GMAT數學精解--算術概述
1.平均數
2.中數
TO CALCULATE THE MEDIAN OF N NUMBERS，FIRSTSGROUPSTHE NUMBERS FROM LEAST TO GREATEST;IF N IS ODD，THE MEDIAN IS DEFINED AS THE MIDDLE NUMBER，WHILE IF N IS EVEN，THE MEDIAN IS DEFINED AS THE AVERAGE OF THE TWO MIDDLE NUMBERS. FOR THE DATA 6, 4, 7, 10, 4, THE NUMBERS, IN ORDER, ARE 4, 4, 6, 7, 10, AND THE MEDIAN IS 6, THE MIDDLE NUMBER. FOR THE NUMBERS 4, 6, 6, 8, 9, 12, THE MEDIAN IS /2 = 7. NOTE THAT THE MEAN OF THESE NUMBERS IS 7.5.
3.眾數：一組數中的眾數是指出現頻率最高的數。
例：THE MODE OF 7，9，6，7，2，1 IS 7。
4.值域：表明數的分布的量，其被定義為最大值減最小值的差。
例：THE RANGE OF1，7，27，27，36 IS 36-= 37。
5.標準方差：
ONE OF THE MOST COMMON MEASURES OF DISPERSION IS THE STANDARD DEVIATION. GENERALLY SPEAKING, THE GREATER THE DATA ARE SPREAD AWAY FROM THE MEAN, THE GREATER THE STANDARD DEVIATION. THE STANDARD DEVIATION OF N NUMBERS CAN BE CALCULATED AS FOLLOWS:
FIND THE ARITHMETIC MEAN ;
FIND THE DIFFERENCES BETWEEN THE MEAN AND EACH OF THE N NUMBERS ;
SQUARE EACH OF THE DIFFERENCES ;
FIND THE AVERAGE OF THE SQUARED DIFFERENCES ;
TAKE THE NONNEGATIVE SQUARE ROOT OF THIS AVERAGE.
NOTICE THAT THE STANDARD DEVIATION DEPENDS ON EVERY DATA VALUE, ALTHOUGH IT DEPENDS MOST ON VALUES THAT ARE FARTHEST FROM THE MEAN. THIS IS WHY A DISTRIBUTION WITH DATA GROUPED CLOSELY AROUND THE MEAN WILL HAVE A SMALLER STANDARD DEVIATION THAN DATA SPREAD FAR FROM THE MEAN.
6.排列與組合
THERE ARE SOME USEFUL METHODS FOR COUNTING OBJECTS AND SETS OF OBJECTS WITHOUT ACTUALLY LISTING THE ELEMENTS TO BE COUNTED. THE FOLLOWING PRINCIPLE OF MULTIPLICATION IS FUNDAMENTAL TO THESE METHODS.
IF A FIRST OBJECT MAY BE CHOSEN IN M WAYS AND A SECOND OBJECT MAY BE CHOSEN IN N WAYS, THEN THERE ARE MN WAYS OF CHOOSING BOTH OBJECTS.
AS AN EXAMPLE, SUPPOSE THE OBJECTS ARE ITEMS ON A MENU. IF A MEAL CONSISTS OF ONE ENTREE AND ONE DESSERT AND THERE ARE 5 ENTREES AND 3 DESSERTS ON THE MENU, THEN 53 = 15 DIFFERENT MEALS CAN BE ORDERED FROM THE MENU. AS ANOTHER EXAMPLE, EACH TIME A COIN IS FLIPPED, THERE ARE TWO POSSIBLE OUTCOMES, HEADS AND TAILS. IF AN EXPERIMENT CONSISTS OF 8 CONSECUTIVE COIN FLIPS, THE EXPERIMENT HAS 28 POSSIBLE OUTCOMES,SWHERESEACH OF THESE OUTCOMES IS A LIST OF HEADS AND TAILS IN SOME ORDER.
階乘：FACTORIAL NOTATION
假如一個大于1的整數N，計算N的階乘被表示為N!，被定義為從1至N所有整數的乘積，
例如：4! = 4321= 24
注意：0! = 1! = 1
排列：PERMUTATIONS
THE FACTORIAL IS USEFUL FOR COUNTING THE NUMBER OF WAYS THAT A SET OF OBJECTS CAN BE ORDERED. IF A SET OF N OBJECTS IS TO BE ORDERED FROM 1ST TO NTH, THERE ARE N CHOICES FOR THE 1ST OBJECT, N-1 CHOICES FOR THE 2ND OBJECT, N-2 CHOICES FOR THE 3RD OBJECT, AND SO ON, UNTIL THERE IS ONLY 1 CHOICE FOR THE NTH OBJECT. THUS, BY THE MULTIPLICATION PRINCIPLE, THE NUMBER OF WAYS OF ORDERING THE N OBJECTS IS
N = N!
FOR EXAMPLE, THE NUMBER OF WAYS OF ORDERING THE LETTERS A, B, AND C IS 3!, OR 6：ABC, ACB, BAC, BCA, CAB, AND CBA.
THESE ORDERINGS ARE CALLED THE PERMUTATIONS OF THE LETTERS A, B, AND C.也可以用P 33表示.
例如：1, 2, 3, 4, 5這5個數字構成不同的5位數的總數為5! = 120
組合：COMBINATION
A PERMUTATION CAN BE THOUGHT OF AS A SELECTION PROCESS IN WHICH OBJECTS ARE SELECTED ONE BY ONE IN A CERTAIN ORDER. IF THESGROUPSOF SELECTION IS NOT RELEVANT AND ONLY K OBJECTS ARE TO BE SELECTED FROM A LARGER SET OF N OBJECTS, A DIFFERENT COUNTING METHOD IS EMPLOYED.
SPECIALLY CONSIDER A SET OF N OBJECTS FROM WHICH A COMPLETE SELECTION OF K OBJECTS IS TO BE MADE WITHOUT REGARD TO ORDER,SWHERES0N . THEN THE NUMBER OF POSSIBLE COMPLETE SELECTIONS OF K OBJECTS IS CALLED THE NUMBER OF COMBINATIONS OF N OBJECTS TAKEN K AT A TIME AND IS CKN.
從N個元素中任選K個元素的數目為:
CKN. = N!/ ! K!
例如：從5個不同元素中任選2個的組合為C25 = 5!/2! 3!= 10
排列組合的一些特性
加法原則：RULE OF ADDITION
做某件事有X種方法，每種方法中又有各種不同的解決方法。例如第一種方法中有Y1種方法，第二種方法有Y2種方法，等等，第X種方法中又有YX種不同的方法，每一種均可完成這件事，即它們之間的關系用OR表達，那么一般使用加法原則，即有：Y1+ Y2+。。。+ YX種方法。
乘法原則：RULE OF MULTIPLICATION
完成一件事有X個步驟，第一步有Y1種方法，第二步有Y2種方法，。。。，第X步有YX種方法，完成這件事一共有Y1 Y2。。。YX種方法。
以上只是GMAT考題中經常涉及到的數學算術方面的問題，今后我們將陸續在新開辟的網上課堂中介紹代數、幾何以及系統的習題、講解，以幫助大家在GMAT數學考試中更好地發揮中國學生的優勢，拿到讓美國人瞠目結舌的成績!
1.平均數
2.中數
TO CALCULATE THE MEDIAN OF N NUMBERS，FIRSTSGROUPSTHE NUMBERS FROM LEAST TO GREATEST;IF N IS ODD，THE MEDIAN IS DEFINED AS THE MIDDLE NUMBER，WHILE IF N IS EVEN，THE MEDIAN IS DEFINED AS THE AVERAGE OF THE TWO MIDDLE NUMBERS. FOR THE DATA 6, 4, 7, 10, 4, THE NUMBERS, IN ORDER, ARE 4, 4, 6, 7, 10, AND THE MEDIAN IS 6, THE MIDDLE NUMBER. FOR THE NUMBERS 4, 6, 6, 8, 9, 12, THE MEDIAN IS /2 = 7. NOTE THAT THE MEAN OF THESE NUMBERS IS 7.5.
3.眾數：一組數中的眾數是指出現頻率最高的數。
例：THE MODE OF 7，9，6，7，2，1 IS 7。
4.值域：表明數的分布的量，其被定義為最大值減最小值的差。
例：THE RANGE OF1，7，27，27，36 IS 36-= 37。
5.標準方差：
ONE OF THE MOST COMMON MEASURES OF DISPERSION IS THE STANDARD DEVIATION. GENERALLY SPEAKING, THE GREATER THE DATA ARE SPREAD AWAY FROM THE MEAN, THE GREATER THE STANDARD DEVIATION. THE STANDARD DEVIATION OF N NUMBERS CAN BE CALCULATED AS FOLLOWS:
FIND THE ARITHMETIC MEAN ;
FIND THE DIFFERENCES BETWEEN THE MEAN AND EACH OF THE N NUMBERS ;
SQUARE EACH OF THE DIFFERENCES ;
FIND THE AVERAGE OF THE SQUARED DIFFERENCES ;
TAKE THE NONNEGATIVE SQUARE ROOT OF THIS AVERAGE.
NOTICE THAT THE STANDARD DEVIATION DEPENDS ON EVERY DATA VALUE, ALTHOUGH IT DEPENDS MOST ON VALUES THAT ARE FARTHEST FROM THE MEAN. THIS IS WHY A DISTRIBUTION WITH DATA GROUPED CLOSELY AROUND THE MEAN WILL HAVE A SMALLER STANDARD DEVIATION THAN DATA SPREAD FAR FROM THE MEAN.
6.排列與組合
THERE ARE SOME USEFUL METHODS FOR COUNTING OBJECTS AND SETS OF OBJECTS WITHOUT ACTUALLY LISTING THE ELEMENTS TO BE COUNTED. THE FOLLOWING PRINCIPLE OF MULTIPLICATION IS FUNDAMENTAL TO THESE METHODS.
IF A FIRST OBJECT MAY BE CHOSEN IN M WAYS AND A SECOND OBJECT MAY BE CHOSEN IN N WAYS, THEN THERE ARE MN WAYS OF CHOOSING BOTH OBJECTS.
AS AN EXAMPLE, SUPPOSE THE OBJECTS ARE ITEMS ON A MENU. IF A MEAL CONSISTS OF ONE ENTREE AND ONE DESSERT AND THERE ARE 5 ENTREES AND 3 DESSERTS ON THE MENU, THEN 53 = 15 DIFFERENT MEALS CAN BE ORDERED FROM THE MENU. AS ANOTHER EXAMPLE, EACH TIME A COIN IS FLIPPED, THERE ARE TWO POSSIBLE OUTCOMES, HEADS AND TAILS. IF AN EXPERIMENT CONSISTS OF 8 CONSECUTIVE COIN FLIPS, THE EXPERIMENT HAS 28 POSSIBLE OUTCOMES,SWHERESEACH OF THESE OUTCOMES IS A LIST OF HEADS AND TAILS IN SOME ORDER.
階乘：FACTORIAL NOTATION
假如一個大于1的整數N，計算N的階乘被表示為N!，被定義為從1至N所有整數的乘積，
例如：4! = 4321= 24
注意：0! = 1! = 1
排列：PERMUTATIONS
THE FACTORIAL IS USEFUL FOR COUNTING THE NUMBER OF WAYS THAT A SET OF OBJECTS CAN BE ORDERED. IF A SET OF N OBJECTS IS TO BE ORDERED FROM 1ST TO NTH, THERE ARE N CHOICES FOR THE 1ST OBJECT, N-1 CHOICES FOR THE 2ND OBJECT, N-2 CHOICES FOR THE 3RD OBJECT, AND SO ON, UNTIL THERE IS ONLY 1 CHOICE FOR THE NTH OBJECT. THUS, BY THE MULTIPLICATION PRINCIPLE, THE NUMBER OF WAYS OF ORDERING THE N OBJECTS IS
N = N!
FOR EXAMPLE, THE NUMBER OF WAYS OF ORDERING THE LETTERS A, B, AND C IS 3!, OR 6：ABC, ACB, BAC, BCA, CAB, AND CBA.
THESE ORDERINGS ARE CALLED THE PERMUTATIONS OF THE LETTERS A, B, AND C.也可以用P 33表示.
例如：1, 2, 3, 4, 5這5個數字構成不同的5位數的總數為5! = 120
組合：COMBINATION
A PERMUTATION CAN BE THOUGHT OF AS A SELECTION PROCESS IN WHICH OBJECTS ARE SELECTED ONE BY ONE IN A CERTAIN ORDER. IF THESGROUPSOF SELECTION IS NOT RELEVANT AND ONLY K OBJECTS ARE TO BE SELECTED FROM A LARGER SET OF N OBJECTS, A DIFFERENT COUNTING METHOD IS EMPLOYED.
SPECIALLY CONSIDER A SET OF N OBJECTS FROM WHICH A COMPLETE SELECTION OF K OBJECTS IS TO BE MADE WITHOUT REGARD TO ORDER,SWHERES0N . THEN THE NUMBER OF POSSIBLE COMPLETE SELECTIONS OF K OBJECTS IS CALLED THE NUMBER OF COMBINATIONS OF N OBJECTS TAKEN K AT A TIME AND IS CKN.
從N個元素中任選K個元素的數目為:
CKN. = N!/ ! K!
例如：從5個不同元素中任選2個的組合為C25 = 5!/2! 3!= 10
排列組合的一些特性
加法原則：RULE OF ADDITION
做某件事有X種方法，每種方法中又有各種不同的解決方法。例如第一種方法中有Y1種方法，第二種方法有Y2種方法，等等，第X種方法中又有YX種不同的方法，每一種均可完成這件事，即它們之間的關系用OR表達，那么一般使用加法原則，即有：Y1+ Y2+。。。+ YX種方法。
乘法原則：RULE OF MULTIPLICATION
完成一件事有X個步驟，第一步有Y1種方法，第二步有Y2種方法，。。。，第X步有YX種方法，完成這件事一共有Y1 Y2。。。YX種方法。
以上只是GMAT考題中經常涉及到的數學算術方面的問題，今后我們將陸續在新開辟的網上課堂中介紹代數、幾何以及系統的習題、講解，以幫助大家在GMAT數學考試中更好地發揮中國學生的優勢，拿到讓美國人瞠目結舌的成績!
