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Question

The frequency distribution:
x A 2A 3A 4A 5A 6A
f 2 1 1 1 1 1

where A is a positive integer, has a variance of 160. Determine the value of A?

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Solution





(x_{i}) \) (f_{i}) \) fixi fix2i
A 2 2A 2A2
2A 1 2A 4A2
3A 1 3A 9A2
4A 1 4A 16A2
5A 1 5A 25A2
6A 1 6A 36A2
Total fi=7 fixi=22A fix2i=92A2

Now, variance σ2=fix2iN(fixiN)2

Where N=fi

Substituting values from the table
and given σ2=160, we get

160=92A27(22A7)2

160=92A27484A249

160=(92×7484)A249

A2=160×49644484=49

A=7 [AZ+]

Hence, the value of A = 7

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