$
|
$75$
|
|
Numbers of workers:
|
$4$
|
$5$
|
$3$
|
$1$
|
$2$
|
Calculate the mean weight. AnswerCalculation of mean:
|
$x_i$
|
$f_i$
|
$x_if_i$
|
|
$60$
|
$4$
|
$240$
|
|
$63$
|
$5$
|
$315$
|
|
$66$
|
$3$
|
$198$
|
|
$72$
|
$1$
|
$72$
|
|
$75$
|
$2$
|
$150$
|
|
Total
|
$\sum\text{f}_\text{i}=15$
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=975$
|
$\therefore\text{Mean weight}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{975}{15}=65\text{kg.}$ View full question & answer→Question 103 Marks
Calculate the mean and median for the following data:
| Marks: |
$10$ |
$11$ |
$12$ |
$13$ |
$14$ |
$16$ |
$19$ |
$20$ |
| Number of Students: |
$3$ |
$5$ |
$4$ |
$5$ |
$2$ |
$3$ |
$2$ |
$1$ |
Using empirical formula, find its mode. AnswerCalculation of Mean:
$\text { mean }=\frac{\sum f_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}} \frac{332}{25}=13.28$
Here, $\mathrm{n}=25$, witch is an odd number. Therefore,
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $13^{\text {th }}$ observation $=13$
Now，
$\Rightarrow \text { Mode }=3 \text { Median }-2 \text { Mean }$
$\Rightarrow \text { Mode }=3(13)-2(13.28)$
$\Rightarrow \text { Mode }=39-26.56$
$\Rightarrow \text { Mode }=12.44$
View full question & answer→Question 113 Marks
Find the mean of the following data:
|
x:
|
$19$ |
$21$
|
$23$
|
$25$
|
$27$ |
$29$ |
$31$
|
|
f:
|
$13$ |
$15$
|
$16$
|
$18$
|
$16$ |
$15$ |
$13$
|
AnswerCalculation of mean:
|
$x_i$
|
$f_i$
|
$x_if_i$
|
|
$19$
|
$13$
|
$247$
|
|
$21$
|
$15$
|
$315$
|
|
$23$
|
$16$
|
$368$
|
|
$25$
|
$18$
|
$450$
|
|
$27$
|
$16$
|
$432$
|
|
$29$
|
$15$
|
$435$
|
|
$31$
|
$13$
|
$403$
|
|
Total
|
$\sum\text{f}_\text{i}=\text{N}=106$
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=2650$
|
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{2650}{106}=25.$ View full question & answer→Question 123 Marks
Find the median of the following data:
$15, 6, 16, 8, 22, 21, 9, 18, 25$
AnswerArranging the data in ascending order, we have:
$6,8,9,15,16,18,21,22,25$
Here, the number of observations, $n=9$ (Odd).
$\Rightarrow$ Median $=$ value of $\frac{\mathbf{n}+1}{2}$ th observation i.e., value of $5^{\text {th }}$ observation $=16$
Hence, the median of the given data is $16$
View full question & answer→Question 133 Marks
The mean weight per student in a group of $7$ students is $55\ kg$. The individual weights of $6$ of them (in kg) are $52, 54, 55, 53, 56$ and $54$. Find the weight of the seventh student.
AnswerWe have, $\text{Mean}=\frac{\text{Sum of the weights of the students}}{\text{Number of students}}$
Let the weight of the seventh student be $x\ kg$.
$\text{Mean}=\frac{52+54+55+53+56+54+\text{x}}{7}$
$55=\frac{52+54+55+53+56+54+\text{x}}{7}$
$\Rightarrow 385 = 324 + x $
$\Rightarrow x = 385 - 324$
$ \Rightarrow x = 61kg.$
Thus, the weight of the seventh student is $61\ kg.$
View full question & answer→Question 143 Marks
Find the mode and median of the data: $13, 16, 12, 14, 19, 12, 14, 13, 14$
By using the empirical relation also find the mean.
AnswerArranging the data in ascending order such that same numbers are put together, we get:
$12,12,13,13,14,14,14,16,19$
Here, $\mathrm{n}=9$
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $5^{\text {th }}$ observation $=14$
Here, $14$ occurs the maximum number of times, i.e., three times. Therefore, $14$ is the mode of the data.
Now,
Mode $=3$ Median $-2$ Mean
$\Rightarrow 14=3 \times 14-2$ Mean
$\Rightarrow 2$ Mean $=42-14=28$
$\Rightarrow$ Mean $=28 \div 2=14$
View full question & answer→Question 153 Marks
The percentage of marks obtained by students of a class in mathematics are: $64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1$. Find their mean.
AnswerWe have,$\text{Mean}=\frac{\text{Sum of the marks obtained}}{\text{Total Number of students}}$
$\Rightarrow\text{Mean}= \frac{64+36+47+23+0+19+81+93+72+35+3+1}{12}$
$\Rightarrow\text{Mean}=\frac{474}{12}=39.5\%$
View full question & answer→Question 163 Marks
The weights (in kg) of $15$ students are: $31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42,$ and $30$. Find the median. If the weight $44kg$ is replaced by $46\ kg$ and $27\ kg$ by $25\ kg$, find the new median.
AnswerArranging the given data in ascending order, we have:
$27,28,29,30,31,32,34,35,36,37,41,42,43,44,45$
Here, the number of observations $n$ is $15$ (odd).
Since the number of observations is odd, therefore.
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $8^{\text {th }}$ observation $=35$
Hence, median $=35 \mathrm{~kg}$.
If $44$ is replaced by $46$ and $27\ kg$ by $25\ kg$ , then the new observations arranged in ascending order are:
$25,28,29,30,31,32,34,35,36,37,41,42,43,45,46$
$\therefore$ New median $=$ value of the $8^{\text {th }}$ observation $=35 \mathrm{~kg}$.
View full question & answer→Question 173 Marks
The mean of $75$ numbers is $35$. If each number is multiplied by $4$, find the new mean.
AnswerLet $x_1, x_2, x_3 \ldots x_{75}$ be $75$ numbers with their mean equal to $35$ . Then,
$\Rightarrow 35=\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots+\mathrm{x}_{75}}{75}$
$\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots+\mathrm{x}_{75}=35 \times 75$
$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots+\mathrm{x}_{75}=2625$
The new numbers are $4 \times 1,4 \times 2,4 \times 3 \ldots 4 \times 75$
Let $M$ be the arithmetic mean of the new numbers. Then,
$\mathrm{M}=\frac{4 \mathrm{x}_1+4 \mathrm{x}_2+4 \mathrm{x}_3+\ldots+4 \mathrm{x}_{75}}{75}$
$\Rightarrow \mathrm{M}=\frac{4\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots+\mathrm{x}_{75}\right)}{75}$
$\Rightarrow \mathrm{M}=\frac{4 \times 2625}{75}$
$\Rightarrow \mathrm{M}=140$
View full question & answer→Question 183 Marks
The mean of $5$ numbers is $18$. If one number is excluded, their mean is $16.$ Find the excluded number.
AnswerLet $x_1, x_2, x_3, x_4$ and $x_5$ be five numbers whose mean is $18$ . Then,
$18=$ Sum of five numbers $\div 5$
$\therefore$ Sum of five numbers $=18 \times 5=90$
Now, if one number is excluded, then their mean is $16$ .
So,
$16=$ Sum of four numbers $\div 4$
$\therefore$ Sum of four numbers $=16 \times 4=64$
The excluded number $=$ Sum of five observations - Sum of four observations.
$\therefore$ The excluded number $=90-64$
$\therefore$ The excluded number $=26$
View full question & answer→Question 193 Marks
The runs scored in a cricket match by $11$ players are as follows:
$6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10$
Find the mean, mode and median of this data.
AnswerArranging the data in ascending order such that same values are put together, we get:
$6,8,10,10,15,15,50,80,100,120$
Here, $\mathrm{n}=11$
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $6^{\text {th }}$ observation $=15$
Here, $10$ occur three times. Therefore, $10$ is the mode of the given data.
Now，
Mode $=3$ Median -$2$ Mean.
$\Rightarrow 10=3 \times 15-2$ Mean.
$\Rightarrow 2$ Mean $=45-10=35$
$\Rightarrow$ Mean $=35 \div 2=17.5$
View full question & answer→Question 203 Marks
The mean of five numbers is $27$. If one number is excluded, their mean is $25$. Find the excluded number.
AnswerWe have, $\text{Mean}=\frac{\text{Sun of the five numbers}}{5}=27$
So, sum of the five numbers $= 5 \times 27 = 135$
Now, The mean of the four numbers $=\frac{\text{Sun of the four numbers}}{4}=25$
So, sum of the four numbers $= 4 \times 25 = 100$
Therefore, the excluded number = Sum of the five number - Sum of the four numbers.
The excluded number $= 135 - 100 = 35$
View full question & answer→Question 213 Marks
The scores in mathematics test (out of $25$) of $15$ students are as follows:
$19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20$
Find the mode and median of this data. Are they same?
AnswerArranging the data in ascending order such that same values are put together, we get:
$5,9,10,12,15,16,19,20,20,20,20,23,24,25,25$
Here, $\mathrm{n}=15$
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $8^{\text {th }}$ observation $=20$
Here, clearly, $20$ occurs most number of times, i.e., 4 times. Therefore, the mode of the given data is $20$
Yes, the median and mode of the given data are the same.
View full question & answer→Question 223 Marks
The daily wages (in Rs) of $15$ workers in a factory are given below:
$200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180$
Prepare the frequency table and find the mean wage.
AnswerThe frequency table for the given data is as follows:
| Wages ($x_i$): |
$130$ |
$150$ |
$180$ |
$200$ |
| Number of workers ($f_i$): |
$2$ |
$4$ |
$6$ |
$3$ |
In order to compute the mean wage, we prepare the following table: Mean wages of the workers:
|
$x_i$
|
$f_i$
|
$x_if_i$
|
|
$130$
|
$2$
|
$260$
|
|
$150$
|
$4$
|
$600$
|
|
$180$
|
$6$
|
$1080$
|
|
$200$
|
$3$
|
$600$
|
|
Total
|
$\sum\text{f}_\text{i}=\text{N}=15$
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=2540$
|
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{2540}{15}=169.33$ View full question & answer→Question 233 Marks
Find the median of the following data: $12, 17, 3, 14, 5, 8, 7, 15$
AnswerArranging the data in ascending order, we have: $3, 5, 7, 8, 12, 14, 15, 17$
Here, the number of observations, $n = 8$ (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$
$\Rightarrow\text{Median}=\frac{\text{Value of 4}^\text{th}\ \text{observation}+\text{Value of 5}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{8+12}{2}=10$
Hence, the median of the given data is $10$
View full question & answer→Question 243 Marks
Find the median of the following data: $83, 37, 70, 29, 45, 63, 41, 70, 34, 54$
AnswerArranging the data in ascending order, we have: $29, 34, 37, 41, 45, 54, 63, 70, 70, 83$
Here, the number of observations, $n = 10$ (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$ $\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$ $\Rightarrow\text{Median}=\frac{45+54}{2}=49.5$
Hence, the median of the given data is $49.5$
View full question & answer→Question 253 Marks
The mean of $5$ numbers is $27$. If one more number is included, then the mean is $25$. Find the included number.
AnswerMean = Sum of five numbers $÷ 5$
$\Rightarrow $ Sum of the five numbers $= 27 \times 5 = 135$
Now, New mean $= 25$
$25 =$ Sum of six numbers $÷ 6$
$\Rightarrow $ Sum of the six numbers $= 25 \times 6 = 150$
The included number = Sum of the six numbers – Sum of the five numbers.
$\Rightarrow $ The included number $= 150 - 135$
$\Rightarrow $ The included number $= 15$
View full question & answer→Question 263 Marks
Following are the weights (in kg) of $10$ new born babies in a hospital on a particular day: $3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6.$ Find the mean $\bar{\text{x}}.$
AnswerWe have, $\overline{\text{X}}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow \overline{\text{X}} = \frac{3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6}{10}$
$\Rightarrow\overline{\text{X}}=\frac{40}{10}$
$\Rightarrow\overline{\text{X}}=4\text{kg.}$
View full question & answer→Question 273 Marks
Find the median of the following data: $25, 34, 31, 23, 22, 26, 35, 29, 20, 32$
AnswerArranging the data in ascending order,
we have: $20, 22, 23, 25, 26, 29, 31, 32, 34, 35$
Here, the number of observations, $n = 10$ (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$ $\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$ $\Rightarrow\text{Median}=\frac{26+29}{2}=27.5$
Hence, the median of the given data is $27.5$
View full question & answer→Question 283 Marks
Find the median of the following data:
$41, 43,127, 99, 71, 92, 71, 58, 57$
AnswerArranging the data in ascending order, we have:
$41,43,57,58,71,71,92,99,127$
Here, the number of observations, $\mathrm{n}=9$ (Odd).
$\therefore$ Median $=$ value of $\frac{9+1}{2}$ th observation i.e., value of $5^{\text {th }}$ observation $=71$
View full question & answer→Question 293 Marks
Find the mode of the following data: $12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14$
AnswerArranging the data in ascending order such that same values are put together,
we get: $10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18$
Here, clearly, $14$ occurs the most number ot times.
Therefore, $14$ is the mode of the given data.
Alternate answer:
Arrenging the data in the form of a frequency table, we get:
Clearly, $14$ has maximum frequency. So, the mode of the given data is $14$ View full question & answer→Question 303 Marks
A die was thrown 20 times and the following scores were recorded:
$5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1$
Prepare the frequency table of the scores on the upper face of the die and find the mean score.
AnswerThe frequency table for the given data is as follows:
|
x:
|
$1$
|
$2$
|
$3$
|
$4$
|
$5$
|
$6$
|
|
f:
|
$2$
|
$5$
|
$1$
|
$4$
|
$6$
|
$2$
|
In order to compute the arithmetic mean, we prepare the following table: Computation of Arithmetic Mean:
|
Scores ($x_i$)
|
Frequency($f_i$)
|
$x_if_i$
|
|
$1$
|
$2$
|
$2$
|
|
$2$
|
$5$
|
$10$
|
|
$3$
|
$1$
|
$3$
|
|
$4$
|
$4$
|
$16$
|
|
$5$
|
$6$
|
$30$
|
|
$6$
|
$2$
|
$12$
|
|
Total
|
$\sum\text{f}_\text{i}=20$
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=73$ |
We have, $\sum\text{f}_\text{i}=20\ \text{and}\ \sum\text{f}_\text{i}\text{x}_\text{i}=73$
$\therefore\text{Mean score} = \frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}\frac{73}{20}=3.65$ View full question & answer→Question 313 Marks
Find the median of the following data: $92, 35, 67, 85, 72, 81, 56, 51, 42, 69$
AnswerArranging the data in ascending order,
we have: $35, 42, 51, 56, 67, 69, 72, 81, 85, 92$
Here, the number of observations, $n = 10$ (Even).
$\Rightarrow\text{Median}=\frac{\text{n}}{2}\text{th}\ \text{observation}+\frac{\text{n}}{2}+1^\text{th}\ \text{observation}$ $\Rightarrow\text{Median}=\frac{\text{Value of 5}^\text{th}\ \text{observation}+\text{Value of 6}^\text{th}\ \text{observation}}{2}$ $\Rightarrow\text{Median}=\frac{67+69}{2}=68$
Hence, the median of the given data is $68$
View full question & answer→Question 323 Marks
The ages (in years) of $50$ students of a class in a school are given below:
|
Age (in years):
|
$14$
|
$15$
|
$16$
|
$17$
|
$18$
|
|
Numbers of students:
|
$15$
|
$14$
|
$10$
|
$8$
|
$3$
|
Find the mean age. AnswerCalculation of mean:
|
$x_i$
|
$f_i$
|
$x_if_i$
|
| $14$ |
$15$ |
$210$ |
|
$15$
|
$14$
|
$210$
|
|
$16$
|
$10$
|
$160$
|
|
$17$
|
$8$
|
$136$
|
|
$18$
|
$3$
|
$54$
|
|
Total
|
$\sum\text{f}_\text{i}=50$
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=770$
|
$\therefore\text{Mean weight}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{770}{50}=15.4\text{ years}$ View full question & answer→Question 333 Marks
Numbers $50, 42, 35, 2x + 10, 2x - 8, 12, 11, 8, 6$ are written in descending order and their median is $25$, find $x$.
AnswerHere, the number of observations $n$ is $9$.
Since $n$ is odd, the median is the $\frac{n+1}{2}$ th observation, i.e., the $5^{\text {th }}$ observation.
As the numbers are arranged in the descending order, we therefore observe from the last.
$\text { Median }=5^{\text {th }} \text { observations. }$
$\Rightarrow 25=2 x-8$
$\Rightarrow 2 x=25+8$
$\Rightarrow 2 x=33$
$\Rightarrow x=\frac{33}{2}$
$\Rightarrow x=16.5$
Hence, $x=16.5$
View full question & answer→Question 343 Marks
The scores in mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
AnswerMode = 20, Median = 20, Yes
View full question & answer→Question 353 Marks
The runs scored in a cricket match by 11 players are as follows: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10. Find the mean, mode and median of this data.
AnswerMode = 10, Median = 15, Mean = 17.5
View full question & answer→Question 363 Marks
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Question 373 Marks
The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.
Question 383 Marks
Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median?
Question 393 Marks
Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Question 403 Marks
If the mean of the following data is 15, find p.
Question 413 Marks
The mean of the following data is 20.6. Find the value of p.
Question 423 Marks
Find the mean of the following data:
| x: | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
| f: | 13 | 15 | 16 | 18 | 16 | 15 | 13 |
Question 433 Marks
Calculate the mean for the following distribution:
Question 443 Marks
The ages (in years) of 50 students of a class in a school are given below:
| Age (in years): | 14 | 15 | 16 | 17 | 18 |
| Numbers of students: | 15 | 14 | 10 | 8 | 3 |
Find the mean age.
Question 453 Marks
The following table shows the weights (in kg) of 15 workers in a factory:
| Weight (in kg): | 60 | 63 | 66 | 72 | 75 |
| Numbers of workers: | 4 | 5 | 3 | 1 | 2 |
Calculate the mean weight.
Question 463 Marks
If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.
Question 473 Marks
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
| Rainfall (in mm) | 0.0 | 12.2 | 2.1 | 0.0 | 20.5 | 5.3 | 1.0 |
(i) Find the range of the rainfall from the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Answer(i) 20.5 mm (ii) 5.87 mm (iii) 5 days
View full question & answer→Question 483 Marks
The enrolment of a school during six consecutive years was as follows:
1555, 1670, 1750, 2019, 2540, 2820
Find the mean enrollment of the school for this period.
Question 493 Marks
The marks (out of 100) obtained by a group of students in science test are 85, 76, 90, 84, 39, 48, 56, 95, 81 and 75. Find the
(i) highest and the lowest marks obtained by the students.
(ii) range of marks obtained.
(iii) mean marks obtained by the group.
Answer(i) 95, 39 marks (ii) 56 marks (iii) 72.9 marks
View full question & answer→Question 503 Marks
The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.
Question 513 Marks
The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.
Question 523 Marks
The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
Question 533 Marks
The percentage of marks obtained by students of a class in mathematics are:
64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.
Question 543 Marks
Following are the weights (in kg) of 10 new born babies in a hospital on a particular day:
$3.4,3.6,4.2,4.5,3.9,4.1,3.8,4.5,4.4,3.6$. Find the mean $\bar{X}$.
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