Question

The measurements (in mm) of the diameters of the head of the screws are given below:

Diameter(in mm)	No of screws
\[30 - 35\]	\[9\]
\[36 - 38\]	\[21\]
\[39 - 41\]	\[30\]
\[42 - 44\]	\[20\]
\[45 - 47\]	\[18\]

Calculate mean diameter of head of a screw by ‘Assumed Mean Method’.

Answer 1

Hint: In this question, we have the set of given observations.we have to find out the mean.
The mean (or average) of observations, is the sum of the values of all the observations divided by the total number of observations.

Formula used: If \[{x_1},{x_2},{x_3},......,{x_n}\] are mid-points or class marks of n class intervals and with respective frequencies \[{f_1},{f_2},{f_3},........,{f_n}\] then the formula for assumed mean method is
\[\overline x = a + \dfrac{{\sum\limits_{i = 1}^n {{f_i}{d_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
Where,
a = assumed mean
\[{f_i}\]= frequency of ${i^{th}}$ class
\[{d_i} = {x_i} - a\] = deviation of ${i^{th}}$ class
\[\sum {{f_i} = n} \] =Total number of observations
\[{x_i}\] = Class mark = $\dfrac{\text{(upper class limit +lower class limit)}}{{2}}$

Complete step-by-step solution:
We have the measurements (in mm) of the diameters of the head of the screws.
We need to find out the mean diameter of head of a screw by ‘Assumed Mean Method’.To find the mean first we need to calculate the midpoint \[\left( {{x_i}} \right)\] of the respective diameters and take the middle value of \[\left( {{x_i}} \right)\] as ‘a’, the assumed mean. Then finding the deviations \[{d_i}\], using the formula and \[{f_i}{d_i}\] value for each case, we can put it in the Assumed mean formula.
Let \[a\] be the assumed mean.
\[a\] is taken as the midpoint of all \[\left( {{x_i}} \right)\].
Let us take \[40\] as assumed mean.
Then \[a = 40\] and the deviations \[{d_i} = {x_i} - a = {x_i} - 40\].

Diameter(in mm)	No. of Screws\[\left( {{f_i}} \right)\]	Mid-point \[\left( {{x_i}} \right)\]	Deviation \[\left( {{d_i}} \right)\]\[ = {x_i} - a\]\[ = {x_i} - 40\]	\[{f_i}{x_i}\]
\[30 - 35\]	\[9\]	\[\dfrac{{30 + 35}}{2} = \dfrac{{65}}{2} = 32.5\]	\[ - 7.5\]	\[9 \times - 7.5 = - 67.5\]
\[36 - 38\]	\[21\]	\[\dfrac{{36 + 38}}{2} = \dfrac{{74}}{2} = 37\]	\[ - 3\]	\[21 \times - 3 = - 63\]
\[39 - 41\]	\[30\]	\[\dfrac{{39 + 41}}{2} = \dfrac{{80}}{2} = 40\]= a	\[0\]	\[30 \times 0 = 0\]
\[42 - 44\]	\[20\]	\[\dfrac{{42 + 44}}{2} = \dfrac{{86}}{2} = 43\]	\[3\]	\[20 \times 3 = 60\]
\[45 - 47\]	\[18\]	\[\dfrac{{45 + 47}}{2} = \dfrac{{92}}{2} = 46\]	\[6\]	\[18 \times 6 = 108\]

Now, the mean of the data is given by,
\[\overline x = a + \dfrac{{\sum\limits_{i = 1}^n {{f_i}{d_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
That is, \[\overline x = 40 + \dfrac{{ - 67.5 + \left( { - 63} \right) + 0 + 60 + 108}}{{9 + 21 + 30 + 20 + 18}} = 40 + \dfrac{{37.5}}{{98}} = 40 + 0.38 = 40.38\]

Hence the mean diameter of the head of a screw is \[40.38\].

Note: Mean
There are several kinds of mean in mathematics, especially in statistics. For a data set, the arithmetic mean, also called the expected value or average, is the central value of a discrete set of numbers specifically, the sum of the values divided by the number of values.
\[{\text{m = }}\dfrac{{{\text{Sum of the terms}}}}{{{\text{Number of terms}}}}\].
In statistics the assumed mean is a method for calculating the arithmetic mean and standard deviation of a data set. It simplifies calculating accurate values by hand. Its interest today is chiefly historical but it can be used to quickly estimate these statistics.