Statistical Tools and Interpretation Test - 3

First arrange the data set in order.

\(1,2,2,4,5,6,7\)

Here the median is the middle number = 4

So, the lower quartile is the median of the lower half of the data.

Then, the middle of the lower half will be \(1,2,2\)

Therefore, the lower quartile is 2.

Relation between Mean, Median and Mode will be

\(\Rightarrow\) Mode \(=(3 \times\) Median \()-(2 \times\) Mean \()\)

\(\therefore\) The required result will be Mode \(=3\) Median - \(2\) Mean

Given: Median = 15

\(8,11,12, x+6,17,18,23\)

\(n=7,\)

since, \(n\) is odd, median should be \(4^{\text {th }}\) term.

So, \(x+6\) should be median.

\(\Rightarrow x+6=15\)

\(\Rightarrow x=9\)

Marks obtained by 10 students in the test:

12, 22, 32, 41, 26, 30, 14, 11, 18, 35

Arranging the marks in ascending order:

11,12,14,18,22,26,30,32,35,41

Since, the number of students are 10, median will be the mean of two middle terms

Median = mean of 5th term and 6th term

Median \(=\frac{22+26}{2}\)

Median \(=24\) marks

The first quartile (Q1) represents the value below which 25% of the data lie. In this case, if 25% of the values are greater than 75, then Q1 must be greater than 75, as it is the point below which 25% of the data fall. So, in a discrete series if \(25 \%\) values are greater than 75, then \({Q}_1>75\).

The given data, mean \(=9 \lambda\) and mode \(=6 \lambda\)

median \(=\frac{\text { mode }+2 \times \text { mean }}{3}\)

\(=\frac{6 \lambda+2 \times 9 \lambda}{3}\)

\(=8 \lambda\)

We first prepare the cumulative frequency table (see above).

We can write the equations as:

\(\Sigma(X-20)=25 \) or \(\Sigma X-\Sigma 20=25\)

\(\Sigma(X-25)=0 \) or \(\Sigma X-\Sigma 25=0\)

\(\Sigma(X-35)=-25 \) or  \(\Sigma X-\Sigma 35=-25\)

Now, we know that \(\Sigma 20=20 n, \Sigma 25=25 n\), and \(\Sigma 35=35 n\), where \(n\) is the number of observations.

Substitute these into the equations:

\(\Sigma X-20 n=25 \cdots (1)\)

\(\Sigma X-25 n=0 \cdots (2)\)

\(\Sigma X-35 n=-25 \cdots (3)\)

Subtracting equation (2) from equation (1) we get,

\((\Sigma X-20 n)-(\Sigma X-25 n) =25-0 \)

\(\Sigma X-20 n-\Sigma X+25 n  =25 \)

\(5 n  =25 \Rightarrow n =5\)

Putting the value of \(n\) in equation 2 

\(\Sigma X-25(5) =0 \)

\(\Sigma X-125  =0 \Rightarrow \Sigma X =125\)

So, the value of \(X\) is:

\(X=\frac{\Sigma X}{n}=\frac{125}{5}=25\)

To find the arithmetic mean (A.M.), we sum all the numbers and divide by the total count of numbers. 

Given that the A.M. of the given numbers is 6:

\(\frac{1+3+5+6+x+10}{6}=6\)

Now, we can solve for \(x\) :

\(1+3+5+6+x+10=6 \times 6\)

\(1+3+5+6+x+10=36\)

\( 25+x=36\)

\(x=36-25 =11\)

So, the value of \(x\) is 11

Median is considered most suitable measure for open end classification because in open end distribution exact values of the data is unknown and it is nearly impossible to calculate the mean, mode and geometric mean of such data.