Can you use #sigma^2=Sigma(x_i-mu)^2p_i# to show that #sigma^2=Sigmax_i^2p_i-mu^2#?

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Answer 1 · 2017-05-01T12:20:33

See proof below

We need

#mu=sumx_1p_i#

#sump_i=1#

Therefore,

#sigma^2=sum(x_i-mu)^2p_i#

#=sumx_i^2p_i-sum2x_ip_imu+summu^2p_i#

#=sumx_i^2p_i-2musumx_ip_i+mu^2sump_i#

#=sumx_i^2p_i-2mu^2+mu^2#

#=sumx_i^2p_i-mu^2#

I hope that this will help!