Four buses carrying 163 high school students arrive to Montreal. The buses carry, respectively, 40, 41, 35, and 47 students. One of the studetns is randomly selected. Let X denote the number of students that were on the bus carrying this randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on his bus. Compute the expectations of X and Y.

E(X) ? Var (X)? E(Y)? Var(Y)?

E(X) = 41.196

Var(X) = 17.975

E(Y) = 40.75

Var(Y) = 9.922

Step-by-step explanation:

40 + 41 + 35 + 47 = 163

The probability that a student was on the bus is proportional to how many students were on the bus. So, for bus a, its = 40/163 ; for bus b, it's = 41/163 ;for bus c, it's 35/163= ;for bus d, it's = 47/163

A) E(X) = 40(40/163) + 41(41/163) + 35(35/163) + 47(47/163)

E(X) = 6715/163

E(X) = 41.196

B) Var(X) = [(40 - 41.196)²•(40/163)] + [(41 - 41.196)²•(41/163)] + [(35 - 41.196)²•(35/163)] + [(47 - 41.196)²•(47/163)]

Var(X) = (1.430416 + 1.575056 + 1343.66456 + 1583.261552)/163

Var(X) = 2929.931584/163

Var(X) = 17.975

C) The bus driver has a 1/4 probability on being on any of the buses. Thus,

E(Y) = 163/4

E(Y) = 40.75

D) Var(Y) = (40.75 - 40)²/4 + (40.75 - 41)²/4 + (40.75-35)²/4 + (40.75-47)²/4

Var(Y) = 0.140625 + 0.015625 + 9.765625

Var(Y) = 9.922

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation: