Respuesta :
Answer: Mean = 4.8 and variance = 5.16
Step-by-step explanation:
Since we have given
Let X be the number of storms occur in next year
Y= 1 if the next year is good.
Y=2 if the next year is bad.
Mean for good year = 3
probability for good year = 0.4
Mean for bad year = 5
probability for bad year = 0.6
So, Expected value would be
[tex]E[x]=\sum xp(x)\\\\=3\times 0.4+5\times 0.6\\\\=1.2+3\\=4.2[/tex]
Variance of the number of storms that will occur.
[tex]Var[x]=E[x^2]-(E[x])^2[/tex]
[tex]E[x^2]=E[x^2|Y=1].P(Y=1)+E[x^2|Y=2].P(Y=2)\\\\=(3+9)\times 0.4+(5+25)\times 0.6\\\\=12\times 0.4+30\times 0.6\\\\=4.8+18\\\\=22.8[/tex]
So, Variance would be
[tex]\sigma^2=22.8-(4.2)^2\\\\=5.16[/tex]
Hence, Mean = 4.8 and variance = 5.16
