Answer:
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:
[tex]\mu=n*p=1600*0.75=1200[/tex]
and its standard deviation as
[tex]\sigma=\sqrt{n*p*(1-p)}=\sqrt{1600*0.75*0.25}=17.32[/tex]
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).
[tex]X=\mu-z*\sigma=1200+2.362*17.32=1159[/tex]
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.