SChand CLASS 9 Chapter 15 Mean, Median and Frequency Polygon Exercise 15(A)

 Exercise 15 A

Question 1

Ans: (a) first $s$ natural numbers are 1,2,3,4,5

$\text { So } \begin{gathered}\text { Mean }=\frac{1+2+3+4+5}{5}=\frac{15}{5}\left\{\bar{x}=\frac{\sum x_{1}}{n}\right\} \\=3\end{gathered}$

(b) first 7 whole numbere are $0,1,2,3,4,5,6$

$\begin{aligned}&\text { So } \text { mean }=\frac{0+1+2+3+4+5+6}{7} \\&\left\{\begin{array}{l}\left\{x=\frac{\sum x_{1}}{n}\right\} \\=\frac{21}{7}=3\end{array}\right.\end{aligned}$

(C) First 4 prime numbers are $2,3,5,7$

$\text { So } \begin{aligned}\text { mean }=\frac{2+3+5+7}{4} &=\frac{17}{4} \quad\left\{\bar{x}=\frac{\sum x_{1}}{n}\right\}\\&=4.25\end{aligned}$

(d) $1.3 \mathrm{~cm}, 5.7 \mathrm{~cm}, 9.8,6.4 \mathrm{~cm}, 6.9 \mathrm{~cm}$

So mean $=\frac{1.3+5.7+9.8+6.4+6.9}{5} \quad\left\{\bar{x}=\frac{\sum x_{1}}{n}\right\}$

$=\frac{30.1}{5}=6.02 \mathrm{~cm}$

(e) Rs. 7, Rs. 19, Rs. 31, Rs 43, Rs. 70

so $\operatorname{mean}=\frac{7+19+31+43+70}{5} \quad\left\{\bar{x}=\frac{\sum x_{1}}{n}\right\}$

$=\frac{170}{5}=$ Rs 34

Question 2

Ans: Score obtained are $80,85,90,71,60,100$

So

$\begin{aligned}&\text { mean }=\frac{80+85+90+71+60+100}{6} \\&=\frac{486}{6}=81\end{aligned}$

Question 3

Ans: Lat 6 batting score $138,144,155,142,167,172$ So mean score

$\begin{aligned}&=\frac{138+144+155+1+2+167+172}{6} \quad\left\{\bar{x}=\frac{\Sigma x_{1}}{n}\right\} \\&=\frac{918}{6}=153\end{aligned}$

Question 4

Ans: Working hours from Monday to  Wednesday

for 3 days $=2 \frac{1}{2}, 3 \frac{1}{4}$ and $2 \frac{3}{4}$

So mean nour $=\frac{2 \frac{1}{2}+3 \frac{1}{4}+2 \frac{3}{4}}{3}=\frac{8 \frac{1}{2}}{3}$

$=\frac{17}{2 \times 3}=\frac{17}{6}$ hours $=2 \frac{5}{6}$ hours

Question 5

Ans: Ayushree's score $=\frac{68,75,70,45,57,77}{6}$

$\left\{\bar{x}=\frac{\sum x}{n}\right\}$

$=\frac{392}{6}=65 \frac{1}{3} \%$

and Ananya's score (in percent)

$=52,87,64,53,74,81,86$

So Her mean $=\frac{52+87+64+53+74+81+86}{7}$

$=\frac{497}{7}=71 \%$

Question 6

Sol: first 10 odd natural number are 

$1,13,5,7,9,11,13,115,17,19$

So

$\begin{aligned}\operatorname{mean} &=\frac{1+3+5+7+9+11+13+15+17+19}{10} \\&=\frac{100}{10}=10\end{aligned}$

Question 7

Sol: mean of 5 herms $=18$

So Then total $=18 \times 5=90$

But sum of given numbers is

$=16+14+x+23+20=73+x$

So $73+x=90 \Rightarrow x=90-73=17$

Hence $x=17$

Question 8

Ans : Mean of 5 terms = 24 

So their total =$24 \times 5=120$

now Sum of mean= $x+x+2+x+4+x+6+x+8$ =5x + 20

So $5 x+20=120 \Rightarrow 5 x=120-20$

$\Rightarrow 5 x=100 \Rightarrow x=\frac{100}{5}=20$

So $x=20$

Question 9

Ans: Madhu practiced on her sitar for 45 minutes , 30 min, 60min, 50min and 20 min 

So her mean practice 

$\frac{45+30+60+50+20}{5} \quad\left\{\bar{x}=\frac{\sum x_{1}}{n}\right\}$

$=\frac{205}{5}=41 \mathrm{~min}$

Question 10

Ans: Marks secured by Nisha in 4 test are = 73, 86,78,75

Let she secured marks in the next fifth test 

Then mean marks secured will be 

$=\frac{73+86+78+75+x}{5}=\frac{312+x}{5}$

But her mean score=80 marks 

So $\frac{312+x}{5}=80 \Rightarrow 312+x=400$

$\Rightarrow x=400-312=88$

Hence her score in the fifth  test = 88 marks 

Question 11

Ans: mean score of a cricketer $=60$ run

No of innings $(n)=10$

So Total runs $=60 \times 10=600$ runs

mean score in 11 innings $=62$

So Total runs $=62 \times 11=682$ run

So No of runs scored in 11 th innings

$=682-600=82 \text { runs }$

Question 12

Ans: (IMAGE TO BE ADDED)

$\begin{aligned} \operatorname{mean}(\bar{x}) &=\frac{\varepsilon f i x i}{\sum f i} \\=\frac{1600}{50} &=32 \mathrm{~kg} \end{aligned}$

Question 13

(IMAGE TO BE ADDED)

So mean $=\frac{\sum f_{i x i ~}}{\sum f_{i}}=\frac{90}{15}=6$

Question 14

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So mean $=\frac{\sum f_{i x i}}{\sum f_{i}}=\frac{71}{200}=0.355$

Question 15

Ans: No. of bogs $(n)=25$

Highest score $=25$

Lowest $\operatorname{score}=16$

So Range $=25-16=9$

Now taking class intervals 

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$\text { mean }=\frac{\sum f x}{\sum f}=\frac{533}{25}=21.32$

So mean mars $=21.32$ marks.