Please answer all parts a-c, the last b should be a C. **Quality Control Analysis of Two Furniture Manufacturers****6) Description:**Two furniture manufacturers, Cardi's and Jordan's, pride themselves in quality control. Both companies manufacture a 4-piece bedroom set. Each piece is manufactured separately, and sets containing the 4 pieces that make up the desk are then packaged together. Quality control at the factory has randomly sampled numerous desk sets and determined the number of damaged pieces in each set.The data collected is shown in the table below:| Damaged Pieces | Cardi's (P(c)) | Jordan's (P(j)) ||----------------|----------------|-----------------|| 0 | 0.75 | 0.25 || 1 | 0.14 | 0.60 || 2 | 0.07 | 0.10 || 3 | 0.03 | 0.02 || 4 | ? | ? |**Questions:****a) Finding the Probabilities to make each Distribution Legitimate:**To make a probability distribution legitimate, the sum of the probabilities for all possible outcomes must equal 1. For Cardi's, we calculate:\[ 0.75 + 0.14 + 0.07 + 0.03 + P(c=4) = 1 \]\[ 0.99 + P(c=4) = 1 \]\[ P(c=4) = 0.01 \]Thus, \[ P(c=4) = 0.01 \]For Jordan's, we calculate:\[ 0.25 + 0.60 + 0.10 + 0.02 + P(j=4) = 1 \]\[ 0.97 + P(j=4) = 1 \]\[ P(j=4) = 0.03 \]Thus, \[ P(j=4) = 0.03 \]**b) Expected Value for Damaged Parts from Jordan's:**The expected value (E) can be calculated using the formula:\[ E = \sum (x \cdot P(x)) \]For Jordan's:\[ E = (0 \cdot 0.25) + (1 \cdot 0.60) + (2 \cdot 0.10) + (3 \cdot 0.02

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Description: Please answer all parts a-c, the last b should be a C. **Quality Control Analysis of Two Furniture Manufacturers****6) Description:**Two furniture manufacturers, Cardi's and Jordan's, pride themselves in quality control. Both companies manufacture a

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a) Find the probabilities that make each distribution legitimate. b) What is the expected value for damaged parts from Jordan's? b) What is the probability Cardi's has more damaged pieces than Jordan's?

a) Find the probabilities that make each distribution legitimate. b) What is the expected value for damaged parts from Jordan's? b) What is the probability Cardi's has more damaged pieces than Jordan's?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Concept explainers

Permutations and Combinations

If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba, cb and ca. It is in a particular order. So, this can be called the permutation of a, b, and c. But if the order does not matter then ab is the same as ba. Similarly, bc is the same as cb and ac is the same as ca. Here the list has ab, bc, and ac alone. This can be called the combination of a, b, and c.

Counting Theory

The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation.

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Question

Please answer all parts a-c, the last b should be a C.

**Quality Control Analysis of Two Furniture Manufacturers**

**6) Description:**
Two furniture manufacturers, Cardi's and Jordan's, pride themselves in quality control. Both companies manufacture a 4-piece bedroom set. Each piece is manufactured separately, and sets containing the 4 pieces that make up the desk are then packaged together. Quality control at the factory has randomly sampled numerous desk sets and determined the number of damaged pieces in each set.

The data collected is shown in the table below:

| Damaged Pieces | Cardi's (P(c)) | Jordan's (P(j)) |
|----------------|----------------|-----------------|
| 0 | 0.75 | 0.25 |
| 1 | 0.14 | 0.60 |
| 2 | 0.07 | 0.10 |
| 3 | 0.03 | 0.02 |
| 4 | ? | ? |

**Questions:**

**a) Finding the Probabilities to make each Distribution Legitimate:**
To make a probability distribution legitimate, the sum of the probabilities for all possible outcomes must equal 1.

For Cardi's, we calculate:

\[ 0.75 + 0.14 + 0.07 + 0.03 + P(c=4) = 1 \]

\[ 0.99 + P(c=4) = 1 \]

\[ P(c=4) = 0.01 \]

Thus, \[ P(c=4) = 0.01 \]

For Jordan's, we calculate:

\[ 0.25 + 0.60 + 0.10 + 0.02 + P(j=4) = 1 \]

\[ 0.97 + P(j=4) = 1 \]

\[ P(j=4) = 0.03 \]

Thus, \[ P(j=4) = 0.03 \]

**b) Expected Value for Damaged Parts from Jordan's:**
The expected value (E) can be calculated using the formula:

\[ E = \sum (x \cdot P(x)) \]

For Jordan's:

\[ E = (0 \cdot 0.25) + (1 \cdot 0.60) + (2 \cdot 0.10) + (3 \cdot 0.02

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