Statistical Tools and Interpretation Test - 2

When the values in a series do not have equal importance, we calculate the weighted mean. In weighted mean calculation, each value in the series is multiplied by its corresponding weight, which reflects its importance or frequency, and then these products are summed together. Finally, the sum is divided by the total weight to obtain the weighted mean.

To calculate the median, all the items of a series have to be arranged in ascending or descending order. Arranging the data in order helps in finding the middle value(s) more easily. If the number of observations is odd, the median is the middle value. If the number of observations is even, the median is the average of the two middle values. Arranging the data in order ensures that you can easily identify these middle values, allowing for accurate calculation of the median.

Arranging the given data in ascending order

\(2,3,3,4,4,4,5,6,8,9,9,10,11,15,19\)

Here, Most frequent data is 4 so

Mode \(=4\)

Total terms in the given data, \((n)=15\) (It is odd)

Median \(=\frac{(n+1) }{ 2}^{\text {th }}\) term when \(n\) is odd

\(= \frac{(15+1)}{2}^{\text {th }}\)term \(=(8)^{\text {th }} \) term \(= 6\)

Now, Range \(=\) Maximum value - Minimum value = 19 - 2 = 17

Mean of Range, Mode and median \(=\frac{\text{(Range + Mode + Median)}}{ 3}\)

\(= \frac{(17+4+6)}{3}=\frac{27}{3}=9\)

So, the mean of the Range, Mode and Median is 9.

Given:

\(10,21,10,12,21,10,21,21,21,12,13,21,12\)

Arranging the given terms in Ascending order,

\(10,10,10,12,12,12,13,21,21,21,21,21,21\)

Most repeated term \(=21\) (6 times)

\(\Rightarrow\)  Mode \(=2 1\)

Total observations \(=10\) (even)

Median \(=\frac{10}{2}=5^{\text {th }}\) observations

and \(\frac{10}{2}+1=6^{\text {th }}\) observations

Median \(=\frac{(5^{\text {th }}+6^{\text {th }}) \text { observations }}{2}\)

\(63=\frac{x+x+2}{2}\)

\(\Rightarrow {x}+1=63\)

\(\Rightarrow x=62\)

Symmetrical distribution occurs when the values of variables occur at regular frequencies and the mean, median and mode occur at the same point.

In symmetrical distribution, Mean = Mode = Median

Percentiles divide the distribution into hundred equal parts. It is a measure used in statistics indicating the value below which a given percentage of observations in a group of observations fall. For example, the 20th percentile is the value (or score) below which 20 percent of the observations may be found. The term percentile and the related term percentile rank are often used in the reporting of scores from norm-referenced tests.

Given observations are \(x, x+2, x+4, x+6, x+8\)

Mean \(=\frac{\text {Sum of obsevations }}{\text {Number of observations }} \)

\( \Rightarrow \frac{x+x+2+x+4+x+6+x+8}{5}=11 \)

\(\Rightarrow \frac{5 x+20}{5}=11 \)

\(\Rightarrow x+4=11 \)

\(\Rightarrow x=7\)

So, given observation are \(7,9,11,13,15\).

So, the required mean of three observations \(=\frac{7+9+11}{3}=\frac{27}{3}=9\)

The numbers arranged in ascending order \(=5,7,8,10,12,13\) and \(N\)

Mean \(=\) sum of the terms/Number of the terms

When \(\mathrm{n}\) is odd, Median \(=\frac{(n+1)}{2}\)

The numbers arranged in ascending order \(=5,7,8,10,12,13\) and \(\mathrm{N}\)

Mean \(=\frac{(5+7+8+10+12+13+\mathrm{N})}{7} =\frac{(55+\mathrm{N})}{7}\)

Now,

Total number of terms \(=7\), which is odd

So, Median \(=\frac{(n+1)}{2}= \frac{(7+1)}{2}=4\)th term

The fourth term \(=10\)

Now,

According to the given condition

\(\Rightarrow\frac{(55+\mathrm{N})}{7}=10\)

\(\Rightarrow 55+\mathrm{N}=70\)

\(\Rightarrow \mathrm{N}=(70-55)\)

\(\Rightarrow \mathrm{N}=15\)

\(\therefore\) The required value of \(\mathrm{N}\) is 15.

In a symmetrical distribution, the median divides the dataset into two equal halves. As a result, the number of observations smaller than the median is indeed the same as the number of observations larger than it. This property holds true for any symmetrical distribution, such as a normal distribution.