Question

the random variable Y, representing the number of nuts in a chocolate bar has the following probability distribution. compute the mean. K= 0,1,2,3,4 P(K)= 1/10,3/10,3/10,2/10,1/10

Answer 1

$$mean=\mu=E(K)=\sum_ik_ip_i$$

$$=0(\dfrac{1}{10})+1(\dfrac{3}{10})+2(\dfrac{3}{10})+3(\dfrac{2}{10})+4(\dfrac{1}{10})$$

$$=\dfrac{19}{10}=1.9$$