Respuesta :
Answer:
a) Binomial distribution B(n=12,p=0.01)
b) P=0.007
c) P=0.999924
d) P=0.366
Step-by-step explanation:
a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.
b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.
[tex]P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107[/tex]
Then,
[tex]P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007[/tex]
c) In this case, the distribution is B(1200,0.01)
[tex]P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924[/tex]
d) In this case, the distribution is B(100,0.01)
We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.
[tex]P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366[/tex]
