**Home** / **logic proof calculator with steps** / Fuzzy logic steps

Total Score 54 out of 70

Score 10 out of 10

Q1. Fuzzy logic steps

1. Fuzzification

Convert numerical, discrete, data into generalized groups called fuzzy member sets Good

2. Rule inference

Use AND / OR to create intersecting and union fuzzy member set. This step identifies all possible combination of fuzzy member sets. Good

3. Rule composition

Creates rules based on the rule inference step Good

4. Defuzzification

Convert fuzzy set rules to a numerical score of confidence in the likelihood of an event. Good

Score 0 out of 10

Q2. ( Answer?

Score 7 out of 10

Q3. Stay with my original choice. Why? Explain your logic.

The probability of your first choice being correct (1/3) never changes (even after the empty is opened). That means that if we keep our first choice, we’ll always be correct only 1 out of 3 times.

Let’s call our doors A, B, and C. Assume the money is behind door A (this example holds for B and C as well by method of exhaustion). We have a P = 1/3 choice of choosing any door at random correctly. If the prize is behind door A here are the possible outcomes of not switching followed by switching:

|1st Choice |Shown |Final Choice |Right |P |

|A |B |A |Yes |0.166666667 |

|A |C |A |Yes |0.166666667 |

|B |C |B |No |0.333333333 |

|C |B |C |No |0.333333333 |

| | | | | |

| | | |Total Yes |0.333333333 |

| | | |Total No |0.666666667 |

| | | | | |

|1st Choice |Shown |Final Choice |Right |P |

|A |B |C |No |0.166666667 |

|A |C |B |No |0.166666667 |

|B |C |A |Yes |0.333333333 |

|C |B |A |Yes |0.333333333 |

| | | | | |

| | | |Total Yes |0.666666667 |

| | | |Total No |0.333333333 |

We see that switching doors makes us twice as likely to choose the correct door as opposed to staying put. Hence, it is better overall that we switch.

Score 8 out of 10

Q4.

Answer 59% (life insurance accept confidence)

IF age is young

AND previous_accepts are several

THEN life_insurance_accept is high

IF age is middle_aged

AND previous_accepts are some

THEN life_insurance_accept is moderate

age is middle_aged 0.10

age is young 0.20

previous_accepts are some 0.20

previous_accepts are several 0.60

Age = 30 years

PA = 5

Rule1: Accept is High

IF Age is Young

AND PA is High

THEN LIA is High

Rule 2: Accept is Moderate

IF Age is Middle_Age

AND PA are Some

THEN LIFA is Moderate

Rule 3: Accept is Low

IF Age is Old

THEN LIA is Low

Step1:

Age is Young = .3

Age is Middle_Aged = .1

Age is Old = Null (Does not intersect that curve)

PA are Few = Null (Does not intersect that curve)

PA are Some = .2

PA are Several = .6

Step2:

Rule1: Accept is High

min(.3,.6) = .3

FIRES!

Rule 2: Accept is Moderate

Min(.1,.2) = .1

FIRES!

Rule 3: Accept is Low

Min (Null) = N/A

DOES NOT FIRE!

Step3:

Clip “Accept is High” to .3

Clip “Accept is Moderate” to .1

Graph of Fuzzy Set is defined by connecting the points: {(25,0),(35,.3),(55,.3),(65,.1),(100,.1)

Step4:

=[.3*(35+45+55)+.1(65+75+85+95)]/[(.2*3)+(.1*4)]

=72.5

Thus we have a confidence value of 72.5% that the customer in question is a likely candidate for the life insurance promotion.

Score 10 out of 10

Q5. Answer 44.55 %

If likes sugar

Then eat cereal X 75/100

If likes sugar

Then does not eat cereal X 40/100

Eats Cereal 30/100

Does not Eat Cereal 70/100

75/100 X 30/100

75/100 X 30/100 + 40/100 X 70/100

= .4455

= 44.55% Very good

Score 9 out of 10

Q6.

If Age 25

AND accepts 3 previous promotion

THEN will accept life insurance promotion

2 instances of age 25

1instance with three previous promotion

1 / 2 X 1 / 2

1 / 2 X 1 / 2 + 1 / 2 X 1 / 2

50% See my computations below

Age = 25 years

Previous accepts = 3

Will accept life insurance promotion = ?

|P(Age=25) |0.142857143 |2/14 |

|P(Age25) |0.857142857 |12/14 |

|P(PA=3) |0.214285714 |3/14 |

|P(PA3) |0.071428571 |11/14 |

|P(LI=Yes) |0.5 |7/14 |

|P(LI=No) |0.5 |7/14 |

|P(Age=25|LI=Yes) |0.285714286 |2/7 |

|P(PA=3|LI=Yes) |0.285714286 |2/7 |

|P(Age=25|LI=No) |0 |0/7 |

|P(PA=3|LI=No) |0.142857143 |1/7 |

|P(LI=Yes|Age=25 & PA=3) = |P(Age=25|LI=Yes) * P(PA=3|LI=Yes) * P(LI=Yes) |

| |P(Age=25|LI=Yes) * P(PA=3|LI=Yes) * P(LI=Yes) + P(Age=25|LI=No) * P(PA=3|LI=No) * P(LI=No) |

|= |1 |

Thus there is a 100% probability that a 25 year old customer who has 3 previous accepts will accept the life insurance promotion. Since this is a finite set we see there is only 1 person who meets such conditions. It so happens that this person accepted the life insurance promotion. This is how we practically obtain the 100% probability figure, 1 for 1.

Score 10 out of 10

Q7.

|Agents: |Most important feature |Reason |

|Filtering |Situated-ness |These agents simply exclude information. |

| | |These agent filter information by using |

| | |key words and phrases; react to a given |

| | |situation or event. |

|Semiautonomous |Adaptivity |These agent are capable of learning, |

| | |functioning proactively and make decision |

| | |based on the circumstance. |

|Find-and-retrieve |Situated-ness |These agent are limited to looking for |

| | |information and within a limited degree |

| | |these agents can make alternative |

| | |decisions however these decisions are |

| | |based on key words, phrases or rules. |

|User |Sociability |These agent are capable of communicating |

| | |with users and other agents. |

|Monitor-and-surveillance |Situated-ness |These agent are limited performing |

| | |predefined task within a predefine |

| | |environment. These agents react within |

| | |these predefined boundaries. |

|Data mining |Situated-ness |These agents are limited to finding trends|

| | |within a database and are incapable of |

| | |function in a proactive manner. |

Very good

How do you prove a proof is correct? Remember: A proof is correct as long as (a) the conclusion of the argument being proven is the last step of the proof, and (b) each step in the proof correctly follows from previous lines via one of the rules of inference correctly applied (and the rule is correctly cited). 1. 1. D ( K 2. K ( ~R