The following observations are arranged in ascending order:
$26,29,42,53,x,x+2,70,75,82,93$
If the median is $65$ , find the value of $x$.
A. $62$
B. $64$
C. $65$
D. $68$
Answer
514.5k+ views
Hint: Here we have been given a set of observations which are arranged in ascending order and there median is given we have to find the value of the unknown variable. Firstly we will write the formula for calculating median then by using that formula we will substitute the value in it. Finally we will solve the equation obtained and get our desired answer.
Complete step-by-step solution:
The observations is given as follows,
$26,29,42,53,x,x+2,70,75,82,93$…..$\left( 1 \right)$
Median is given as follows,
Median $=65$…..$\left( 2 \right)$
Now as we know median is calculated as follows,
Median $={{\dfrac{\left( n+1 \right)}{2}}^{th}}$ term (If total number of observation is odd)…..$\left( 3 \right)$
Median $=\dfrac{{{\dfrac{n}{2}}^{th}}\,term+{{\left( \dfrac{n}{2}+1 \right)}^{th}}term}{2}$ (If total number of observation is even)……$\left( 4 \right)$
On counting the number of terms in equation (1) we get,
$n=10$
So as the number of observation is even we will use formula in equation (4) as follows,
Median $=\dfrac{{{\left( \dfrac{10}{2} \right)}^{th}}term+{{\left( \dfrac{10}{2}+1 \right)}^{th}}term}{2}$
Median $=\dfrac{{{5}^{th}}term+{{6}^{th}}term}{2}$…$\left( 5 \right)$
Now from equation (1) the ${{5}^{th}}$ and ${{6}^{th}}$ terms are,
$26,29,42,53,\underline{x,x+2},70,75,82,93$
Put the value in equation (5) from equation (2) and above we get,
$\Rightarrow 65=\dfrac{x+x+2}{2}$
$\Rightarrow 65=\dfrac{2x+2}{2}$
Solving further we get,
$\Rightarrow 65\times 2=2x+2$
$\Rightarrow 130-2=2x$
Divide both sides by $2$
$\Rightarrow \dfrac{128}{2}=\dfrac{2x}{2}$
$\Rightarrow 64=x$
So we got the value of $x$ as $64$ .
Hence the correct option is (B).
Note: Median is the middle value of the given data set when we arrange them in ascending or descending order. It is one of the measures of central tendency the other two measures are the mean and mode. A median is that number that separates the higher half of the population by the lower half of the population in the given probability distribution. The formula to calculate the median when the number of observations is even or when they are odd is different and that should be kept in mind while solving these problems.
Complete step-by-step solution:
The observations is given as follows,
$26,29,42,53,x,x+2,70,75,82,93$…..$\left( 1 \right)$
Median is given as follows,
Median $=65$…..$\left( 2 \right)$
Now as we know median is calculated as follows,
Median $={{\dfrac{\left( n+1 \right)}{2}}^{th}}$ term (If total number of observation is odd)…..$\left( 3 \right)$
Median $=\dfrac{{{\dfrac{n}{2}}^{th}}\,term+{{\left( \dfrac{n}{2}+1 \right)}^{th}}term}{2}$ (If total number of observation is even)……$\left( 4 \right)$
On counting the number of terms in equation (1) we get,
$n=10$
So as the number of observation is even we will use formula in equation (4) as follows,
Median $=\dfrac{{{\left( \dfrac{10}{2} \right)}^{th}}term+{{\left( \dfrac{10}{2}+1 \right)}^{th}}term}{2}$
Median $=\dfrac{{{5}^{th}}term+{{6}^{th}}term}{2}$…$\left( 5 \right)$
Now from equation (1) the ${{5}^{th}}$ and ${{6}^{th}}$ terms are,
$26,29,42,53,\underline{x,x+2},70,75,82,93$
Put the value in equation (5) from equation (2) and above we get,
$\Rightarrow 65=\dfrac{x+x+2}{2}$
$\Rightarrow 65=\dfrac{2x+2}{2}$
Solving further we get,
$\Rightarrow 65\times 2=2x+2$
$\Rightarrow 130-2=2x$
Divide both sides by $2$
$\Rightarrow \dfrac{128}{2}=\dfrac{2x}{2}$
$\Rightarrow 64=x$
So we got the value of $x$ as $64$ .
Hence the correct option is (B).
Note: Median is the middle value of the given data set when we arrange them in ascending or descending order. It is one of the measures of central tendency the other two measures are the mean and mode. A median is that number that separates the higher half of the population by the lower half of the population in the given probability distribution. The formula to calculate the median when the number of observations is even or when they are odd is different and that should be kept in mind while solving these problems.
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