Question

This year's graduating class at Uniform University contains exactly 900 seniors. Each senior has a random number of parents (0, 1, or 2) who will attend the graduation ceremony, and it happens that each of those three possibilities is equally likely, with probability 1/3 each, and the numbers of parents attending the ceremony for different seniors are independent random variables The seating area for parents in the auditorium has 900 seats. Use a Normal approximation to approximate the probability that all of the parents who attend wil be able to be seated. What if the seating area has 925 seats? What is the minimum number of seats that can ensure that with probabty 0.95 of the parents who attend be able to be seated?

Answer 1

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

Answer 2

We have the minimum number of seats required for all parents' attendance to be seat is 941 at a probability of 0.95.

What is a normal distribution?

It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.

1, 2, 3, .....900 is the number of parents that are going to attend the ceremony.

A random number of parents (0, 1, 2)

The probability is 1/3.

$\rm E(X) = 0 * \dfrac{1}{3} + 1*\dfrac{1}{3}+2*\dfrac{1}{3} = 1\\\\\\E(X^2) = 0^2*\dfrac{1}{3}+1^2*\dfrac{1}{3}+2^2*\dfrac{1}{3}= \dfrac{5}{3}$

Then the variance will be

$\rm Var = E(X^2) -[E(X)]^2\\\\Var = \dfrac{5}{3} -1 \\\\Var = \dfrac{2}{3}$

∑Xi with n = 900 will have to be the number of parents attending this ceremony for all seniors

$\rm E \left| S \right| = 900* 1 \\\\E \left| S \right| = 900$

And

$\rm Var \left| S \right| = 900 * \dfrac{2}{3} \\\\Var \left| S \right| = 600$

Then by the normal distribution, we have

$\rm \mu = E(S) = 900\\\\\sigma ^2 = Var (S) = 600$

For a seating area with 900 seats then the probability of the parents present will have seats.

$\rightarrow \dfrac{S - \mu }{\sqrt{\sigma}}\\\\\\\rightarrow \dfrac{900-900}{\sqrt{600}}\\\\\\\rightarrow 0$

For a seating area with 925 seats then the probability of the parents present will have seats.

$\rm \rightarrow \dfrac{S - \mu }{\sqrt{\sigma}}\\\\\\\rightarrow \dfrac{925-900}{\sqrt{600}}\\\\\\\rightarrow P(z \leq 1.0208) \\\\\rightarrow 0.84623$

From the standard normal distribution table, we have

$\rm p(z\leq 1.644845) = 0.95\\\\\dfrac{S-\mu}{\sqrt{\sigma }} \leq 1.644845= 0.95\\\\S \leq \mu + 1.644845 \sqrt{\sigma }= 0.95\\\\S=900 + 1.644845*\sqrt{600}\\\\S= 941$

We have the minimum number of seats required for all parents' attendance to be seat is 941 at a probability of 0.95.

More about the normal distribution link is given below.

https://brainly.com/question/12421652