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Maths-

General

Easy

Question

A random variable has the following distribution.

Then for the values, A = K, B = Mean, C = Variance, the ascending order is

A, B, C
B, A, C
C, B, A
C, A, B

hint Hint:

In probability, sum of all the events = 1 and
$w e space k n o w space t h a t space M e a n space equals space begin inline style stack sum x subscript i P with blank below end style$ and
$V a r i a n c e space f o r m u l a begin inline style stack sum x subscript i squared P space minus space mu squared with blank below end style$

The correct answer is: B, A, C

Given :

In probability, sum of all the events = 1
0.1 + 2K + 3K + 7K + 0.2 + 0.1 = 1
12K + 0.4 = 1
12K = 0.6
K = $1 over 20$ = 0.05
$w e space k n o w space t h a t space M e a n space equals space begin inline style stack sum x subscript i P with blank below end style$
Find mean by substituting values from the table and adding
$mu space equals space left parenthesis negative 3 right parenthesis left parenthesis 0.1 right parenthesis space plus space left parenthesis negative 2 right parenthesis left parenthesis 2 K right parenthesis space plus space left parenthesis negative 1 right parenthesis left parenthesis 3 K right parenthesis space plus space left parenthesis 0 right parenthesis left parenthesis 7 K right parenthesis space plus space left parenthesis 1 right parenthesis left parenthesis 0.2 right parenthesis mu space equals space left parenthesis negative 3 right parenthesis left parenthesis 0.1 right parenthesis space plus space left parenthesis negative 2 right parenthesis left parenthesis 2 cross times 0.05 right parenthesis space plus space left parenthesis negative 1 right parenthesis left parenthesis 3 cross times 0.05 right parenthesis space plus space left parenthesis 0 right parenthesis left parenthesis 7 cross times 0.05 right parenthesis space plus space left parenthesis 1 right parenthesis left parenthesis 0.2 right parenthesis S o l v i n g space a b o v e mu space equals space minus 0.25$
$V a r i a n c e space f o r m u l a begin inline style stack sum x subscript i squared P space minus space mu squared with blank below end style$
Find variance by substituting values from the table and adding
$sum x subscript i squared P space equals space left parenthesis negative 3 right parenthesis squared left parenthesis 0.1 right parenthesis space plus space left parenthesis negative 2 right parenthesis squared left parenthesis 2 cross times 0.05 right parenthesis space plus space left parenthesis negative 1 right parenthesis squared left parenthesis 3 cross times 0.05 right parenthesis space plus space left parenthesis 0 right parenthesis squared left parenthesis 7 cross times 0.05 right parenthesis space plus space left parenthesis 1 right parenthesis squared left parenthesis 0.2 right parenthesis sum x subscript i squared P equals space 2.1 C space equals space sum x subscript i squared P space minus space mu squared C space equals space 2.1 space minus space left parenthesis negative 0.25 right parenthesis squared C space equals space 2.0375$
-0.25 < 0.5<2.0375
Thus, B < A<C

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