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2 years ago

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A certain bookstore chain has two stores, one in San Francisco and one in Los Angeles. It stocks three kinds of books: hardcover

, softcover, and plastic (for infants). At the beginning of January, the central computer showed the following books in stock.
Hard Soft Plastic
San Francisco 1,000 3,000 6,000
Los Angeles 1,000 6,000 3,000

Its sales in January were as follows: 700 hardcover books, 1,200 softcover books, and 2,000 plastic books sold in San Francisco, and 400 hardcover, 200 softcover, and 500 plastic books sold in Los Angeles. The bookstore chain actually maintained the same sales figures for the first 6 months of the year. Each month, the chain restocked the stores from its warehouse by shipping 600 hardcover, 1,500 softcover, and 1,500 plastic books to San Francisco and 500 hardcover, 500 softcover, and 500 plastic books to Los Angeles.

Required:
a. Use matrix operations to determine the total sales over the 6 months, broken down by store and type of book.
b. Use matrix operations to determine the inventory in each store at the end of June.

1 answer:

victus00 [196]2 years ago

3 0

Answer:

Answer:

a. Sales from January to June: Matrix B6

Hard Soft Plastic

San Francisco 4,200 7,200 12,000

Los Angeles 2,400 1,200 3,000

b) Ending Inventory: Matrix D:

Hard Soft Plastic

San Francisco 400 4,800 3,000

Los Angeles 1,600 7,800 3,000

Step-by-step explanation:

a) Data and Calculations:

Stock on January 1: Matrix A

Hard Soft Plastic

San Francisco 1,000 3,000 6,000

Los Angeles 1,000 6,000 3,000

Sales in January: Matrix B

Hard Soft Plastic

San Francisco 700 1,200 2,000

Los Angeles 400 200 500

Restocking: Matrix C

Hard Soft Plastic

San Francisco 600 1,500 1,500

Los Angeles 500 500 500

Total Sales over the first 6 months =

Matrix B * 6 = Matrix B6

Sales in January: Matrix B

Hard Soft Plastic

San Francisco 700 1,200 2,000

Los Angeles 400 200 500

* 6

=

Sales from January to June: Matrix B6

Hard Soft Plastic

San Francisco 4,200 7,200 12,000

Los Angeles 2,400 1,200 3,000

Matrix C6 = Matrix C * 6

=

Restocking: Matrix C6

Hard Soft Plastic

San Francisco 3,600 9,000 9,000

Los Angeles 3,000 3,000 3,000

Inventory at the end of June =

Matrix A + Matrix C6 - Matrix B6

= Matrix D

Stock on January 1: Matrix A

Hard Soft Plastic

San Francisco 1,000 3,000 6,000

Los Angeles 1,000 6,000 3,000

+

Restocking: Matrix C6

Hard Soft Plastic

San Francisco 3,600 9,000 9,000

Los Angeles 3,000 3,000 3,000

-

Sales from January to June: Matrix B6

Hard Soft Plastic

San Francisco 4,200 7,200 12,000

Los Angeles 2,400 1,200 3,000

Ending Inventory: Matrix D:

Hard Soft Plastic

San Francisco 400 4,800 3,000

Los Angeles 1,600 7,800 3,000

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a) 13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) 17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) 17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) 75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

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Step-by-step explanation:

For each building, there are only two possible outcomes after a earthquake. Either they have been damaged, or they have not. The probability of a building being damaged is independent from other buildings. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

15% of all buildings have been structurally compromised.

This means that $p = 0.15$

20 buildings

This means that $n = 20$

a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756$

17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?

Either they find 4 or less, or they find more than 4. The sum of the probabilities of these events is 1. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0388 + 0.1368 + 02293 + 0.2428 + 0.1821 = 0.8298$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8298 = 0.1702$

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$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X = 5) = C_{20,5}.(0.15)^{5}.(0.85)^{15} = 0.1028$

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2293 + 0.2428 + 0.1821 + 0.1028 = 0.7570$

75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?

The expected value of the binomial distribution is:

$E(X) = np$

So

$E(X) = 20*0.15 = 3$

The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

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