A consumer organization inspecting new cars found that many had appearance defects (dents, scratches, paint chips etc.). No car had more than three defects. Here is the probability distribution for the number of appearance defects in a new car. X = number of defects 0 1 2 3 P(x) 0.58 0.24 0.13 0.05 Find the expected number of appearance defects in a new car. Enter a number in decimal form. Do not round.

Answer:0.65Step-by-step explanation:When we have a finite number of events  [tex]x_1,x_2,...,x_n[/tex] with its corresponding probabilities [tex]p_1,p_2,...,p_n[/tex] with [tex]p_1+p_2+...+p_n=1[/tex]  then the expected value of the distribution is given by [tex]x_1.p_1+x_2.p_2+...+x_n.p_n[/tex]  In our case, we have [tex] x_1=0,x_2=1,x_3=2,x_4=3[/tex] with probabilities [tex] p_1=0.58,p_2=0.24,p_3=0.13,p_4=0.05[/tex] So, the expected value is E = 0*0.58+1*0.24+2*0.13+3*0.05 = 0.65