Question

The fruit vendor has apples packed in the boxes. Each of the boxes contains a different number of apples. The following was the distribution of apples according to the number of boxes. Find the mean number of apples kept in a packing box using the direct method.

Number of apples	30-33	33-36	36-39	39-42	42-45	45-48
Number of boxes	60	120	150	80	50	90

(a) 39.5
(b) 41.5
(c) 38.5
(d) 35.5

Answer 1

Hint: In the above problem, we are asked to find the mean number of apples kept in a box using the direct method. For that, first of all, we are going to find ${{x}_{i}}$ where “i” corresponds to a different number of boxes. Now, to find ${{x}_{i}}$, we are going to add the lower and upper limit of the number of apple distribution intervals then we will divide it by 2. Now, we are going to name each of the different numbers of boxes as ${{f}_{i}}$ where “i” represents the counting number and it takes value from 1 to 6. Now, we are going to multiply ${{f}_{i}}$ by ${{x}_{i}}$ corresponding to different numbers of boxes and then this multiplication over different numbers of boxes. To find the mean, we are going to divide this addition by the addition of all the different numbers of boxes.

Complete step-by-step solution:
The apple distribution table is given below:

Number of apples	30-33	33-36	36-39	39-42	42-45	45-48
Number of boxes	60	120	150	80	50	90

Let us name the number of boxes as ${{f}_{i}}$ in the above table.

Number of apples	30-33	33-36	36-39	39-42	42-45	45-48
Number of boxes $\left( {{f}_{i}} \right)$	60	120	150	80	50	90

Now, we are going to find ${{x}_{i}}$ corresponding to different numbers of boxes which we will calculate by adding the lower and upper limit of the apples distribution interval and then divide this addition by 2.
The “i” in ${{x}_{i}}$for interval 30-33 is 1 so the value of ${{x}_{1}}$ is equal to:
$\begin{align}
  & \dfrac{30+33}{2} \\
 & =\dfrac{63}{2}=31.5 \\
\end{align}$
Now, the value of ${{x}_{2}}$ for interval 33-36 is equal to:
$\begin{align}
  & \dfrac{33+36}{2} \\
 & =\dfrac{69}{2}=34.5 \\
\end{align}$
Similarly, you can find the value of ${{x}_{i}}$ for the remaining intervals.

Number of apples	30-33	33-36	36-39	39-42	42-45	45-48
${{x}_{i}}$	31.5	34.5	37.5	40.5	43.5	46.5
Number of boxes $\left( {{f}_{i}} \right)$	60	120	150	80	50	90

Now, we are going to multiply ${{f}_{i}}$ by ${{x}_{i}}$ in the above table as follows:

Number of apples	30-33	33-36	36-39	39-42	42-45	45-48
${{x}_{i}}$	31.5	34.5	37.5	40.5	43.5	46.5
Number of boxes $\left( {{f}_{i}} \right)$	60	120	150	80	50	90
${{x}_{i}}{{f}_{i}}$	1890	4140	5625	3240	2175	4185

The summation of ${{x}_{i}}{{f}_{i}}$ over different number of boxes is equal to:
$\begin{align}
  & \sum\limits_{i=1}^{6}{{{x}_{i}}}{{f}_{i}}=1890+4140+5625+3240+2175+4185 \\
 & \Rightarrow \sum\limits_{i=1}^{6}{{{x}_{i}}}{{f}_{i}}=21255 \\
\end{align}$
The summation of $\left( {{f}_{i}} \right)$ over different values of “i” from 1 to 6 is equal to:
$\begin{align}
  & \sum\limits_{i=1}^{6}{{{f}_{i}}}=60+120+150+80+50+90 \\
 & \Rightarrow \sum\limits_{i=1}^{6}{{{f}_{i}}}=550 \\
\end{align}$
Now, we are going to divide $\sum\limits_{i=1}^{6}{{{x}_{i}}}f$ by $\sum\limits_{i=1}^{6}{{{f}_{i}}}$ we will get the mean number of apples.
$\begin{align}
  & \dfrac{\sum\limits_{i=1}^{6}{{{x}_{i}}}f}{\sum\limits_{i=1}^{6}{{{f}_{i}}}}=\dfrac{21255}{550} \\
 & \Rightarrow \dfrac{\sum\limits_{i=1}^{6}{{{x}_{i}}}f}{\sum\limits_{i=1}^{6}{{{f}_{i}}}}=38.64 \\
\end{align}$
Hence, the correct option is (c).

Note: You can check the mean value that we got above is correct or not. The mean value which you got in the above solution is 38.64. And you can find that this value which is 38.64 lying in one of the marks intervals lies in the marks interval given in the above problem and that interval is 36-39.
Now, if you got the mean as 3940 then you will know that this is not the correct mean because none of the intervals contain 3940.