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Q.

Find the mean deviation about the mean for the following data:

Marks obtained10-2020-3030-4040-5050-6060-7070-80
Number of Students23814832

OR

Find standard deviation by short cut method for the following distribution:

Classes30-4040-5050-6060-7070-8080-9090-100
Frequency371215832
Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2023Subjective· 3mImportance★★★★★
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The mean deviation about the mean for this frequency distribution is 1010.

First find the class midpoints xix_i: 15,25,35,45,55,65,7515, 25, 35, 45, 55, 65, 75, with frequencies fif_i: 2,3,8,14,8,3,22, 3, 8, 14, 8, 3, 2 and N=∑fi=40N = \sum f_i = 40.

Compute ∑fixi\sum f_i x_i: 2(15)+3(25)+8(35)+14(45)+8(55)+3(65)+2(75)2(15)+3(25)+8(35)+14(45)+8(55)+3(65)+2(75)

=30+75+280+630+440+195+150=1800= 30+75+280+630+440+195+150 = 1800

Mean xˉ=180040=45\bar{x} = \dfrac{1800}{40} = 45.

Now find ∣xi−xˉ∣|x_i - \bar{x}| for each class: 30,20,10,0,10,20,3030, 20, 10, 0, 10, 20, 30.

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