Answer:
A) P(Z > 5000) = 0.0322
B) P( Y = 2 or 3) ≅ 0.9032
Step-by-step explanation:
From the given information;
Suppose the sales for the first week are denoted by X and the sales for the second week are denoted by Y.
Then;
X & Y are independent and they follow a normal distribution.
i.e.
[tex]XY \sim N(\mu,\sigma^2)[/tex]
[tex]XY \sim N(2200,230^2)[/tex]
If we set Z to be equal to X+Y
Then, [tex]Z \sim N(2 \times 2200,2 \times 230^2)[/tex] since two normal distribution appears normal
[tex]Z \sim N(4400,105800)[/tex]
So;
[tex]P(Z > 5000) = 1 - P( Z< \dfrac{x = \mu}{\sqrt{\sigma}})[/tex]
[tex]P(Z > 5000) = 1 - P( Z< \dfrac{5000-4400}{\sqrt{105800}})[/tex]
[tex]P(Z > 5000) = 1 - P( Z< \dfrac{600}{325.2691})[/tex]
[tex]P(Z > 5000) = 1 - P( Z< 1.844626495)[/tex]
[tex]P(Z > 5000) = 1 - P( Z< 1.85)[/tex]
From the Z - tables;
P(Z > 5000) = 1 - 0.9678
P(Z > 5000) = 0.0322
B)
Let Y be the random variable that obeys the Binomial distribution.
Y represents the numbers of weeks in the next 3 weeks where the gross weekly sales exceed $2000
Thus;
[tex]Y \sim Bin(3,p)[/tex]
where;
[tex]p = 1 - P( Z < \dfrac{2000-2200}{230})[/tex]
[tex]p = 1 - P( Z < \dfrac{-200}{230})[/tex]
p = 1 - P( Z < - 0.869565)
From the Z - tables;
p = 1 - 0.1924
p = 0.8076
Now;
P(Y ≥ 2) = P(Y = 2) + P( Y =3 )
Using the formula
[tex]P(X = r ) = ^nC_r \times p^r \times q ^{n-r}[/tex]
[tex]P( Y = 2 \ or \ 3) =^ 3C_2 \times 0.8076^2 \times ( 1- 0.8076) ^ {3-2} + ^ 3C_3 \times 0.8076^3 \times ( 1- 0.8076) ^ {3-3}[/tex]
[tex]P( Y = 2 \ or \ 3) =\dfrac{3!}{2!(3-2)!} \times 0.8076^2 \times ( 0.1924) ^ 1 + \dfrac{3!}{3!(3-3)!}\times 0.8076^3 \times ( 0.1924) ^ {0}[/tex]
[tex]P( Y = 2 \ or \ 3) =0.3764600911 +0.526731063[/tex]
P( Y = 2 or 3) = 0.9031911541
P( Y = 2 or 3) ≅ 0.9032
