Exercise 15 B
Question 1
Ans : 2,3,5,7,9
Here h=5 which is odd
If the median is the middle most term
So median = $\frac{n+1}{2}$ th term $=\frac{5+1}{2}$ th
$=3$ rd term which is 5
so median $=5$
Question 2
Sol: Here n=8 which is even
If the median is the middle most term
median = mean of $\frac{n}{2}$ th $\frac{n+1}{2}$ th
$=\frac{1}{2}(4+h+5+h$ term $)$
$=\frac{1}{2}(16+20)=\frac{36}{2}=18$
Hence median $=18$
Question 3
Ans: Here number of terms n= 7 which is odd
Arrange in Ascending order,
$33,44,148,51,60,61,63$
If median is the middle most term
So median $=\frac{n+1}{2} \mathrm{th}$ term $=\frac{7+1}{2} \mathrm{th}$ term
=4th term which is 51
Hence median =51
Question 4
Ans: Arranging is ascending order
$8,10,11,13,116,17,19,22,25,31$
Here number of terms (n) = 10 which is even if the median is the middle most term
So median $=\frac{1}{2}\left[\frac{n}{2}\right.$ th term $+\left(\frac{n}{2}+1\right)$ th term $]$
$=\frac{1}{2}\left[\frac{10}{2}\right.$ th term $+\left(\frac{10}{2}+1\right)$ th term $]$
$=\frac{1}{2}[5$ th term $+6$ th term $]$
$=\frac{1}{2}[16+17]=\frac{33}{2}=16.5$
So median $=16.5$
Question 5
Ans: First 10 prime number are $2,3,5,7,11,13,17,19,23,29$
So Here n= 10 which is even
So median = $=\frac{1}{2}\left[\frac{n}{2}\right.$ th term $+\left(\frac{n}{2}+1\right)$ th term $]$
$\begin{aligned}=& \frac{1}{2}\left[\frac{10}{2} \text { th term }+\left(\frac{10}{2}+1\right) \text { th term }\right] \\=& \frac{1}{2}[5 \text { th term }+6 \text { th term }] \\=& \frac{1}{2}[11+13]=\frac{24}{2}=12 \\ & \text { So median }=12 \end{aligned}$
Question 6
Ans: No of scores (n) = 13 which is odd
35, 28 , 13 ,17 20 ,30,19,29,11,10,29,23 ,18,25,17
Arranging in ascending order
$10,11,13,17,17,18,19,20,23,25,28,29,29,30,35$
$\begin{aligned} \text { median } &=\frac{n+1}{2} \text { th rerm }=\frac{15+1}{2} \text { th }=\frac{16}{2} \text { th } \\ &=8 \text { th term } \end{aligned}$
Which is 20
So median =20 marks
Question 7
Ans: Here $n=7$ which is odd
Arranging in descending order.
$5,15,25,35,45,55,60$
So median $=\frac{n+1}{2}$ th term $=\frac{7+1}{2}$ th term
$=4$ th term which is 35
So median marks $=35$
Question 8
Ans: No. of obervations $=16$ which is even
$\begin{aligned}&\text { So median }=\frac{1}{2}\left[\frac{n}{2} \text { th lerm }+\left(\frac{n}{2}+1\right) \text { thterm }\right] \\&\left.=\frac{1}{2}\left[\frac{16}{2} \text { th term }+\frac{16}{2}+1\right) \text { th term }\right] \\&=\frac{1}{2}[8 \text { th lerm }+9 \text { th term }] \\&=\frac{1}{2}(25+27)=\frac{52}{2}=26 \\&\text { So median }=26\end{aligned}$
Question 9
Ans: The observation are given in ascending order $8.9112,18(x+2),(x+1), 30,31,34.39$
Here $n=10$ which is even
$\text { So } \begin{aligned}\text { median } &=\frac{1}{2}\left[\frac{n}{2} \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }\right] \\&\left.=\frac{1}{2}\left[\frac{10}{2} \text { th term }+\frac{10}{2}+1\right] \text { th term }\right] \\&=\frac{1}{2}[5 th \text { term }+6 \text { th term }] \\&=\frac{1}{2}[x+2+x+4] \\&=\frac{1}{2}[2 x+6]=x+3\end{aligned}$
But median is given $=24$
$\text { So } x+3=24 \Rightarrow x=24-3=21$
Question 10
Ans: (i) The data is given ascending order $18,20,25,126,30, x, 37,38,39,48$ Which are 10 an even number
$\begin{aligned}\text { So } \text { median } &=\frac{1}{2}\left[\frac{n}{2} \mathrm{th} \text { term }+\left(\frac{n{2}+1\right) \text { th term }\right] \\&=\frac{1}{2}\left[\frac{10}{2} \text { th term }+\left(\frac{10}{2}+1\right) \text { th term }\right] \\&=\frac{1}{2}[5 \text { th term }+6 \text { th term }] \\&=\frac{1}{2}[30+x]=\frac{30+x}{2}\end{aligned}$
But median is given $=35$
So $\begin{aligned} \frac{30+x}{2} &=35 \Rightarrow 30+x=70 \\ \Rightarrow x &=70-30=40 \end{aligned}$
(ii) By replacing 48 by 28 the terms will be
$18,20,25,26,28,130,37,38,39,40$
New median $=\frac{1}{2}$ [5th term $+6$ th term $]$
$=\frac{1}{2}[28+30]$
$=\frac{58}{2}=29$
Hence new median $=29$
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