Download presentation

Presentation is loading. Please wait.

Published byDante Harmison Modified over 8 years ago

1
x012 P(x)0.20.60.2 A probability distribution is as shown. If it is repeated and the 2 distributions combined then the following would be the table of potential results Value 012 0 1 2 0 2 42 13 3 0.2x0.2 2 1 Prob’ 012 0 1 2 0.6x0.60.2x0.6 0.6x0.2 0.2x0.6 0.2x0.2 0.6x0.2

2
Prob’012 00.040.120.04 10.120.360.12 20.040.120.04 y01234 P(y)0.040.240.440.240.04 yP(y)00.240.880.720.16 y 2 P(y)00.241.762.160.64 E(y)= y P(y)] =0+0.24+0.88+0.72+ 0.16 = 2.00 2 = y 2 P(y) – = 0 + 0.24 + 1.76 + 2.16 + 0.64 - 2 2 2 = 4.8 – 4 = 0.8

3
y01234 y - -2012 (y- ) 2 41014 P(y)0.040.240.440.240.04 (y- ) 2 P(y) 0.160.240 0.16 E(y)= y P(y) =0+0.24+0.88+0.72+0.16 = 2.00 2 = (y – P(y) = 0.16 + 0.24 + 0 + 0.24 + 0.16 2 = 0.8 Alternatively using the other formula for Variance

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google