Statistics help?

Past car sales experiences has shown that 20% of all cars sold are black. if 25 recent sales of car are random


selected find the probabilities








a) Wat is the expected number of black cars in our sample?


b) wat is the variances of number of black cars in our sample?


c_ exactly 6 of the cars are black


d) between 1 and 4 cars are inclusive are black


e) no more than 3 cars are black


f) more than 20 of the cars are not black?





i dont get this math question. can someone please help me and explain it

Statistics help?
Let X be the number of black cars. X has the binomial distribution with n = 25 trials and success probability p = 0.20





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.





the mean of the binomial distribution is n * p


the variance of the binomial distribution is n * p * (1 - p)





a) the mean is 25 * 0.20 = 5





b) variance is 25 * .2 * .8 = 4





c) P(X = 6) = 0.1633459





d) P(1 ≤ X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)


= 0.02361183 + 0.07083550 + 0.13576804 + 0.18668105


= 0.4168964





e) more than 20 are not black is the same as less than five being black.





P(X %26lt; 5)


= P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)


= 0.186681050 + 0.135768036 + 0.070835497 + 0.023611832 + 0.003777893


= 0.4206743
Reply:a) E(X) = 25*0.2 = 5


b) Var(X) = 25*0.2*0.8 = 4


c) P(X=6) = 25choose6 (0.8)^19 (0.2)^6 = 0.1633


d) P(1 %26lt;= X %26lt;= 4) = the sum for x = (1,2,3,4) ... = 0.4169


e) P(X%26lt;= 3) = sum for x = (0,1,2,3) = 0.2340


f) P(X %26lt; 5) = 0.4207
Reply:The number of cars that are black follows a binomial probability distribution (see link below) with parameters p=0.20 and n=25 and all of the answers follow from its properties.





a) expected number of black cars = np = 25(0.20) = 5





b) variance = np(1-p) = 25(0.20)(0.80) = 4





c) P(6 black) = 25!/(6!(25-6)!) 0.20^6 0.80^(25-6) = 0.163346





Note: you can get all these probabilities easily from various sources. See 2nd link below, for example





d) P(between 1 and 4 black) = P(1 black ) + ... + P(4 black)





e) etc.





f) etc.