Anew cinema is under construction and the total number of the seats is to be determined.

assume that in this area, 1600 people will go to cinema everyday and for the probability for each person to choose this cinema is 3/4. the following are the requirements for determining the seat quantity q:

1) q is as large as possible
2) the probability of "there are over 200 empty seats in one day" is no more than 0.1

hint: central limit theorem

a) 1359 seats

Step-by-step explanation:

We have a population of 1600 people. There is a probability p=0.75 of choosing this cinema for every individual.

We could treat this problem as a binomial distribution problem, but because of n=1600 being bigger enough, we can treat this a normal distribution.

The mean of this distribution can be estimated as:

and its standard deviation as

In the standard normal distribution (mean=0 and sd=1), the value z in which the probability P(x<z)=0.1 is z=-2.362.

This z-value (z=-2.362) is equivalent to X=1159 in our distribution X=N(1200;17.32).

Then, can be said that there is a probability P=0.1 of having less than 1159 people in the cinema.

If this probability correspond to a situation with 200 empty seats, the minimum amount of seats needed is 1159+200=1359 seats.

i dont know

step-by-step explanation:

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !