Question

Find the value of k for probability distribution of random variable X is given by х 0 1 2 3 P(X) 2k 3k 4k 5k​

Answer 1

Answer:

As we know that

∑Pi=1

∴k+2k+3k+4k+5k=1

⇒k=151

Thus the value of k is 151

X=xi P(X=xi) XP X2P 1 k k k 22k 4k8k33k 9k27k44k16k64k55k25k125k  ∑XP=55k=55×151=3.66∑X2P=225k225×151=15

Mean of the given data (xˉ)=∑XP=3.66

Variance of the given data =∑X2P−xˉ2=