如果没有数字重复，你如何找到模式？

在统计学中，对于给定的数据分布，众数是最常出现的值或数字。它代表出现次数最多的值或整数。但是，这里的 t 可能是也可能不是给定数据集的模态值。这是因为给定的数据集可能具有重复或非重复值。除此之外，给定的数据集可能具有一种、两种或多种模式。数据集会影响众数的值。

模式公式

$Mode=L+h\frac{(f_m-f_1)}{(f_m-f_1)+(f_m-f_2)}$

Here, we have,

L – lower limit of the modal class, 

h – size of the class interval, 

fm – frequency of the modal class, 

f1 – frequency of the class preceding the modal class, and 

f2 –  frequency of the class succeeding the modal class.

未分组数据的众数公式

在未分组数据的情况下，数据分布首先按升序或降序排列。然后将重复值连同它们的频率一起描述。对应于最高频率的观察被称为给定数据的模态值。

分组数据的众数公式

分组数据的众数公式由下式给出，

$Mode=L+h\frac{(f_m-f_1)}{(f_m-f_1)+(f_m-f_2)}$

Here, we have, 

L – lower limit of the modal class,

h – size of the class interval,

f_m – frequency of the modal class,

f₁ – frequency of the class preceding the modal class and

f₂ – frequency of the class succeeding the modal class.

如何找到模式？

未分组数据的模式

未出现在组中的数据称为未分组数据。为了说明，我们举了一个例子，假设有一家服装公司生产冬衣。显示的频率分布表中提到了以下带有衬衫的表格数据以及尺寸：

Size of the winter coat	38	39	40	42	43	44	45
Total number of shirts	33	11	22	55	44	11	22

因为，很明显，42 号的频率最高。因此，冬季大衣尺码的众数为 42。

未分组数据的众数计算不同于分组数据的众数。

分组数据模式

以下步骤对应于分组数据的模式计算：

步骤 1：计算对应于最大频率的类间隔。该值也称为模态类。

第 2 步：通过从相应的下限中减去上限来计算类的大小。

步骤 3：使用众数公式计算众数的值：

$Mode=L+h\frac{(f_m-f_1)}{(f_m-f_1)+(f_m-f_2)}$

如果没有数字重复，你如何找到模式？

解决方案：

For a data distribution that has no repeated numbers, there exists no mode. To prove this, let us assume a data distribution given by : A, B, C, D and E. 

Constructing the frequency distribution table for the given set of observations, we have, 

We can clearly observe that each of the observation is repeated just once, thereby having a frequency equivalent to 1. Hence, there is no maximum occurrence in this data set and hence, no mode. 

示例问题

问题 1. 找出给定未分组数据的模式。

Observation	A	B	C	D	E
Frequency	1	1	1	1	1

解决方案：

Here as to find the mode of this ungrouped data

Observe the bike color with the highest frequency

As we can see that ‘Silver’ color bike has the highest frequency 

Therefore,

The mode is 55.

问题2.计算下面给出的分组数据的模式

Bike color	Red	Blue	Green	Black	Golden	Silver	Grey
Number of cars sold	15	25	42	32	35	55	22

解决方案：

Here we have to find the mode of marks

First find the maximum frequency

f_m = 16

The corresponding class interval to f_m is 30-40

The lower limit of this class ‘L’ is 30

Size of the class interval = 10

Frequency of the preceding class f₁ = 8

Frequency of the succeeding class f₂ = 12

Substitute the value in the mode formula

i.e,

$Mode=L+h\frac{(f_m-f_1)}{(f_m-f_1)+(f_m-f_2)}\\ =30+10\times\frac{16-8}{(16-8)(16-12)}\\ =30+10\times\frac{8}{8\times4}\\ =30+\frac{80}{32}\\ =30+2.5\\ =32.5$

问题3.计算以下数据的众数

15, 14, 19, 25, 26, 58, 109, 15, 14, 59, 58, 15, 17, 14, 19, 20, 25, 26, 109, 15, 109, 25, 59, 14, 17, 15

解决方案：

To find the mode of the following data

First arrange the data in the ascending order

14, 14, 14, 14, 15, 15, 15, 15, 15, 17, 17, 19, 19, 20, 25, 25, 25, 26, 26, 58, 58, 59, 59, 109, 109, 109

The repeating numbers in the data

14 = 4 times

15 = 5 times

17 = 2 times

19 = 2 times

25 = 3 times

26 = 2 times

58 = 2 times

59 = 2 times

109 = 3 times

Here as we can see that 15 occurs most of the times

Therefore,

Mode of the data is 15.

问题 4. 找出以下频率分布的模式

Marks obtained	Number of Students
10-20	4
20-30	8
30-40	16
40-50	12

解决方案：

Highest frequency f_m = 66

Lower level of frequency (L) = 30

Modal class = 30-40

Frequency of the interval class preceding f₁ = 40

Frequency of the class succeeding f₂ = 55

Width of the class h = 10

Thus,

$Mode=L+h\frac{(f_m-f_1)}{(f_m-f_1)+(f_m-f_2)}\\ =30+10\times\frac{66-40}{(66-40)(66-55)}\\ =30+10\times\frac{16}{16\times11}\\ =30+\frac{10}{11}\\ =30+0.90\\ =30.90$

问题5.找到数据的模式

14, 15, 5, 3, 18, 19, 5, 16, 25, 33, 5, 3, 14, 18

解决方案：

To find the mode first arrange the data in the ascending order

3, 5, 5, 5, 14, 14, 15, 16, 18, 18, 19, 25, 33 

Here as we can observe that 5 is occurring maximum times

Therefore,

The mode of the data is 5.