GMAT数学算术基本知识查看

　　据留学360介绍，本文总结了GMAT数学算术概念，希望GMAT考生在备考GMAT过程中不要因为忽视细节而错失一分，任何一门考试只有从细节入手才有可能摘得高分，所以请广大考生注意GMAT数学算术基本概念的掌握。

　　1.平均数(average or arithmetic mean)

　　2.中数(median)

　　To calculate the median of n numbers，first order the 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 (6+8 )/2 = 7. Note that the mean of these numbers is 7.5.

　　3.众数(mode)：一组数中的众数是指出现频率最高的数。

　　例：the mode of 7，9，6，7，2，1 is 7。

　　4.值域(range)：表明数的分布的量，其被定义为最大值减最小值的差。

　　例：the range of–1，7，27，27，36 is 36-(-1)= 37。

　　5.标准方差(standard deviation)：

　　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:

　　(1)find the arithmetic mean ;

　　(2)find the differences between the mean and each of the n numbers ;

　　(3)square each of the differences ;

　　(4)find the average of the squared differences ;

　　(5)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.排列与组合

　　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 5×3 = 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, where each of these outcomes is a list of heads and tails in some order.

　　☆阶乘：factorial notation

　　假如一个大于1的整数n，计算n的阶乘被表示为n!，被定义为从1至n所有整数的乘积，

　　例如：4! = 4×3×2×1= 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-1) (n-2)…( 3) (2) (1) = 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.

　　Pkn = n!/ (n-k)!

　　例如：1, 2, 3, 4, 5这5个数字构成不同的5位数的总数为5! = 120

　　☆组合：combination

　　A permutation can be thought of as a selection process in which objects are selected one by one in a certain order. If the order of 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, where 0≤k≤n . 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.

　　从n个元素中任选k个元素的数目为:

　　Ckn. = n!/ (n-k)! k!

　　例如：从5个不同元素中任选2个的组合为C25 = 5!/2! 3!= 10