
The AM of n observations is\[\overline x \] . If the sum of \[n - 4\;\]observations is K, then the mean of the remaining 4 observations is
A. \[\dfrac{{\overline x - K}}{4}\]
B. \[\dfrac{{n\overline x - K}}{{n - 4}}\]
C. \[\dfrac{{n\overline x - K}}{4}\]
D. \[\dfrac{{n\overline x - (n - 4)K}}{4}\]
Answer
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Hint: We are given arithmetic mean of some observations and also the sum of some of the numbers. To find the solution we will first consider the numbers as \[{x_1},{x_2},{x_3},...................{x_n}\] and put them into values of K and \[\overline x \]. Then we would exclude the last 4 terms to get their sum and also the average.
Complete step by step answer:
We will start by saying,
Let n observationsbe, \[{x_1},{x_2},{x_3},...................{x_n}\]
Now, we have their arithmetic mean as, \[\overline x \]
So,
\[\dfrac{{{x_1} + {x_2} + {x_3} + ................... + {x_n}}}{n} = \overline x \]........eq(1)
And also given, the sum of \[n - 4\;\] observations is K.
Then we also get,
\[{x_1} + {x_2} + {x_3} + ................... + {x_{n - 4}} = K\].........eq(2)
Equation 1 can be written as,
\[\dfrac{{{x_1} + {x_2} + {x_3} + ................... + {x_{n - 4}} + {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n}}}{n} = \overline x \]
On substituting the value of equation (2) we get,
\[ \Rightarrow\dfrac{{K + {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n}}}{n} = \overline x \]
On Simplifying we get,
\[ \Rightarrow K + {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n} = n\overline x \]
On subtracting K from both sides we get,
\[ \Rightarrow {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n} = n\overline x - K\]
Hence, the mean of last four observation is,
\[\dfrac{{{x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n}}}{4} = \dfrac{{n\overline x - K}}{4}\]
Hence, option (c) is correct.
Note: The arithmetic mean gives us the idea of simply the mean or the average (when the context is clear), is the sum of a collection of numbers divided by the count of numbers in the collection. The collection is often a set of results of an experiment or an observational study, or frequently a set of results from a survey.
Complete step by step answer:
We will start by saying,
Let n observationsbe, \[{x_1},{x_2},{x_3},...................{x_n}\]
Now, we have their arithmetic mean as, \[\overline x \]
So,
\[\dfrac{{{x_1} + {x_2} + {x_3} + ................... + {x_n}}}{n} = \overline x \]........eq(1)
And also given, the sum of \[n - 4\;\] observations is K.
Then we also get,
\[{x_1} + {x_2} + {x_3} + ................... + {x_{n - 4}} = K\].........eq(2)
Equation 1 can be written as,
\[\dfrac{{{x_1} + {x_2} + {x_3} + ................... + {x_{n - 4}} + {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n}}}{n} = \overline x \]
On substituting the value of equation (2) we get,
\[ \Rightarrow\dfrac{{K + {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n}}}{n} = \overline x \]
On Simplifying we get,
\[ \Rightarrow K + {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n} = n\overline x \]
On subtracting K from both sides we get,
\[ \Rightarrow {x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n} = n\overline x - K\]
Hence, the mean of last four observation is,
\[\dfrac{{{x_{n - 1}} + {x_{n - 2}} + {x_{n - 3}} + {x_n}}}{4} = \dfrac{{n\overline x - K}}{4}\]
Hence, option (c) is correct.
Note: The arithmetic mean gives us the idea of simply the mean or the average (when the context is clear), is the sum of a collection of numbers divided by the count of numbers in the collection. The collection is often a set of results of an experiment or an observational study, or frequently a set of results from a survey.
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