Question

The weights of 50 fruits were recorded as given below. Calculate the mean weight, to the nearest gram by step-deviation method.

$$\begin{array}{cccccccc}\text{Weight in grams}& \text{80-85}& \text{85-90}& \text{90-95}& \text{95-100}& \text{100-105}& \text{105-110}& \text{110-115}\\ \text{Number of fruits}& \text{5}& \text{8}& \text{20}& \text{10}& \text{12}& \text{8}& \text{4}\end{array}$$

• A. 96
• B. 95
• C. 99
• D. 100

Answer 1

The correct option is A 96

$$\begin{array}{cccccc}\text{Weight}& f_i& \text{Mid- value}& d=x_i-a& \text{u =}\frac{x_i-a}{h}& f_i{u}_i\\ \text{80-85}& \text{5}& \text{82.5}& \text{-15}& \text{-3}& \text{-15}\\ \text{85-90}& \text{8}& \text{87.5}& \text{-10}& \text{-2}& \text{-16}\\ & & & & & \\ \text{90-95}& \text{10}& \text{92.5}& \text{-5}& \text{-1}& \text{-10}\\ & & & & & \\ \text{95-100}& \text{12}& \text{97.5 = a}& \text{0}& \text{0}& \text{0}\\ & & & & & \\ \text{100-105}& \text{8}& \text{102.5}& \text{5}& \text{1}& \text{8}\\ & & & & & \\ \text{105-110}& \text{4}& \text{107.5}& \text{10}& \text{2}& \text{8}\\ & & & & & \\ \text{110-115}& \text{3}& \text{112.5}& \text{15}& \text{3}& \text{9}\end{array}$$

From the table we have
$\sum f_i=50$ and $\sum f_i{u}_i=-16$

$\therefore Mean=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$
$=97.5+\frac{-16}{50}\times 5=95.9$

$=96\text{g (nearest gram)}$