The gross weekly sales at a certain super market are a Gaussian random with mean $2200 and standard deviation $230. Assume that the sales from week to week are independent.
A) Find the probability that the gross sales over the next two weeks exceed $5000.
B) Find the probability that the gross weekly sales exceed $2000 in at least 2 of the next 3 weeks.

Answers

Answer 1

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


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Answers

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Answers

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Answers

Answer:

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Step-by-step explanation:

1/2 x 4 = 2 + 4 = 6

Answer:

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Pre-Algebra

Order of Operations: BPEMDAS

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

Let’s solve your equation step by step.

Step 1: Multiply both sides by x.

x

2

=

8

x

3

6

x

+

6

x

2

(

8

x

3

6

x

+

6

)

=

8

x

3

6

x

+

6

(

8

x

3

6

x

+

6

)

(Subtract -8x^3-6x+6 from both sides)

8

x

3

+

x

2

+

6

x

6

=

0

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x

=

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Answers

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The price of fuel may increase due to demand and decrease due to overproduction. Marco is studying the change in the price of two types of fuel, A and B, over time.


The price f(x), in dollars, of fuel A after x months is represented by the function below:


f(x) = 2.27(0.88)x


Part A: Is the price of fuel A increasing or decreasing and by what percentage per month? Justify your answer. (5 points)


Part B: The table below shows the price g(m), in dollars, of fuel B after m months:



m (number of months) 1 2 3 4

g(m) (price in dollars)3.44 3.30 3.17 3.04



Which type of fuel recorded a greater percentage change in price over the previous month? Justify your answer.

Answers

Answer:

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Step-by-step explanation:

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A.

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  f(x) = (initial value) × (growth factor)^x

Comparing parts of the equation, we see ...

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This tells us the rate of change in fuel price is -12% per month.

B.

To find the percentage change, we can use the formula ...

  percent change = ((new value) / (old value) -1) × 100%

Applied to the first two table values, we have ...

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