Answer:
Then, the distribution for x="number of red marbles seen in 900 draws" is a normal distribution with mean = 810 and standard deviation = 9.
Step-by-step explanation:
As there are draws with replacement, it would have been modeled by a binomial distribution, with n=900 and p=9/10=0.9.
But because n is too large to keep binomial calculus simple, it can be approximated by a normal distribution, with parameters:
[tex]\mu=np=900*0.9=810\\\\\sigma=\sqrt{np(1-p)}=\sqrt{900*0.9*0.1} =\sqrt{81}=9[/tex]
Then, the distribution for x="number of red marbles seen in 900 draws" is a normal distribution with mean = 810 and standard deviation = 9.
