Question

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes:
50−52
53−55
56−58
59−61
62−64
Number of boxes:
15
110
135
115
25
Find the mean number of mangoes kept in a packing box.

Answer 1

STEP DEVIATION METHOD:

Step deviation method is used in the cases where the deviation from the assumed mean 'A' are multiples of a common number. If the values of ‘di’ for each class is a multiple of ‘h’ the calculation become simpler by taking ui= di/h = (xi - A )/h

Here, h is the class size of each class interval.

★★ Find the class marks of class interval. These class marks would serve as the representative of whole class and are represented by xi.  

★★ Class marks (xi)  = ( lower limit + upper limit) /2

★★ We may take Assumed mean 'A’ to be that xi which lies in the middle of x1 ,x2 …..xn

MEAN = A + h ×(Σfiui /Σfi) , where ui =  (xi - A )/h

[‘Σ’ Sigma means ‘summation’ ]

FREQUENCY DISTRIBUTION TABLE IS IN THE ATTACHMENT  

★★ Here,the frequency table is given in inclusive form. So we first change it into exclusive form by  subtracting and adding h/2 to the lower and upper limits respectively of each class ,where ‘h’ denotes the difference of lower limit of a class and upper limit of the previous class.

Here, h/2 = ½ = 0.5  

From the table : Σfiui = 25 ,  Σfi = 400

Let the assumed mean, A = 57,  h = 3

MEAN = A + h ×(Σfiui /Σfi)

Mean = 57 + 3 (25/400)

= 57 + 3 (1/16)

= 57 + 3/16

= 57 + 0.187

= 57.187

Mean = 57.19  (approximate)

Hence, the Mean number of mangoes kept in a packing box is 57.19  (approximate)

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

➡️Solution:

To solve this problem by direct method it has a complex calculation,so Using Assume Mean Method:

Since there is a gap in consecutive classes,so apply true class limit.

➡️True class lower limit =

$lower \: limit \: - 0.5$
True class upper limit
$= upper \: limit \: + 0.5$
Let us first complete the table:

Assume that Mean A= 57

$\begin{table}[] \begin{tabular}{|l|l|l|l|l|} \cline{1-5} \begin{tabular}[c]{@{}l@{}}True Class \\ Limit\end{tabular} & Freq. & \begin{tabular}[c]{@{}l@{}}Class \\ Mark\end{tabular} & d_{i}=x_{i}-A & d_{i}f_{i} \\ \cline{1-5} 49.5-52.5 & 15 & 51 & -6 & -90\\ \cline{1-5}52. 5-55.5 & 110 & 54 & -3 &-330 \\ \cline{1-5} 55.5-58.5 & 135 & 57 & 0 &0 \\ \cline{1-5} 58.5-61.5 & 115 & 60 & 3&345 \\ \cline{1-5} 61.5-64.5 & 25 & 63 &6 &150 \\ \cline{1-5} & \Sigma=400 & & & \Sigma=75\\ \cline{1-5} \end{tabular} \end{table}$

$\bar x = A + \frac{\Sigma d_{i}f_{i}}{\Sigma f_{i}} \\ \\ \bar x = 57 + \frac{75}{400} \\ \\ \bar x = 57 + 0.19 \\ \\ \bar x = 57.19 \\ \\ \bar x = 57.2$
Mean number of mangoes kept in a packing box are approx 57.

Hope it helps you.