Download App

Maths 3 Marks Question

Stastics · CBSE STD 10 (CBSE - English Medium). 18 questions.

Questions

3 Marks Question

Take a timed test

18 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Heights of 50 students of class X of a school are recorded and following data is obtained:
Height (in cm):130-135135-140140-145145-150150-155155-160
Number of Students:411127106

Find the median height of the students.
Answer
Height (in cm)No. of Students (f)Cumulative Frequency (cf)
130-13544
135-1401115
140-1451227→ Median Class
145-150734
150-1551044
155-160650
N = $sigma$f = 50
Since, $N=50$ is an even number.
So, $\frac{N}{2}=\frac{50}{2}=25$
and median class is $140-145$.
$
l=140, h=5, c=15, f=12 \quad \text { (given) }
$
Now, Median
Area of $\triangle A B C=\frac{1}{2} \times B C \times A E$
$
\begin{aligned}
\text { In } \triangle D B C & =l+h\left(\frac{\frac{N}{2}-c}{f}\right) \\
& =140+5\left(\frac{25-15}{12}\right)=140+\left(\frac{5 \times 10}{12}\right) \\
& =140+4.167 \\
& =144.167
\end{aligned}
$
Hence, median height of the students of 144.167 cm .
View full question & answer→
Question 23 Marks
Find the mean of the following data using assumed mean method:
Class:0-55-1010-1515-2020-25
Frequency:87101312
Answer
Class IntervalMid- point (x)Frequency (f)d = x - Afd
0-52.58-10-80
5-107.57-5-35
10-1512.5 = A1000
15-2017.513565
20-2522.51210120
Sigma*f = 50Sigma*fd = 70


Here, assumed mean, $A=12.5$
Now,
$
\begin{aligned}
\text { Mean } & =A+\frac{\Sigma f d}{\Sigma f}=12.5+\frac{70}{50} \\
& =12.5+1.4=13.9
\end{aligned}
$
View full question & answer→
Question 33 Marks
The mean of the following frequency distribution is $25$. Find the value of $f.$
Class: $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
Frequency: $5$ $18$ $15$ $f$ $6$
Answer
Given, mean $= 52$
Class Interval Mid$-$point $(xi)$ Frequency$(fi)$ $fixi$
$0-10$ $5$ $5$ $25$
$10-20$ $15$ $18$ $270$
$20-30$ $25$ $15$ $375$
$30-40$ $35$ $F$ $35f$
$40-50$ $45$ $6$ $270$
    $\sum f_i=44+f$ $ \sum f_i x_i =940 \\ +35 f$
$\therefore \text { Mean }=\frac{\sum f_i x_i}{\sum f_i}$
$\Rightarrow 25=\frac{940+35 f}{44+f}$
$\Rightarrow 25 \times 44+25 f=940+35 f$
$\Rightarrow 10 f=1100-940$
$\Rightarrow 10 f=160$
$\Rightarrow f=16$
View full question & answer→
Question 43 Marks
Calculate the mode of the following distribution:
Class:10-1515-2020-2525-3030-35
Frequency:472081
Answer
Here, the maximum frequency is 20 and the Corresponding class is $20-25$, So, modal class is $20-25$
We have, Lower class boundary of modal group (l) $=20$
group width $( h )=25-20=5$
Frequency of the modal group $\left(f_1\right)=20$
Frequency of the group before the modal group $\left(f_0\right)=7$
Frequency of the group after the modal group ( $f_2$ ) $=8$
$
\begin{aligned}
\text { Mode } & =l+\frac{f_1-f_2}{2 f_{l}-f_0-f_2} \times h \\
= & 20+\frac{20-8}{2 \times 20-7-8} \times 5 \\
= & 20+\frac{12}{40-15} \times 5=20+\frac{12}{25} \times 5 \\
= & 20+2.4=22.4
\end{aligned}
$
View full question & answer→
Question 53 Marks
Find the mean of the data by step deviation method.
$C.I$ $15-25$ $25-35$ $35-45$ $45-55$ $55-65$ $65-75$ $75-85$ $85-95$
Frequency $6$ $11$ $7$ $4$ $4$ $2$ $1$ $10$
Answer
We will first find the mid values and take middle value as assumed mean.
Here, let assumed mean $(a) = 60$ and $h =$ upper limit $-$ lower limit $= 10$
$C.I$ $x_i=\frac{u . l+l . l}{2}$ $fi$ $di = xi - a$ $u_i=\frac{d_i}{h}$ $fiui$
$15 - 25$ $20$ $6$ $-40$ $-4$ $-24$
$25 - 35$ $30$ $11$ $-30$ $-3$ $-33$
$35 - 45$ $40$ $7$ $-20$ $-2$ $-14$
$45 - 55$ $50$ $4$ $-10$ $-1$ $-4$
$55 - 65$ $60$ $4$ $0$ $0$ $0$
$65 - 75$ $70$ $2$ $10$ $1$ $2$
$75 - 85$ $80$ $1$ $20$ $2$ $2$
$85 - 95$ $90$ $10$ $30$ $3$ $
$
    $\sum f_i=45$     $\sum f_i u_i=-41$
Mean $\bar{x}=a+h\left[\frac{\sum f_i u_i}{\sum f_1}\right]$
$\bar{x}=60+10\left[\frac{-41}{45}\right]$
$\bar{x}=60+10(-0.91)$
$\bar{x}=60-9.1, \bar{x}=50.9$
Therefore, the mean of the given data is $50.9 .$
View full question & answer→
Question 63 Marks
Find the mean of the following distribution:
Literacy Rate (in %)30-4040-5050-6060-7070-8080-90
Number of odes6710683
Answer
Literacy Rate (in %)Number of odes (frequency)(fi)Cumulative frequency
30-4066
40-5076+7=13
50-601013+10 = 23
60-70623+6= 29
70-80829+8=37
80-90337 + 3 = 40

Mode $=l+\left(\frac{f_1-f_0}{2 \times f_1-f_0-f_2}\right) \times h$
Where, $l=$ Lower class limit of the modal class $h=$ Class size
$f_1=$ stands for the frequency of the modal class.
$f_0=$ Frequency of the class preceding the modal class.
$f_2=$ Frequency of the class succeeding the modal class.
Modal class is the class with highest frequency.
Thus,
Modal class $=50-60$
Mode $=50+\left(\frac{10-7}{2 \times 10-7-6}\right) \times 10$
Mode $=50+\left(\frac{3}{20-13}\right) \times 10$
Mode $=50+\left(\frac{3}{7}\right) \times 10$
Mode $=54.29$
Thus the modal literacy rate is $54.29 \%$.
View full question & answer→
Question 73 Marks
Find the mean of the following distribution:
Class0-66-1212-1818-2424-30
Frequency7510122
Answer
Mean can be found using the step-deviation method,
Class intervalMid value (xi)di = xi - 15$u_i=\frac{\left(x_i-15\right)}{6}$Frequency fifiui
0-63-12-27-14
6-129-6-15-5
12-181500100
18-2421611212
24-302712224
$\sum f_i=36$$\sum f_i u_i=-3$


Hear, $a=15$ and the class interval $(h)=6$
The mean of the data is given by,
$
\bar{x}=a+\left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h=15+\left(\frac{-3}{36}\right) \times 6=15-\frac{1}{2}=14.5
$
Thus the mean of the following distribution is 14.5 .
View full question & answer→
Question 83 Marks
A school conducted a test $($of $100$ marks$)$ in English for students of class $X$. The marks obtained by students are shown in the following table:
Age $($in years$)$ $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$
Number of persons $1$ $2$ $4$ $15$ $15$ $25$ $15$ $10$ $2$ $1$
Evaluate modal marks.
Answer
To find the modal marks look for the group that has the highest frequency.
This is because the mode is the number that comes up the most times.
From the given table it can be observed that there are $25$ students who have obtained the marks in the range $50-60$.
Therefore the modal class is $50-60$.
The formula to estimate the Mode is:
$\text { Mode }=L+\frac{\left(f_m-f_{m-1}\right)}{\left(f_m-f_{m-1}\right)+\left(f_m-f_{m+1}\right)} \times w$
Where, $L=$ the lower boundary of the modal group $=50$
$f_{ m -1}=$ the frequency of the group before the modal group $= 15$
$f_{ m }=$ the frequency of the modal group $=25$
$f_{m+1}=$ the frequency of the group after the modal group $=15$
$w =$ the group width $=10$
$\text { Mode }=50+\frac{(25-15)}{(25-15)+(25-15)} \times 10, $
$\text { Mode }=50+\frac{10}{10+10} \times 10$
$\Rightarrow \text { Mode }=50+5=55$
Hence, the modal marks are $55$.
View full question & answer→
Question 93 Marks
The following table gives the ages of 1000 persons who visited a shopping centre on Sunday:
Age (in years)0-1010-2020-3030-4040-5050-6060-70
Number of persons105222220138102113100

Find the mean number of the people who visited a shopping centre on Sunday
Answer
Age (in years)Class marks (xi)Number of persons (fi)fixi
0-105105525
10-20152223330
20-30252205500
30-40351384830
40-50451024590
50-60551136215
60-70651006500


It is given that there are 1000 persons.
Therefore $\sum f_i=1000$
$
\begin{aligned}
\begin{aligned}
\sum f_i x_i=525+3330+5500+4830+4590
+6215+6500=31490
\end{aligned} \\
\text { Therefore mean }=\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{31490}{1000}=31.490
\end{aligned}
$
Hence, the mean number of the people who visited a shopping centre on Sunday is 31 rounded to nearest whole number.
View full question & answer→
Question 103 Marks
In a hospital, age record of diabetic patients was recorded as follows:
Age $($in years$)$ $0-15$ $15-30$ $30-45$ $45-60$ $60-75$
Number of patients $5$ $20$ $40$ $50$ $25$
Find the median age.
Answer
Age $($in years$)$ Number of patients $(fi)$ Cumulative frequency
$0-15$ $5$ $5$
$15-30$ $20$ $25$
$30-45$ $40$ $65$
$45-60$ $50$ $115$
$60-75$ $25$ $140$
Median $m=L+\left(\frac{\frac{N}{2}-F}{f_m}\right) C$
Here, $L =$ lower boundary of median class
$N =$ Total frequency
$F =$ Cumulative frequency before median class
$C =$ Class size
$fm =$ frequency of median class,
$\frac{N}{2}=70$, so median class is $45-60$.
$f_{m}=50, C=1,5, F=65, L=45 $ and $ N=140$
Now substituting these values,
$m=45+\left(\frac{\frac{140}{2}-65}{50}\right) 15$
$=45+\left(\frac{70-65}{50}\right) 15$
$=45+\frac{75}{50}=45+1.5=46.5$
Hence median age is $46.5$ years.
View full question & answer→
Question 113 Marks
In annual examination, marks (out of 90) obtained by students of class IX in mathematics are given below:
Marks0-1515-3030-4545-6060-7575-90
Number of students24520910

Find the mean marks of the student.
Answer
To find the mean of the given data, we can find the midpoint of the given class.
MarksNumber of students (fi)xifixi
0 - 1527.515
15 - 30422.590
30 - 45537.5187.5
45 - 602052.51050
60 - 75967.5607.5
75 - 901082.5825

Here total number of students $=50$
Total weight $=\sum f_i x_i=2775$
Mean $=\frac{\sum f_i x_i}{\sum f_i}=\frac{2775}{50}=55.5$
Mean marks of the students 55.5.
View full question & answer→
Question 123 Marks
Find the median of the following data:
Marks20-3030-4040-5050-6060-7070-8080-90
Number of Students51525207810
Answer
MarksNumber of studentsCumulative Frequency
20 - 3055
30 - 40155 + 15 = 20
40 - 502525 + 20 = 45
50 - 602020 + 45 = 65
60 - 7077 + 65 =72
70 - 8088 + 72 = 80
80 - 901010 + 80 = 90

From the table, it can be observed that $n=90$
$
\Rightarrow \frac{n}{2}=\frac{90}{2}=45
$
Also, cumulative frequency(cf) just greater than $\frac{n}{2}$ (i.e. 45 ) is 65 , which belongs to the interval $50-60 \backslash \backslash 192.168 .0 .59 \backslash$ scan $\backslash$ Hasan sir\04 Carbon and its Compounds (47-66) (a)
Folder
Therefore, median class $50-60$
Lower limit(l) of median class $=50$.
Class size (h) $=10$
Frequency $\left(f_{ i }\right)$ of median class $=20$
Cumulative frequency (cf) of class preceding the median class $=45$
Median $=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h=50+\left(\frac{45-45}{20}\right) \times 10$
Median $=50+0=50$
Hence, the median is 50 .

View full question & answer→
Question 133 Marks
Find the mean of the following data:
Classes5-1515-2525-3535-4545-5555-6565-75
Frequency61016152487
Answer
Using direct method0:
ClassesFrequency (fi)Class Mark(xi)fixi
5 - 1561060
15 - 251020200
25 - 351630480
35 - 451540600
45 - 5524501200
55 - 65860480
65 - 75770490
TotalΣfi = 86Σfixi = 3510

The sum of the values in the last column gives $\sum f_i x_i$. So, the mean is given by
$\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{3510}{86}=40.81$
Hence, the mean of data is 40.81 .
View full question & answer→
Question 143 Marks
Compute the arithmetic mean for the following data:
Marks ObtainedNumber of students
Below 1014
Below 2022
Below 3037
Below 4058
Below 5067
Below 6075
Answer
The given frequency table is of less than type represented with upper class limits. Therefore, the class intervals with their respective cumulative frequency can be defined as below:
Marks ObtainedNumber of students(f1)Cumulative frequency (cf)XiFixi
0-101414570
10-2022-14=82215120
20-3037-22=153725375
30-4058-37=215835735
40-5067-58=96745405
50-6075-6=787555440
Total(n)75Σfiχi =2145

The mean is given by the formula: Mean $=\frac{\sum f_i x_i}{\sum f_i}=\frac{2145}{75}=28.6$
Therefore, mean $=28.6$.
View full question & answer→
Question 153 Marks
Calculate the median for the following distribution:
Marks ObtainedNumber of students
Below 106
Below 2015
Below 3029
Below 4041
Below 5060
Below 6070
Answer
The given frequency table is of less than type represented with upper class limits. Therefore, the class intervals with their respective cumulative frequency can be defined as below:
Marks ObtainedNumber of students (f)Cumulative frequency (cf)
0 - 1066
10 - 2015 – 6 = 915
20 - 3029 - 15 = 1429
30 - 4041 - 29 = 1241
40 - 5060 - 41 = 1960
50 - 6070 - 60 = 1070
Total(n)70

From the table, it can be observed that $n=70$ $\Rightarrow \frac{n}{2}=\frac{70}{2}=35$
Also, cumulative frequency(cf) just greater than $\frac{n}{2}$ (i.e. 35 ) is 41 , which belongs to the interval $30-40$. Therefore, median class $=30-40$
Lower limit(l) of median class $=30$.
Class size (h) $=10$
Frequency $\left(f_t\right)$ of median class $=12$
Cumulative frequency ( $c f$ ) of class preceding the median class $=29$
$
\begin{aligned}
\text { Median } & =l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\
& =30+\left(\frac{35-29}{12}\right) \times 10 \\
& =30+\left(\frac{6}{12}\right) \times 10 \\
& =30+5=35
\end{aligned}
$
Hence, the median is 35 .
View full question & answer→
Question 163 Marks
The mean of the following frequency distribution is $52.$ Find the missing frequency.
$C.I.$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency $5$ $3$ $4$ $f$ $2$ $6$ $13$
Answer
Construct the following table:
Class Interval Frequency $(fi)$ $Xi$ $Fi x xi$
$10-20$ $5$ $15$ $75$
$20-30$ $3$ $25$ $75$
$30-40$ $4$ $35$ $140$
$40-50$ $F$ $45$ $45f$
$50-60$ $2$ $55$ $110$
$60-70$ $6$ $65$ $390$
$70-80$ $13$ $75$ $975$
$\text { Mean }=\frac{\sum f_i x_i}{\sum f}$
$52=\frac{1765+45 f}{33+f}$
$\Rightarrow 1716+52 f=1765+45f$
$\Rightarrow 7 f=49$
$\Rightarrow 52(33+f)=1765+45 f$
Dividing both sides by $7,$
$f=\frac{49}{7}=7$
Hence, the missing frequency is $7.$
View full question & answer→
Question 173 Marks
$200$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows
No. of letters $1-5$ $5-10$ $10-15$ $15-20$ $20-25$
No. of surnames $20$ $60$ $80$ $32$ $8$
Evaluate the median of it
Answer
Construct the following table:
Class Interval Frequency Cumulative Frequency
$1-5$ $20$ $20$
$5-10$ $60$ $80$
$10-15$ $80$ $160$
$15-20$ $32$ $192$
$20-25$ $8$ $200$
$\frac{N}{2}=\frac{200}{2}=100$
So, the median class is $10-15$.
Median $=l+\left(\frac{\frac{N}{2}-c f}{f}\right) \times h$
Here $c f=$ cumulative frequency of class preceding median class,
$f=$ frequency of median class,
$l=$ lower limit of median class and
$h=$ size of class.
$l=10, c f=80, f=80$ and $h=5$
Median $=10+\frac{100-80}{80} \times 5$
Median $=10+1.25=11.25$
Hence, median is $11.25$ .
View full question & answer→
Question 183 Marks
Find the mean of the following frequency distribution using assumed mean method
Classes2-88-1414-2020-2626-32
Frequency f:6312118
Answer
Construct the following table:
ClassesFiXidi = xi - Adi × fi
2-865-12-72
8-14311-6-18
14-201217 = A$0$0
20-261123666
26-328291296

Here, $A =17$
Mean $=A+\frac{\sum f_i d_i}{\sum f_i}$
$\sum f_i d_i=72$ and $\sum f=40$
Mean $=17+\frac{72}{40}$
Mean $=17+1.8=18.8$
Hence, the mean is 18.8 .
View full question & answer→
Generate Paper FreeAll Stastics topics
3 Marks Question - Maths STD 10 Questions - Vidyadip