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Maths 3 Marks Question

Mean, Median, Mode of Grouped Data · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 9 questions.

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Question 13 Marks
What is the cumulative frequency of the modal class of the following distribution?
Class
3-6
6-9
9-12
12-15
15-18
18-21
21-24
Frequency
7
13
10
23
4
21
16
Answer
Class interval
Frequency
Cumulative frequency
3-6
7
7
6-9
13
20
9-12
10
30
12-15
23
53
15-18
4
57
18-21
21
78
21-24
16
94
Here, maximum frequency = 23
Hence, Modal class is 12-15.
Thus, the cumulenve frequency of modal dass is 53.
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Question 23 Marks
Find the mode of the given data:
Class interval
0-20
20-40
40-60
60-80
Frequency
15
6
18
10
Answer
As the dess 40-60has the maximum frequency,
so it is the modal class.
$\therefore\text{x}_{\text{k}}=40,\ \text{f}_{\text{k}}=18,\ \text{f}_{\text{k}-1}=6$ and $\text{f}_{\text{k}+1}=10$
Now,
Mode $=\text{x}_{\text{k}}+\text{h}\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1}}\Big\}$
$=40+20\Big\{\frac{18-6}{2(18)-6-10}\Big\}$
$=40+20\times\frac{14}{20}$
$=40+12$
$=52$
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Question 33 Marks
In a frequency distribution table with 12 classes, the class width is 2.5 and the lowest class boundary is 8.1, then what is the upper class boundary of the highest class?
Answer
Number of desses = 12
Class-width = 2.5
Lowest class boundary = 8.1
Thus, Upper class boundary of the highest class
= Lowest dass brundary + Class-width, x Number of classes
= 8.1 + 2.5 × 12
= 8.1 + 30
= 38.1
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Question 43 Marks
The observations 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 are arranged in ascending order. What is the value of x if the median of the data is 63?
Answer
Observation on ascending order:
$29, 32, 48, 50, x, x +2, 72, 78, 84, 95$
$\Rightarrow$ Niumber of observations = N = 10 (even)
$\Rightarrow\text{Median}=\frac{\text{Value of}\Big(\frac{\text{N}}{2}\Big)^{\text{th}}\text{observation+Value of}\Big(\frac{\text{N}}{2}+1\Big)^{\text{th}}\text{observation}}{2}$
$\Rightarrow63=\frac{\text{Value of}\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation+value of}\Big(\frac{10}{2}+1\Big)^{\text{th}}\text{observation}}{2}$
$\Rightarrow126=$ Value of $5^{\text {th }}$ observation + value of $6^{\text {th }}$ observation
$\Rightarrow126=\text{x}+(\text{x}+2)$
$\Rightarrow126=2\text{x}+2$
$\Rightarrow2\text{x}=124$
$\Rightarrow\text{x}=62$
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Question 53 Marks
??The following frequency distribution gives the monthly consumption of electricity of 64 consumers of locality.
Monthly consumption (in units)
65-85
85-105
105-125
125-145
145-165
165-185
Number of consumers
4
5
13
20
14
8
Form a 'more than type' cumulative frequency distribution.
Answer
Monthly consumption (in units)
Cumulative frequency
More than 65
60 + 4 = 64
More than 85
55 + 5 = 60
More than 105
42 + 13 = 55
More than 125
22 + 20 = 42
More than 145
8 + 14 = 22
More than 165
8
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Question 63 Marks
Compute the mean of following data:
Class
1-3
3-5
5-7
7-9
Frequency
12
22
27
19
Answer
We have
Class
Frequency $(f_i$​​​​​​​)
Class mark ($x_i​​​​​​​$)
$f_ix_i$
1-3
12
2
24
3-5
22
4
88
5-7
27
6
162
7-9
19
8
152
Total
$\sum\text{f}_\text{i}=80$
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=426$
$\therefore$ Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{426}{80}=5.325$
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Question 73 Marks
If the median of $\frac{\text{x}}{5},\ \frac{\text{x}}{4},\ \frac{\text{x}}{2},\ \text{x}$ and $\frac{\text{x}}{3},$ where x > 0, is 8, find the value of x.
Answer
Data in ascending order:
$\frac{\text{x}}{5},\ \frac{\text{x}}{4},\ \frac{\text{x}}{2},\ \text{x}$
⇒ Numberof observations - 5 (odd)
⇒ Median = value of $\Big(\frac{5+1}{2}\Big)^{\text{th}}$ observation
⇒ 8 = Value of $3^{rd}$ observation
$\Rightarrow8=\frac{\text{x}}{3}$
$\Rightarrow\text{x}=24$
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Question 83 Marks
The median of 19 observations is 30. Two more observations are made and the values of these are 8 and 32. Find the median of 21 observations taken together.
Answer
Total number of observations $= N =19$ (odd)
$\Rightarrow$ Median velue of $\left(\frac{ N +1}{2}\right)^{\text {th }}$ observation Value of $10^{\text {th }}$ observetion
Given, medtan $=30$
$\Rightarrow 10^{\text {th }}$ observation 1 s 30 .
Now, two values 8 and 32 are added.
Since $8<30$ and $32>30$, each one $d$ these two will go on either side of median.
Hence, median is not affected.
$\Rightarrow$ Median $=30$
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Question 93 Marks
??The following table gives the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
Marks obtained (in per cent)
11-20
21-30
31-40
41-50
51-60
61-70
71-80
Number of students
141
221
439
529
495
322
153
  1. Convert the given frequency distribution into the continuous form.
  2. Find the median class and write its class mark.
  3. Find the modal class and write its cumulative frequency.
Answer
  1. Continuous frequency distribution:
Marks obtained (in percent)
Number of students
Cumulative frequency
10.5-20.5
141
141
20.5-30.5
221
362
30.5-40.5
439
801
40.5-50.5
529
1330
50.5-60.5
495
1825
60.5-70.5
322
2147
70.5-80.5
153
2300
  1. $\text{N}=2300\Rightarrow\frac{\text{N}}{2}=1150$
The cumulabve frequency Just greater than 1150 is 1330.

Hence, median dass is 40.5-50.5.

$\therefore$ class-mark of median class $=\frac{40.5+50.5}{2}=\frac{91}{2}=45.5$
  1. Here, maximum frequency = 529
Hence, Modal dess is 40.5-50.5

Thus, the cumulative frequency of modal class is 1330.
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