Question

The mean of the $100$ observation is $45$.It was later found that two observations $19$ and $31$ are incorrectly recovered as $91$ and $13$.Find the correct mean. $(PS)$. the $A.M$ of $10$ number is $20$ and $A.M$ of $30$number is $60$ , then Find the $A.M$ of the combined data.$(PS)$.Find the mean of the number from $1$ to $20$that are divisible by $3?$

Answer 1

Hint: Given the cost, numbers in this statement, so it is simple to calculate to add up all the numbers, then divide by how many numbers there are. When divided by the number of items in that list, the total of all the amounts of a group is classified as the Arithmetic Mean, then find each statement solution with the use of a particular formula.

Formula used:
Mean formula ${\text{Mean = }}\dfrac{{{\text{Sum}}\,{\text{of All}}\,{\text{data}}\,{\text{Points}}}}{{{\text{Number}}\,{\text{of}}\,{\text{data}}\,{\text{Points}}}}$
${\text{Mean = Assumed}}\,{\text{Mean + }}\dfrac{{{\text{Sum}}\,{\text{of All}}\,{\text{data}}\,{\text{Points}}}}{{{\text{Number}}\,{\text{of}}\,{\text{data}}\,{\text{Points}}}}$
Arithmetic Mean Formula$ = \dfrac{{{x_{1 + }}{x_2} + {x_3}........ + {x_n}}}{n}$
Where, $n$ is the number of total terms,

Complete step-by-step answer:
The mean of the $100$ observation is $45$.
Two number incorrect is $91$ and $13$
Correct number is $19$ and $31$
Mean $ = 45$
Now we find the Mean,
${\text{Mean = }}\dfrac{{{\text{Sum}}\,{\text{of All}}\,{\text{data}}\,{\text{Points}}}}{{{\text{Number}}\,{\text{of}}\,{\text{data}}\,{\text{Points}}}}$
Where,
${\text{Number}}\,{\text{of}}\,{\text{Data}}\,{\text{Points}} = 100$
$\dfrac{{{\text{Sum}}\,{\text{of All}}\,{\text{data}}\,{\text{Points}}}}{{100}} = 45$
Substituting the given value in equation
$\dfrac{x}{{100}} = 45$
On simplifying,
$\Rightarrow$ $x = 4500$
The two number $91 + 13 = 104$ is incorrect
The sum of Mean and incorrect are
We get,
Again, we add the correct number $19 + 31 = 50$
Let the sum of mean incorrect number and correct number,
Here,
$\Rightarrow$ $4396 + 50 = 4446$
Now we know that sum of all data points is $4446$
${\text{Mean = }}\dfrac{{{\text{Sum}}\,{\text{of All}}\,{\text{data}}\,{\text{Points}}}}{{{\text{Number}}\,{\text{of}}\,{\text{data}}\,{\text{Points}}}}$
Then, substituting the given equation,
$\Rightarrow$ ${\text{Mean}} = \dfrac{{4446}}{{100}}$
On simplifying given equation
$\Rightarrow$ $Mean = 44.46$
Now we take,
Let the Sum of number of first data be $'x'$ and second data be $'y'$ ,
then the $A.M$ of $10$ number is $20$
substituting the given equations
$\Rightarrow$ $\dfrac{x}{{10}} = 20$
On simplifying,
$\Rightarrow$ $x = 200$
$A.M$ of $30$number is$60$
Substituting the given equations
$\dfrac{y}{{30}} = 60$
On simplifying,
$y = 1800$
Therefore,$A.M$ of combined data,
$ \Rightarrow \dfrac{{x + y}}{{10 + 30}}$
Substituting the value in above equations
$ \Rightarrow \dfrac{{1800 + 200}}{{40}}$
$ \Rightarrow \dfrac{{2000}}{{40}}$
Thus, the value of $A.M$ combined data is $50$
Now we find the mean of the number from $1$ to $20$ that are divisible by $3$
We know the numbers between $1$ and $20$ which are divisible by $3$ are $3,6,9,12,15,18$
$A.M = \dfrac{{{x_{1 + }}{x_2} + {x_3}........ + {x_n}}}{n}$
Mean of these numbers are substituting the above equation is
$ \Rightarrow \dfrac{{3 + 6 + 9 + 12 + 15 + 18}}{6}$
On simplifying,
$10.5$
Hence, the mean of the number from $1$ to $20$ that are divisible by $3$ is $10.5$

Note:The Average is different from an Arithmetic Mean. In several ways, indicators of central tendency or averages are used and form the basis of statistics. The ratio of the sum of observations to the number of observations is the formula for measuring the mean.