Answer:
The expectation the instructor to send each day
E(X) = 3.65
Step-by-step explanation:
Given data x : 0 1 2 3 4 5
P(X=x) : 0.05 0.05 0.1 0.1 0.4 0.3
The given data satisfy the two conditions
I) Given all probabilities are p₁(x) ≥0
ii) sum of all probabilities is equal to one
0.05 + 0.05 + 0.1 + 0.1 + 0.4 +0.3 =1
Therefore given data is discrete probability distribution
Expectation :-
suppose a random variable X assumes the values [tex]x_{1}, x_{2},..x_{n}[/tex] with respective probabilities [tex]p_{1}, p_{2},..p_{n}[/tex] , then the Expectation or Expected value of X , denoted by E(X) = ∑pi xi
[tex]E(X) = 0X0.05+1X 0.05 +2X 0.1 +3X 0.1 +4X0.4 +5X 0.3[/tex]
on simplification, we get
E(X) = 3.65