Question 15 Marks
Calculate man of the following: $4, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 11, 3$
Answer
| $x$ |
$f$ |
$fx$ |
| $3$ |
$1$ |
$3$ |
| $4$ |
$1$ |
$4$ |
| $6$ |
$3$ |
$18$ |
| $7$ |
$4$ |
$28$ |
| $8$ |
$2$ |
$16$ |
| $9$ |
$2$ |
$18$ |
| $11$ |
$1$ |
$11$ |
| Total |
$14$ |
$98$ |
$\therefore$ Mean$=\frac{\sum f x}{\sum f}$
Or
$=\frac{98}{14}=7 \text {. }$ View full question & answer→Question 25 Marks
Construct a frequency distribution table from the given cumulative frequency distribution showing the weights of $750$ students in a school:
| Weight $($in $kg)$ |
$c.f.$ |
| More Than$ 25$ |
$750$ |
| More Than $30$ |
$640$ |
| More Than $35$ |
$615$ |
| More Than $40$ |
$485$ |
| More Than $45$ |
$370$ |
| More Than $50$ |
$220$ |
| More Than $55$ |
$124$ |
| More Than $60$ |
$49$ |
| More Than $65$ |
$24$ |
| More Than $70$ |
$0$ |
$a$. Find the number of students whose weight lie in the interval $40-45;b$. Find the interval which has the most number of students. AnswerFrequency distribution table is as follows:
| Weight $($in $kg)$ |
$c.f.$ |
| $25 - 30$ |
$110 (750 - 640)$ |
| $30 - 35$ |
$25 (640 - 615)$ |
| $35 - 40$ |
$130 (615 - 485)$ |
| $40 - 45$ |
$115 (485 - 370)$ |
| $45 - 50$ |
$150 (370 - 220)$ |
| $50 - 55$ |
$96 (220 - 124)$ |
| $55 - 60$ |
$75 (124 - 49)$ |
| $60 - 65$ |
$25 (49 - 25)$ |
| $65 - 70$ |
$24 (24 - 0)$ |
| $70 - 75$ |
$0$ |
$a$. The number of students whose weight lie in the interval $40 - 45$ is $115.$
$b$. The interval $45 - 50$ has the most number of students. View full question & answer→Question 35 Marks
From the cumulative frequency distribution given below, construct a frequency distribution table:
| Marks |
$c.f.$ |
| Less than $10$ |
$10$ |
| Less than $20$ |
$18$ |
| Less than $30$ |
$32$ |
| Less than $40$ |
$45$ |
| Less than $50$ |
$50$ |
Answer
| Class |
$c.f.$ |
Frequency |
| $0 - 10$ |
$10$ |
$10$ |
| $10 - 20$ |
$18$ |
$18 - 10 = 8$ |
| $20 - 30$ |
$32$ |
$32 - 18 = 14$ |
| $30 - 40$ |
$45$ |
$45 - 32 = 13$ |
| $50 - 60$ |
$50$ |
$50 - 45 = 5$ |
View full question & answer→Question 45 Marks
The electricity bills of $45$ houses in a particular locality are given below. Tabulate the given data and present it as a cumulative frequency table with one of the classes being $300 - 450:784, 567, 890, 231, 150, 458, 356, 762, 386, 824, 525, 663, 724, 841, $$315, 641, 156, 715, 156, 317, 814, 547, 879, 456, 463, 664, 175, 584, 515, 487, 871, 511, 522, 454, 247, $$819, 412, 326, 445, 311, 321, 545, 344, 266, 351.$
Answer
| Class |
Tally Marks |
Frequency |
Cum Frequency |
| $150 - 300$ |
$IIII, II$ |
$7$ |
$7$ |
| $300 - 450$ |
$IIII, IIII ,I$ |
$11$ |
$18$ |
| $450 - 600$ |
$IIII, IIII, III$ |
$13$ |
$31$ |
| $600 - 750$ |
$IIII, II$ |
$7$ |
$38$ |
| $750 - 900$ |
$IIII ,II$ |
$7$ |
$45$ |
View full question & answer→Question 55 Marks
The table given below shows the ages of patients being treated in a hospital. Construct a cumulative frequency distribution table for the same:
| Age |
No. of patients |
| $10 - 20$ |
$90$ |
| $20 - 30$ |
$50$ |
| $30 - 40$ |
$60$ |
| $40 - 50$ |
$80$ |
| $50 - 60$ |
$50$ |
| $60 - 70$ |
$30$ |
Answer
| Age |
No. of patients |
Cum. Frequency |
| $10 - 20$ |
$90$ |
$90$ |
| $20 - 30$ |
$50$ |
$140$ |
| $30 - 40$ |
$60$ |
$200$ |
| $40 - 50$ |
$80$ |
$280$ |
| $50 - 60$ |
$50$ |
$330$ |
| $60 - 70$ |
$30$ |
$360$ |
View full question & answer→Question 65 Marks
Prepare a cumulative frequency distribution table of the marks scored by $60$ students in a test are given below:
| Marks |
No. of students |
| $0 - 10$ |
$4$ |
| $10 20$ |
$15$ |
| $20 - 30$ |
$21$ |
| $30 - 40$ |
$12$ |
| $40 50$ |
$8$ |
Answer
| Marks |
No. of students |
Cum. frequency |
| $0 - 10$ |
$4$ |
$4$ |
| $10 - 20$ |
$15$ |
$19$ |
| $20 - 30$ |
$21$ |
$40$ |
| $30 - 40$ |
$12$ |
$52$ |
| $40 - 50$ |
$8$ |
$60$ |
View full question & answer→Question 75 Marks
If the class intervals of a frequency distribution are $5 - 12, 13 - 20, 21 - 28, 29 - 36, 37 - 44, 45 - 52$ and $53 - 60$, find the following:$(i)$ The class limits and class boundaries of $21 - 28(ii)$ The class size and the class mark of the class interval $45 - 52.(iii)$ Find the true class limits of all the class intervals.
Answer$(i)$ Here, the lower limit is $21$ and the upper limit is $28 .$
The actual lower limit $=21-0.5=20.5$
The actual upper limit $=28+0.5=28.5$
$\therefore$ The class boundaries are $20.5$ and $28.5 .$
$(ii)$ For the class $45 - 52,$
The actual class limiits are $45-0.5=44.5$ and $52+0.5=52.5$
$\therefore$ The class size of this class
$=52.5-44.5$
$=8$
$\therefore$ The class mark of this class
$=\frac{1}{2}(44.5+52.5)$
$=48.5 .$
$(iii)$ As the classes are exclusive, so the true class limits are the same as the class limits.
| Class |
True class limits |
| $5 - 12$ |
$4.5 - 12.5$ |
| $13 - 20$ |
$12.5 - 20.5$ |
| $21 - 28$ |
$20.5 - 28.5$ |
| $29 - 36$ |
$28.5 - 36.5$ |
| $37 - 44$ |
$36.5 - 44.5$ |
| $45 - 52$ |
$44.5 - 52.5$ |
| $53 - 60$ |
$52.5 - 60.5$ |
View full question & answer→Question 85 Marks
The number of goals scored by Arsenal Football Club in the English Premier League in the season $2007$ were:$1, 2, 1, 3, 2, 5, 1, 6, 4, 4, 2, 3, 5, 6, 4, 2, 2, 3, 4, 1, 0, 5, 0, 5, 3, 2, 3, 4, 4, 1, 1, 2, 4, 3. 1$. $4$ Arrange these data in a distance frequency distribution table and answer the following:$(i)$ What is the range of the number of goals scored by $\text{AFC}$?$(ii)$ How many times did $\text{AFC}$ score $3$ or more than $3$ goals?$(iii)$ Which variatie has the highest frequency?
AnswerThe discrete frequency distribution table is a below:
| No. Of goals |
Tally Marks |
Frequency |
| $0$ |
$II$ |
$2$ |
| $1$ |
$IIII,II$ |
$7$ |
| $2$ |
$IIII, II$ |
$7$ |
| $3$ |
$IIII, I$ |
$6$ |
| $4$ |
$IIII, III$ |
$8$ |
| $5$ |
$IIII$ |
$4$ |
| $6$ |
$II$ |
$2$ |
$(i)$ Maximum goals scored $= 6$
Minimum goals scored $= 0$
$\therefore $ Range of the goals scored $= 6 - 0 = 6$
$(iii)$ No. of times $\text{AFC}$ scored $3$ or more goals $= 6 + 8 + 4 + 2 = 20$
$(iii)$ The variate which has highest frequency is $4.$ View full question & answer→Question 95 Marks
Observe the given frequency table to answer the following:
| Class Interval |
$20 - 24$ |
$25 29$ |
$30 - 34$ |
$35 - 39$ |
$40 - 44$ |
$45 - 49$ |
| Frequency |
$6$ |
$12$ |
$10$ |
$15$ |
$9$ |
$2$ |
$a$. The true class limits of the fifth class.;$b$. The size of the second class.;$c$. The class boundaries of the fourth class.$d$. The upper and lower limits of the sixth class.$e$. The class mark of the third class. Answer$a$. Fifth class : $40-44$
Since classes are inclusive, we have
Adjustment factor
$ =\frac{25-24}{2}$
$=\frac{1}{2}$
$=0.5 $
$\therefore$ True lower limit of $5^{\text {th }}$ class $=40-0.5=39.5$
True Upper limit of $5^{\text {th }}$ class $=44+0.5=44.5$.
$b$. Size of class $25-29=29.5-24.5=5$.
$c$. True lower limit of $4^{\text {th }}$ class $=35-0.5=34.5$True Upper limit of $4^{\text {th }}$ class $=39+0.5=39.5$
$\therefore$ Class boundaries of $4^{th}$ class are $34.5$ and $39.5 .$
$d$. Lower limit of $6^{\text {th }}$ class $=45-0.5=44.5$
Upper limit of $6^{\text {th }}$ class $=49+0.5=49.5$.
$e$. True lower limit of $3^{\text {rd }}$ class $=30-0.5=29.5$
True Upper limit of $3^{\text {rd }}$ class $=34+0.5=34.5$
$ \therefore$ Class mark
$=\frac{29.5+34.5}{2}$
$=32 . $
View full question & answer→Question 105 Marks
Find the actual $($or true$)$ lower and upper class limits and class$-$marks $($or mid values$)$ of the following classes: $2.1 - 4.0, 4.1 - 6.0$ and $6.1 - 8.0.$
AnswerClasses: $2.1-4.0,4.1-6.0$ and $6.1-8.0$
Since classes are inclusive, we have
Adjustment factor
$=\frac{4.1-4.0}{2}$
$=\frac{0.1}{2}$
$=0.05 $
| True Lower Limit |
True Upper Limit |
Class - Mark |
| $2.1 - 0.05 = 2.05$ |
$4.0 + 0.05 = 4.05$ |
$3.05$ |
| $4.1 - 0.05 = 4.05$ |
$6.0 + 0.05 = 6.05$ |
$5.05$ |
| $6.1 - 0.05 = 6.05$ |
$8.0 + 0.05 = 8.05$ |
$7.05$ |
View full question & answer→Question 115 Marks
Find the class boundaries and class marks of the following classes:$55 - 59, 60 - 64, 65 - 69, 70 - 74, 75 - 79, 80 - 84, 85 - 89, 90 - 94$ and $95 - 99.$
AnswerFor the class $55-59$
The actual lower limit $=55-0.5=54.5$
The actual upper limit $=59+0.5=59.5$
$\therefore$ The class boundaries are $54.5$ and $59.5$
$\therefore$ The class mark $=\frac{1}{2}(54.5+59.5)=57$
| Class |
Class Boundaries |
Class Mark |
| $55 - 59$ |
$54.5 - 59.5$ |
$57$ |
| $60 - 64$ |
$59.5 - 64.5$ |
$62$ |
| $65 - 69$ |
$64.5 - 69.5$ |
$67$ |
| $70 - 74$ |
$69.5 - 74.5$ |
$72$ |
| $75 - 79$ |
$74.5 - 79.5$ |
$77$ |
| $80 - 84$ |
$79.5 - 84.5$ |
$82$ |
| $85 - 89$ |
$84.5 - 89.5$ |
$87$ |
| $90 - 94$ |
$89.5 - 94.5$ |
$92$ |
| $95 - 99$ |
$94.5 - 99.5$ |
$97$ |
View full question & answer→Question 125 Marks
The mark obtained by the students in a class test are given below:$31, 12, 28, 45, 32, 16, 49, 12, 18, 26, 34, 39, 29, 28, 25, 46, 32, 13, 14, 26, 25,$$ 34, 23, 23, 25, 45, 33, 22, 18, 37, 26, 19, 20, 30, 28, 38, 42, 21, 36, 19, 20, 40, 48, 15, 46, 26, 23, 33, 47, 40.$Arrange the above marks in classes each with a class size of $5$ and answer the following:$(i)$ what is the highest score?$(ii)$ What is the lowest score?$(iii)$ What is the range?$(iv)$ If the pass mark is $20$, how many students failed;$(v)$ How many students got $40$ or more marks?
Answer
| CLASS |
TALLY MARKS |
FREQUENCY |
| $11 - 15$ |
$IIII$ |
$5$ |
| $16 - 20$ |
$IIII, II$ |
$7$ |
| $21- 25$ |
$IIII, III$ |
$8$ |
| $26 - 30$ |
$IIII, IIII$ |
$9$ |
| $31 - 35$ |
$IIII, II$ |
$7$ |
| $36 - 40$ |
$IIII ,I$ |
$6$ |
| $41 - 45$ |
$III$ |
$3$ |
| $46 - 50$ |
$IIII$ |
$5$ |
| |
Total |
$50$ |
$(i)$ The highest score is $49.$
$(ii)$ The lowest score is $12.$
$(iii)$ Range $= 49 - 12 = 37.$
$(iv)$ Given, pass marks is $20$
So, all the students in the class $11 - 15$ and $16 - 20$ must have failed expect for the students with score $20.$
$\therefore $ Number of such students $= 5 + 7 - 2 = 10$
$(v)$Number of students scoring above 40 is the sum total of students in the classes $41 - 45$ and $46 - 50$
i.e. $3 + 5 = 8$
Number of students scoring exactly $40 = 2$
$\therefore $ Number of students scoring $40$ or more marks $= 8 + 2 = 10.$ View full question & answer→Question 135 Marks
Construct a grouped frequency table from the following data of the daily wages earned by $60$ labourers in a company. Take each class size as $7.25, 26, 34, 48, 39, 16, 55, 28, 37, 42, 45, 55, 28, 54, 53, 18, 35, 47, 44, 28, 55, 45, 39, 54, 21, 49, 45, 38, 29, 53, $$48, 44, 15, 28, 14, 32, 15, 44, 14, 15, 16, 41, 33, 52, 29, 34, 51, 22, 19, 37, 44, 25, 48, 38, 24, 52, 51, 42, 32, 27.$
AnswerMinimum value of variate $=14$
Maximum value of variate $=55$
$\therefore$ Range
$=55-14$
$ =41$
Class size$=7$
$\therefore$ No. Of class intervals
$=\frac{41}{7}$
$=6$
| Class |
Tally Marks |
Frequency |
| $14 - 21$ |
$IIII, IIII$ |
$9$ |
| $211 - 28$ |
$IIII ,II$ |
$7$ |
| $28 - 35$ |
$IIII, IIII I$ |
$11$ |
| $35 - 42$ |
$IIII, III$ |
$8$ |
| $42 - 49$ |
$IIII, IIII III$ |
$13$ |
| $49 - 56$ |
$IIII ,IIII II$ |
$12$ |
| |
Total |
$60$ |
View full question & answer→Question 145 Marks
The class marks of a frequency distribution are $: 27, 32, 37, 42, 47, 52, 57, 62, 67, 72$ and $77.$ Find the class size and true class limits.
AnswerThe class marks are uniformly spread.
$\therefore$ The class size is the difference between any two consecutive class marks.
Class size
$ =32-27$
$=5$
The lower limit of the first class
$ =27-\frac{5}{2}$
$=27-2.5$
$=24.5$
The upper limit of the first class
$ =27+\frac{5}{2}$
$=27+2.5$
$=29.5 $
Thus, the first class interval is $24.5-29.5$
Similarly, we can find the class limits of all the class marks given.
| Class Marks |
Class limits |
| $27$ |
$24.5 - 29.5$ |
| $32$ |
$29.5 - 34.5$ |
| $37$ |
$34.5 - 39.5$ |
| $42$ |
$39.5 -- 44.5$ |
| $47$ |
$44.5 - 49.5$ |
| $52$ |
$49.5 - 54.5$ |
| $57$ |
$54.5 - 59.5$ |
| $62$ |
$59.5 - 64.5$ |
| $67$ |
$64.5 - 69.5$ |
| $72$ |
$69.5 - 74.5$ |
| $77$ |
$74.5 - 79.5$ |
View full question & answer→Question 155 Marks
The class marks of a frequency distribution are $: 15, 25, 35, 45, 55, 65$ and $75.$ Determine the class limits.
AnswerThe class marks are uniformly spread.
The class size is the difference between anty two consecutive class marks.
Class size
$=25-15 $
$=10$
The lower limit of the first class
$=15-\frac{10}{2} $
$=15-5 $
$=10$
The upper limit of the first class
$=15+\frac{10}{2} $
$=15+5 $
$=20$
Thus, the first class interval is $10-20$
Similarly, we can find the class limits of all the class marks given.
| Class Marks |
Class limits |
| $15$ |
$10 - 20$ |
| $25$ |
$20 - 30$ |
| $35$ |
$30 - 40$ |
| $45$ |
$40 - 50$ |
| $55$ |
$50 - 60$ |
| $65$ |
$60 - 70$ |
| $75$ |
$70 - 80$ |
View full question & answer→