First of all, one needs to understand that this is another method of finding the Mean of a list of data. Sometimes the data is presented in a form which makes the calculation of the mean by the usual method impossible. Such a situation would involve a frequency table where the data is presented as an interval. For example, a question might be, “A person rolls a 1, 2 or 3 five times, then rolls a 4, 5 or 6 three times. What is the mean roll value?” Since we do not know exactly how many times each number has been rolled, we need another method to solve this problem. This would be an occasion to use an Assumed Mean. To solve the problem one chooses an arbitrary number which is close to the mean, in this case we could use 6 as the assumed value. Since it is an arbitrary choice, we could also use 5, or 2. It doesn’t matter.
Next we set up a table as shown below
Frequency (f) Mid-Value (x) d = x – A fd A
5 2 -4 -20 6
3 5 -1 -3 8 -23 fd/f A + (fd/f) -2.875 3.125
We find the mid-value of each interval, and subtract the Assumed value (A) in this case 6, then we multiple that value (-4) times the frequency (5) to yeild -20. Continue down the table. In the end, we add up our frequencies (8) and f x d (-23), then divide fd by d. We then add this value to the assumed mean (6 + (-2.875)) and this is our mean 3.125
Statistics – Assumed Mean Method
Like its, name in this method we assume a number as the mean of the given data.
Then we subtract this assumed mean from all class marks. Hence find di.
Let a is our assumed mean and x1, x2, x3 …….xn are the class marks. Then
d1 = x1 – a
d2 = x2 – a
d3 = x3 – a
dn = xn – a
Collectively we can write
di = xi – a
After that we calculate
∑fi di = f1 d1 + f2 d2 + f3 d3 + ——- + fn dn
Since di = xi – a
This is the formula for calculating mean with assumed mean method. The method
will be more clear with following example
Note : You can assume any number as assumed mean (a) but the general choice is
any xi which lies in the middle of range of xi.
Calculate the mean of following data with the help of assumed
|C.I.||Xi||fi||d = xi – a||fidi|
|50-60||55 ? a||3||0||0|
|`sum` fi=80||`sum` fidi= 1560|
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Let a = 55
= 55 + (1560 / 80)
= 55 + 19.5