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الدرس 1 Ch:- 5……
Probability
P=The number of Possible cassesTotal number of casses
==============================
EX1:-
BLUE=12 RED=3 GREEN=5
What is the Probability:-
P(B) OR P(R)
P(B) OR P(G)
P(B) OR P(R) OR P(G)
SOLVE:-
P=PB+ PR
P= 1220+ 320= 1520
P=PB+ P(G)
P=1220+ 520= 1720
P=PB+ PR+ P(G)
P= 1220+ 320+520 = 2020=1
بأختصار (( الاحتمالات ))
بيعطيك بالسؤال الاحتمالات المطلوبه
فقط ضع الاحتمال المطلوب في البسط
ومجموع الأحتمالات في المقام مهم (( OR )) معناها +

الدرس 2 Ch:- 5……
General Rule of Addition:-
PA or B=PA+ PB- PA and B
==============================
EX:- total | Female | Male | SEXCOLLEGE | 8 | 2 | 6 | BUSINESS | 12 | 5 | 7 | LAW | 20 | 7 | 13 | total |

IF we draw one person WHAT is the Probability:-
بما انه قال " شخص واحد " اذا الــ and تعني الـــ (( تقاطع ))
1) - To be Male?
2) – To be study in BUSINESS COLLEGE?
3) – To be Female and study in LAW COLLEGE?
4) – To be Female or study in LAW COLLEGE?
5) – To be study in BUSINESS COLLEGE or LAW COLLEGE?
6) – To be Male or study in BUSINESS COLLEGE?
7) – To be Male and Female?
SOLVE:-
1 – PMale
= 1320
2 – PBUSINESS
=820
3 – PFemale and LAW
= 520
4 -
PFemale or LAW=PFemale+ PLAW- PFemale and LAW = 720+ 1220 – 520= 1420
5 –
PBUSINESS or LAW=PBUSINESS+ PLAW- PBUSINESS and LAW = 820+ 1220 – 020= 2020=1
6 –
PMale or BUSINESS=PMale+ PBUSINESS- PMale and BUSINESS = 1320+ 820 – 620= 1520
7 – PMale and Female
= 020=0
هالقاعده تستخدم في حال أعطانا بالسؤال جدول ^_^
نطلع الـــ ToTaL
^^^^
(( and )) معناهآ ((تقاطع بالجدول))
بدون الجدول تصير ((*))
شرط اذا كان بالسؤال أكثر من 1 (( 2 PERSON )) زي المثال التالي ^_^

الدرس 3 Ch:- 5……
Special Rule of Multiplication:-
PA and B= PA*PB
EX:-
total | Female | Male | SEXCOLLEGE | 8 | 2 | 6 | BUSINESS | 12 | 5 | 7 | LAW | 20 | 7 | 13 | total |

IF we select 2 person(( WITH RETURN)) WHAT is the probability:-
بما أنه قال " أكثر من شخص " اذا الــ and تعني الــ (( الضرب ))
1) - First one Male and Second one Male?
تعني احتمال يكون الشخصين الاول رجل (( ضرب )) الثاني رجل 2) – Both of them are LAW?
تعني احتمال يكون الشخصين كلاهما يدرسان القانون ... اذا قانون (( ضرب )) قانون
3) – Both of them the same Meager?
تعني احتمال يكون الشخصين في التخصص نفسه ... ما حدد اذا تشمل كل التخصصات
قانون (( ضرب )) قانون + اداره (( ضرب )) اداره
4) – The First one is Male and second one is Female?
احتمال يكون الشخصين الاول رجل (( ضرب )) الثاني فتاة
5) – First one is Male and second one is Male or Female?
تعني احتمال يكون الشخصين الاول رجل ((ضرب)) الثاني فتاة + الاول رجل ((ضرب)) فتاة

SOLVE:-
1 – P=PMale*PMale
= 1320* 1320= 169400
==============================
2 – P=PLAW*PLAW
= 1220* 1220= 144400
==============================
3 – P=PBUSINESS*PBUSINESS+ PLAW*PLAW
= 820 * 820 + 1220 * 1220= 64400 + 144400= 208400
==============================
4 – P=PMale* pFemale
= 1320 * 720= 91400
==============================
5 – P=PMale* pMale+ PMale*PFemale = 1320 * 1320 + 1320 * 720= 260400 ==============================
EX2:-
total | Female | Male | SEXCOLLEGE | 8 | 2 | 6 | BUSINESS | 12 | 5 | 7 | LAW | 20 | 7 | 13 | total |

IF we select 2 person(( WITH OUT RETURN ))WHAT is the probability:-
نفس طريقة المثال اللي قبل الاختلاف أنه " بدون ارجاع "
أي عندما تسحب احتمال لا ترجعه وبالمثال يتضح المقال
1) - First one Male and Second one Male?
2) – Both of them are LAW?
3) – Both of them the same Meager?
4) – The First one is Male and second one is Female?
5) – First one is Male and second one is Male or Female?

SOLVE:-
1 – P=PMale*PMale
= 1320* 1219= 156380
المظلل هو التطبيق على " بدون ارجاع "
وهو انك تطرح واحد من البسط و المقام كل ما سحبت من نفس المطلوب ^_^
==============================
2 – P=PLAW*PLAW
= 1220* 1119= 132380
==============================
3 – P=PBUSINESS*PBUSINESS+ PLAW*PLAW
= 820 * 719 + 1220 * 1119= 188760
==============================
4 – P=PMale* pFemale
= 1320 * 719= 91380
==============================
5 – P=PMale* pMale+ PMale*PFemale = 1320 * 1219 + 1320 * 719= 247760
هنا بتشوف سحب الرجل 3 مرات
بس الاول نزل كامل ثم الثاني يطرح البسط والمقام ثم بعد + يرجع كما كان ((انتبه لا تغلط ))

==============================

الدرس 4 Ch:- 5……
General Rule of Multiplication:-
PA∕B= P(A and B)P(B)
==============================
Ex:- total | Female | Male | SEXCOLLEGE | 8 | 2 | 6 | BUSINESS | 12 | 5 | 7 | LAW | 20 | 7 | 13 | total |
1 – To be Male GIVEN study in LAW COLLEGE?
أول ما تشوف كلمة GIVEN"" علطول تطبق على القانون (( تآخذ تقاطع الرجل مع القانون تقسمها على القانون ))
PMale∕Law= P(Male and Law)P(Law)= 7201220= 712
==============================

الدرس 5 Ch:- 5……
Bayes' Theorem:-
PAi∕B=PAi P(BAi)PAi P(BAi)
==============================
EX:-
Three factories producing the same commodity, where the first factory will produces 0.20 of the total production of the commodity and the second factory produces 0.35 of the total production of commodity, and the third factory produces 0.45 of the total production of the commodity. If the defective production rates in the three factories on the order of 2%, 2. 5%, 3% . If the unite chosen at random from the production of this item and found it flawed, what is the probability to be
1 – The production of the first factory
2 – The production of the second factory
3 – The production of the third factory
SOLVE:-
PA1=0.20
PA2=0.35
PA3=0.45
PBA1=0.02
PBA2=0.025
PBA3=0.03
هذا لازم تكون حافظ القانون وبيجيك السؤال نفسه بالضبط
كل اللي عليك تستخرج المعطيات زي ما هي فوق بالاحمر و الازرق

PAi P(BAi) | P(BAi) | P(Ai) | P Factory | 0.004 | 0.02 | 0.20 | 1 | 0.00872 | 0.025 | 0.35 | 2 | 0.0135 | 0.03 | 0.45 | 3 | 0.02625 | | | Total |

PAi∕B=PAi P(BAi)PAi P(BAi)
PA1∕B=0.0040.02625=0.154
PA2∕B=0.008720.02625=0.333
PA3∕B=0.01350.02625=0.514

مشروح بالالوان ^_^

==============================
طبق ع الاسئلة التاليه بالكتاب
11 & 13 & 15 & 17 & 63 ( a,b,c ) & 65 & 66 = صفحة 154

الدرس 1 Ch:- 6……
Mean of a probability distribution:- μ=Ex= x .P(x)
Variance of a probability distribution:- σ2=Ex2- [Ex]2
علشان توجد الــ Ex2
Ex2= x2. P(x)
وفي حال طلب الــ Standard <<<<<< نــآآخذ الجذر لناتج الــ Variance
==============================
Ex:- P(x) | x | 0.5 | 6 | 0.33333 | 8 | 0.25 | 10 | 0.125 | 4 |
Find (Compute):-
1 – Mean /// 2 – Variance /// 3 – Standard
SOLVE:-
x2 . P(x) | x2 | x . P(x) | P(x) | x | 18 | 36 | 3 | 0.5 | 6 | 8 | 64 | 1 | 0.33333 | 8 | 25 | 100 | 2.5 | 0.25 | 10 | 2 | 16 | 0.5 | 0.125 | 4 | 53 | | 7 | | 28 |

1 – Mean:- μ=Ex= x .P(x) μ=Ex= 3+1+2.5+0.5=7
==============================
2 – Variance:-
Ex2= x2. P(x)
Ex2= 53 σ2=Ex2- [Ex]2 σ2=53-72=4 ==============================
3 – Standard:-
X=4 =2

==============================
طبق ع الاسئلة التاليه بالكتاب 1 & 2 = صفحة 189

الدرس 2 Ch:- 6……
Binomial probability formula:-
Px= n Cx . πx . (1- π)n-x
Mean of a Binomial distribution:- μ= n . π
Variance of a Binomial distribution:- σ2=n . π .(1- π)
==============================
Ex:- n = 6 /// π=0.60 /// (1- π)=0.40

المعطيات بتجيك بالسؤال وفيه احتمال يعطيك الـــ π=0.60 بس وأنت تطلع
الـــ (1- π) << طبعا علشان تكمل 1 (( اطرح 1 – 0.60 ))
تصير0.40 والعكس صحيح ^_^
وقبل لا تحل شف النسبهـ الأكبر لمين
لــلــ π اذا تحل على احتمالية النجآآح العايشين يهمونك
أو
لــلــ (1- π) اذا تحل على احتمالية الفشل الميتين يهمونك العايشين و الميتين في السؤال التالي ^_^

What is the probability:-
1 – "ALL" of them will "LIFE"?
2 – "ALL" of them will "DIE"?
3 – "HALF" of them will "LIFE"?
4 – "AT LEAST HALF" of them will "LIFE"?
5 – "2" at the must will "LIFE"?
SOLVE:-
1 -
Px= n Cx . πx . (1- π)n-x
Px=6= 6 C6 . (0.60)6 . (0.40)6-6=0.0467
عددهم كــآمل 6 وقال كلهم عايشين ... اذا Px=6
==============================
2 -
Px= n Cx . πx . (1- π)n-x
Px=0= 6 C0 . 0.600 . 0.406-0=0.0041
عددهم كــآمل 6 وقال كلهم ماتو كم يبقى عايش ؟؟ صفر ... اذا Px=0
==============================
3 -
Px= n Cx . πx . (1- π)n-x
Px=3= 6 C3 . 0.603 . 0.406-3=0.2765
==============================
4 –
P=Px=3+Px=4+Px=5+ Px=6
P=0.2765+0.31104+0.186624+ 0.0467=0.8208
Px=4= 6 C4 . 0.604 . 0.406-4=0.31104
Px=5= 6 C5 . 0.605 . 0.406-5=0.186624
يقول " كحد أدنى " نصهم عاشين يعني أقل شيء 3
اذا من 3 وطالع الى الــ 6 اذا Px=3+Px=4+Px=5+ Px=6
بما اننا طلعنا الــ Px=6& Px=3 من قبل
اذا نحتاج نوجد الــ Px=4 & Px=5

==============================

5 –
P=Px=2+Px=1+Px=0
P=0.13824+0.36864+0.0041=0.51098
Px=2= 6 C2 . 0.602 . 0.406-2=0.13824
Px=1= 6 C1 . 0.601 . 0.406-1=0.036864
يقول " بالكثير " 2 عايشين يعني 2 أعلى شيء
اذا من 2 ونازل الى الــ صفر
اذا P=Px=2+x=1+Px=0
بما اننا طلعنا الــ Px=0 من قبل
اذا نحتاج نوجد الــx=1 & Px=2
==============================
Ex2:- n = 6 /// π=0.60 /// (1- π)=0.40
Find:-
1) - The mean 2) - Standard
SOLVE:-
1 - μ= n . π μ= 6 .0.60= 3.6
==============================
2 - σ2=n . π .(1- π) σ2=6 .0.60.0.40=1.44
بما انه طالب الـــ Standard نآآخذ الجذر لــ Variance
فتحله على قانون الـــ Variance ثم تطلع الـــ Standard σ2=6 .0.60.0.40= 1.44=1.2
==============================

طبق ع الاسئلة التاليه بالكتاب
9 & 10 & 13 & 14 & 15 & 16 & 19 & 20 & 21 = صفحة 197 & 199

الدرس 3 Ch:- 6……
Poisson probability distribution:-
Px= μx . e-μx!
Mean of a Binomial distribution:- μ= n . π
==============================
Ex:-
In Poisson distribution n = 200 /// π=0.02 a )- What is the probability that x=2 b )- What is the probability that x ≤2 c )- What is the probability that x >2
SOLVE:-
A - μ= n . π μ= 200 . 0.02=4
Px= μx . e-μx!
Px=2= 42 . e-42!= 0.1465
==============================
B -
Px ≤2=Px=2+x=1+Px=0
الاشــآره تقول بأن X أصغر أو يساوي الــ 2
اذا نطلع احتمال Px=2 & x=1 & Px=0
Px=1= 41 . e-41!= 0.0732
Px=0= 40 . e-40!= 0.0183
Px ≤2=Px=2+x=1+Px=0
Px ≤2=0.1465+0.0732+0.0183=0.2381
==============================
C – Px>2=1- Px≤2 Px>2=1- 0.2381=0.7619
الاشــآره تقول بأن X أكبر من الــ 2 فــ من المستحيل أنك تطلع 2 ,3,4,5,6,7,8,………
لذلك حل على أن الاشاره عكس((طبعا اليساوي غير مهم))وبما اننا حليناها قبل فقط تطرح 1

==============================

طبق ع الاسئلة التاليه بالكتاب
31 & 32 & 33 & 35 = صفحة 220

الدرس 1 Ch:- 7……
Normal distribution:-
Z=X-μσ

==============================
الشرح في شرح الشابتر 7
الرابط التالي بنفس الموضوع بالمنتدى
==============================
طبق ع الاسئلة التاليه بالكتاب
10 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22
صفحة 229 & 232 & 235

الدرس 1 Ch:- 13……
Correlation Coefficient:- r=n . xy- x . yn x2- (x)2 n y2- (y)2
==============================
Ex:- Sales | Price | 5 | 10 | 4 | 12 | 3 | 14 | 2 | 16 |

SOLVE:- y2 | x2 | xy | Sales( Y) | Price (X) | 25 | 100 | 50 | 5 | 10 | 16 | 144 | 48 | 4 | 12 | 9 | 196 | 42 | 3 | 14 | 4 | 256 | 32 | 2 | 16 | 54 | 686 | 172 | 14 | 52 |

r=n . xy- x . yn x2- (x)2 n y2- (y)2 r=4 . 172-52.144 696-522 4 54-142= -4080 20= -4040= -1
( S)
==============================

طبق ع الاسئلة التاليه بالكتاب
1 & 2 & 3 (a,c,e) & 5 (a,c,e)
صفحة 463
الدرس 2 Ch:- 13……
Linear Regression:-
Y=a+b x b=n . xy- x . yn x2- (x)2
أو
b=rSxSy a=yn-b xn
==============================
Ex:- y | x | 4 | 4 | 6 | 5 | 5 | 3 | 7 | 6 | 7 | 10 |

SOLVE:- x2 | x . y | y | x | 16 | 16 | 4 | 4 | 25 | 30 | 6 | 5 | 9 | 15 | 5 | 3 | 36 | 42 | 7 | 6 | 100 | 70 | 7 | 10 | 186 | 173 | 29 | 28 |

b=n . xy- x . yn x2- (x)2 b=5 . 173-28(29)5 1862-(28)2= 53146=0.363 a=yn-b xn a=295-0.363 285=3.7672
Y=a+b x
Y=3.7672+0.363 x
==============================
طبق ع الاسئلة التاليه بالكتاب
13 & 14 & 15 & 16 & 17 (b,e,f) & 19 (a,b)
صفحة 472

الدرس 1 Ch:- 16……
Linear Trend (( Time series )):-
Y=a+b t b=n . ty- t . yn t2- (t)2 a=yn-b tn
==============================
Ex:-
The following table shows the profits of a company (in millions of Riyals) for a period of 5 years. 1430 | 1429 | 1428 | 1427 | 1426 | Years | 10 | 8 | 4 | 5 | 3 | profits |

1 – Determine the least squares equation.
2 – What are the estimated profits for 1431?
3 – What are the estimated profits for 1432?
SOLVE:-
t2 | t .y | t | Profits | Year | 1 | 3 | 1 | 3 | 1426 | 4 | 10 | 2 | 5 | 1427 | 9 | 12 | 3 | 4 | 1428 | 16 | 32 | 4 | 8 | 1429 | 25 | 50 | 5 | 10 | 1430 | 55 | 107 | 15 | | Total | | | 6 | | 1431 | | | 7 | | 1432 |

b=n . ty- t . yn t2- (t)2 b=5 . 107-15(30)5 552-(15)2=1.7 a=yn-b xn a=305-1.7 155=0.9
Y=a+b x
Y=0.9+1.7 t
==============================
2 -
Y=0.9+1.7 6=11.1
كيف اوجدنا 6 عن طريق الجدول سوينا اضافه وبما انه 1430 = 5 <<<< نضيف 1431 = 6 <<<< ونضيف 1432 = 7
وفي حال طلب 1426 مثلا فكم بتساوي الــ t = 3
==============================
3 -
Y=0.9+1.7 7=12.8

==============================

طبق ع الاسئلة التاليه بالكتاب
3 & 4 & 5 & 6
صفحة 609

الدرس 1 Ch:- 16……
SIMPLE INDEX:-
P=p0pt*100
SIMPLE AVERAGE INDEX:-
P=p0pt*100
LASPEYRES INDEX:-
P=ptq0p0q0*100
PAASCHE INDEX:-
P=ptqtp0qt*100
Fisher's ideal index:-
Fisher's ideal index=Laspeyres'index*(Paasche's index)
Ex:-
Below are the prices of toothpaste (9 oz), shampoo (7 oz), cough tablets (package of 100), and antiperspirant (2 oz) for August 2000 and August 2008. Also included are the quantity purchased. Use August 2000 as the base. August 2008 | August 2000 | Item | Quantity | Price | Quantity | Price | | 6 | 3.35 | 6 | 2.49 | toothpaste | 5 | 4.49 | 4 | 3.29 | shampoo | 3 | 4.19 | 2 | 1.59 | cough tablets | 4 | 4.49 | 3 | 1.79 | antiperspirant |

Solve:- ptqt | p0qt | ptq0 | p0q0 | August 2008 | August 2000 | Item | | | | | qt | pt | q0 | p0 | | 20.1 | 14.94 | 20.1 | 14.94 | 6 | 3.35 | 6 | 2.49 | toothpaste | 22.45 | 16.45 | 17.96 | 13.16 | 5 | 4.49 | 4 | 3.29 | shampoo | 12.57 | 4.77 | 8.38 | 3.18 | 3 | 4.19 | 2 | 1.59 | cough tablets | 9.96 | 7.16 | 7.47 | 5.37 | 4 | 4.49 | 3 | 1.79 | antiperspirant | 65.08 | 43.32 | 53.91 | 36.65 | | 14.52 | | 9.16 | |

1 –
P=p0pt*100
P toothpaste=2.493.35*100=134.5%
P shampoo=3.294.49*100=136.5%
P cough tablets=1.594.19*100=263.3%
P antiperspirant=1.794.49*100=139.1%
2 –
P=p0pt*100
P=14.529.16*100=158.5%
3 –
P=ptq0p0q0*100
P=53.9136.65*100=147.09%
4 –
P=ptqtp0qt*100
P=65.0843.32*100=150.23%
5 –
Fisher's ideal index=Laspeyres'index*(Paasche's index)
Fisher's ideal index=147.09*150.23=148.64%

طبق ع الاسئلة التاليه بالكتاب
6 & 7 & 8
صفحة 576

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