मराठी

Find the Mean Variance and Standard Deviation of the Following Probability Distribution Xi : A B Pi : P Q Where P + Q = 1 . - Mathematics

Advertisements
Advertisements

प्रश्न

Find the mean variance and standard deviation of the following probability distribution 

xi : a b
pi : p q
where p + q = 1 .
बेरीज
Advertisements

उत्तर

xi pi pixi pixi2
a p ap a2p
b q bq b2q
    `∑`pixi = ap + bq `∑`pixi2=a2p+b2q
 

\[\text{ Now } , \]
\[\text{ Mean }  = \sum p_i x_i = ap + bq\]
\[\text{ Variance}  = \sum p_i {x_i}2^{}_{} - \left( \text{Mean } \right)^2 = a^2 p + b^2 q - \left( ap + bq \right)^2 \]
\[ = a^2 p + b^2 q - a^2 p^2 - b^2 q^2 - 2abpq\]
\[ = a^2 p - a^2 p^2 + b^2 q - b^2 q^2 - 2abpq\]
\[ = a^2 p\left( 1 - p \right) + b^2 q\left( 1 - q \right) - 2abpq\]
\[ = a^2 pq + b^2 qp - 2abpq ............... \left( \because p + q = 1 \right)\]
\[ = pq\left( a^2 + b^2 - 2ab \right)\]
\[ = pq \left( a - b \right)^2 \]
\[\text{ Step Deviation } = \sqrt{\text{ Variance} }\]
\[ = \sqrt{pq \left( a - b \right)^2}\]
\[ = \left| a - b \right|\sqrt{pq}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 32: Mean and Variance of a Random Variable - Exercise 32.2 [पृष्ठ ४३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 32 Mean and Variance of a Random Variable
Exercise 32.2 | Q 3 | पृष्ठ ४३
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×