Answer:
941
Step-by-step explanation:
= 1, 2, 3...900 is the number of parents that are going to attend the ceremony.
we are given a random number of parents (0 1 2)
each of these random numbers have a probability of 1/3
so we multiply the numbers by their probability
E(X) = 0*1/3 + 1*1/3 +2*1/3
= 1
E(X²) = 0²*1/3 + 1²*1/3 + 2²*1/3
= 5/3
we calculate the variance
= E(X²)-[E(X)²
= 5/3 - 1²
= 2/3( got this by taking the lcm)
∑Xi with n = 900 will have to be the number of parents attending this ceremony for all seniors
E|S| = 900 * 1
= 900
var |S| = 900*2/3
=600
we are usng normal distribution to solve this
μ = E(S) = 900
σ² = var(S) = 600
for a seating area wth 900 seats now we are to find the probability that all the parents present will have seats
s-μ/√σ
900 -900/√600
= 0
p(z≤0)
= 0.5
for a seating area wth 925 seats we are to fnd probability that all parents will be seated
= s-μ/√σ
= 925-900/√600
= p(z≤1.0208)
= 0.84623
going to the table of standard normal distribution we have
p(z≤1.644845)= 0.95
s-μ/σ≤1.644845 = 0.95
we cross multply
s-μ≤1.644845σ
we make s subject of the formula
s ≤ μ+1.644845σ = 0.95
we now μ = 900 and
σ = √600
900+1.644845*√600
=900 +40.388
=940.388
= 941
we have that the minimum number of seat requred for all parents n attendance to be seat is 941 at a probability of 0.95
