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

  1. A, B, C    
  2. B, A, C    
  3. C, B, A    
  4. C, A, B    

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

    Book A Free Demo

    +91

    Grade*

    Related Questions to study