During the twelve months of the year 1998, an executive charged 4, 1, 5, 6, 3, 5, 1, 0, 5, 6, 4, and 3 business luncheons at the Wardlaw club. What was the mean monthly number of luncheons charged by the executive? If the mean is \[x\] then find \[100x\].
Answer
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Hint: For solving this problem we use the general formula of mean because the given data does not depend on the class interval. The formula of mean s given as \[\text{mean}=\dfrac{\text{sum of luncheons in each month}}{\text{number of months}}\]
As we get the mean from the above formula we can find \[100x\] by multiplying 100 to mean.
Complete step-by-step solution:
We are given that in the 12 months of 1998, the number of luncheons charged by the executive is
Let us find the sum of all luncheons charged in all months as
\[\begin{align}
& \Rightarrow sum=4+1+5+6+3+5+1+0+5+6+4+3 \\
& \Rightarrow sum=43 \\
\end{align}\]
Here, we know that there are a total of 12 months.
Let us assume that the mean is \[x\].
We know that the formula of mean is given as
\[\text{mean}=\dfrac{\text{sum of luncheons in each month}}{\text{number of months}}\]
Now, by substituting the required values in the above formula of mean we get
\[\begin{align}
& \Rightarrow x=\dfrac{43}{12} \\
& \Rightarrow x=3.5833 \\
& \Rightarrow x\simeq 3.58 \\
\end{align}\]
Therefore the mean of luncheons charged by the executive is 3.58.
Now, let us multiply the mean with 100 to get a value of \[100x\]
By multiplying mean by 100 we get
\[\begin{align}
& \Rightarrow 100x=100\times 3.58 \\
& \Rightarrow 100x=358 \\
\end{align}\]
Therefore, if the mean is \[x\] then the value of \[100x\] is 358.
Note: This is a very simple problem which is just based on a simple formula of the mean. Mistake while solving the problem can be done only while finding the sum of luncheons in all months. Due to fast solving or reading students may miss one or two quantities and solve the problem that results in the wrong answer. Considering all the 12 months is important.
As we get the mean from the above formula we can find \[100x\] by multiplying 100 to mean.
Complete step-by-step solution:
We are given that in the 12 months of 1998, the number of luncheons charged by the executive is
| Month | Number of luncheons |
| January | 4 |
| February | 1 |
| March | 5 |
| April | 6 |
| May | 3 |
| June | 5 |
| July | 1 |
| August | 0 |
| September | 5 |
| October | 6 |
| November | 4 |
| December | 3 |
Let us find the sum of all luncheons charged in all months as
\[\begin{align}
& \Rightarrow sum=4+1+5+6+3+5+1+0+5+6+4+3 \\
& \Rightarrow sum=43 \\
\end{align}\]
Here, we know that there are a total of 12 months.
Let us assume that the mean is \[x\].
We know that the formula of mean is given as
\[\text{mean}=\dfrac{\text{sum of luncheons in each month}}{\text{number of months}}\]
Now, by substituting the required values in the above formula of mean we get
\[\begin{align}
& \Rightarrow x=\dfrac{43}{12} \\
& \Rightarrow x=3.5833 \\
& \Rightarrow x\simeq 3.58 \\
\end{align}\]
Therefore the mean of luncheons charged by the executive is 3.58.
Now, let us multiply the mean with 100 to get a value of \[100x\]
By multiplying mean by 100 we get
\[\begin{align}
& \Rightarrow 100x=100\times 3.58 \\
& \Rightarrow 100x=358 \\
\end{align}\]
Therefore, if the mean is \[x\] then the value of \[100x\] is 358.
Note: This is a very simple problem which is just based on a simple formula of the mean. Mistake while solving the problem can be done only while finding the sum of luncheons in all months. Due to fast solving or reading students may miss one or two quantities and solve the problem that results in the wrong answer. Considering all the 12 months is important.
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