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.
![XY \sim N(\mu,\sigma^2)](https://tex.z-dn.net/?f=XY%20%5Csim%20N%28%5Cmu%2C%5Csigma%5E2%29)
![XY \sim N(2200,230^2)](https://tex.z-dn.net/?f=XY%20%5Csim%20N%282200%2C230%5E2%29)
If we set Z to be equal to X+Y
Then,
since two normal distribution appears normal
![Z \sim N(4400,105800)](https://tex.z-dn.net/?f=Z%20%5Csim%20N%284400%2C105800%29)
So;
![P(Z > 5000) = 1 - P( Z< \dfrac{x = \mu}{\sqrt{\sigma}})](https://tex.z-dn.net/?f=P%28Z%20%3E%205000%29%20%3D%201%20-%20P%28%20Z%3C%20%5Cdfrac%7Bx%20%3D%20%5Cmu%7D%7B%5Csqrt%7B%5Csigma%7D%7D%29)
![P(Z > 5000) = 1 - P( Z< \dfrac{5000-4400}{\sqrt{105800}})](https://tex.z-dn.net/?f=P%28Z%20%3E%205000%29%20%3D%201%20-%20P%28%20Z%3C%20%5Cdfrac%7B5000-4400%7D%7B%5Csqrt%7B105800%7D%7D%29)
![P(Z > 5000) = 1 - P( Z< \dfrac{600}{325.2691})](https://tex.z-dn.net/?f=P%28Z%20%3E%205000%29%20%3D%201%20-%20P%28%20Z%3C%20%5Cdfrac%7B600%7D%7B325.2691%7D%29)
![P(Z > 5000) = 1 - P( Z< 1.844626495)](https://tex.z-dn.net/?f=P%28Z%20%3E%205000%29%20%3D%201%20-%20P%28%20Z%3C%201.844626495%29)
![P(Z > 5000) = 1 - P( Z< 1.85)](https://tex.z-dn.net/?f=P%28Z%20%3E%205000%29%20%3D%201%20-%20P%28%20Z%3C%201.85%29)
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;
![Y \sim Bin(3,p)](https://tex.z-dn.net/?f=Y%20%5Csim%20Bin%283%2Cp%29)
where;
![p = 1 - P( Z < \dfrac{2000-2200}{230})](https://tex.z-dn.net/?f=p%20%3D%201%20-%20P%28%20Z%20%3C%20%5Cdfrac%7B2000-2200%7D%7B230%7D%29)
![p = 1 - P( Z < \dfrac{-200}{230})](https://tex.z-dn.net/?f=p%20%3D%201%20-%20P%28%20Z%20%3C%20%5Cdfrac%7B-200%7D%7B230%7D%29)
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
![P(X = r ) = ^nC_r \times p^r \times q ^{n-r}](https://tex.z-dn.net/?f=P%28X%20%3D%20r%20%29%20%3D%20%5EnC_r%20%5Ctimes%20p%5Er%20%5Ctimes%20q%20%5E%7Bn-r%7D)
![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}](https://tex.z-dn.net/?f=P%28%20Y%20%3D%202%20%5C%20or%20%5C%20%203%29%20%3D%5E%203C_2%20%5Ctimes%200.8076%5E2%20%5Ctimes%20%28%201-%200.8076%29%20%5E%20%7B3-2%7D%20%2B%20%5E%203C_3%20%5Ctimes%200.8076%5E3%20%5Ctimes%20%28%201-%200.8076%29%20%5E%20%7B3-3%7D)
![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}](https://tex.z-dn.net/?f=P%28%20Y%20%3D%202%20%5C%20or%20%5C%20%203%29%20%3D%5Cdfrac%7B3%21%7D%7B2%21%283-2%29%21%7D%20%5Ctimes%200.8076%5E2%20%5Ctimes%20%28%200.1924%29%20%5E%201%20%2B%20%5Cdfrac%7B3%21%7D%7B3%21%283-3%29%21%7D%5Ctimes%200.8076%5E3%20%5Ctimes%20%28%20%200.1924%29%20%5E%20%7B0%7D)
![P( Y = 2 \ or \ 3) =0.3764600911 +0.526731063](https://tex.z-dn.net/?f=P%28%20Y%20%3D%202%20%5C%20or%20%5C%20%203%29%20%3D0.3764600911%20%2B0.526731063)
P( Y = 2 or 3) = 0.9031911541
P( Y = 2 or 3) ≅ 0.9032
Answer:
4
Step-by-step explanation:
The answer is 4
Step-by-step explanation:
5^7+5^6=93750
93750÷6=1525