9.2 经验规则

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  经验规则

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  If the price per pound of USDA Choice Beef is normally distributed with a mean of $4.85/lb and a standard deviation of $0.35/lb, what is the estimated probability that a randomly chosen sample (from a randomly chosen market) will be between $5.20 and $5.55 per pound?
  ::如果USDA选择牛肉每磅的价格通常平均分配为4.85美元/磅,标准差为0.35/磅/磅,则随机选择的样本(从随机选择的市场上)每磅在5.20美元至5.55美元之间的估计概率是多少?

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  The Empirical Rule
  ::经验规则

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  This lesson on the Empirical Rule is an extension of the previous lesson “Understanding the Normal Distribution”. In the prior lesson, the goal was to develop an intuition of the interaction between decreased probability and increased distance from the mean. In this lesson, we will practice applying the Empirical Rule to estimate the specific probability of occurrence of a sample based on the range of the sample, measured in standard deviations.
  ::关于经验规则的这一教训是以前“了解正常分布”的教训的延伸,在以前的教训中,目标是对概率下降和与平均值距离增加之间的相互作用形成直觉,在这个教训中,我们将运用经验规则来根据抽样范围(按标准偏差衡量)估计抽样的具体发生概率。

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  The graphic below is a representation of the Empirical Rule:
  ::下面的图示是经验规则的缩写:

  

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  The graphic is a rather concise summary of the ‘vital statistics’ of a Normal Distribution. Note how the graph resembles a bell? Now you know why the normal distribution is also called a “ bell curve ”.
  ::图形是正常分布的`vital 统计 ' 的简明摘要。 注意图表与铃声的相似之处。 现在你知道为什么正常分布也被称为“ 钟曲线 ” 了。

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  • 50% of the data is above, and 50% below, the mean of the data
    ::50%的数据高于数据平均值,50%低于数据平均值
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  • Approximately 68% of the data occurs within 1 SD of the mean
    ::大约68%的数据发生在平均值1SD以内
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  • Approximately 95% occurs within 2 SD’s of the mean
    ::大约95%发生在平均2SD范围内
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  • Approximately 99.7% of the data occurs within 3 SD’s of the mean
    ::大约99.7%的数据发生在3SD的平均值范围内。

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  It is due to the probabilities associated with 1, 2, and 3 SD’s that the Empirical Rule is also known as the “ $\begin{array}{r}68-95-99.7\text{rule}\end{array}$ ”.
  ::经验规则也称为 " 68-95-99.7规则 " ,是因为与1、2和3SD有关的可能性。

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  Calculating Probability 
  ::计算概率

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  1. If the diameter of a basketball is normally distributed, with a mean $\begin{array}{r}(\mu )\end{array}$ of 9″, and a standard deviation $\begin{array}{r}(\sigma )\end{array}$ of 0.5″, what is the probability that a randomly chosen basketball will have a diameter between 9.5″ and 10.5″?
  ::1. 如果篮球直径通常分布,平均(微)9,标准偏差(0.5)0.5,随机选择的篮球直径在9.5至10.5之间的可能性有多大?

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  Since the  $\begin{array}{r}\sigma ={0.5}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$ and the $\begin{array}{r}\sigma =9^{\mathrm{\prime }\mathrm{\prime }}\end{array}$ , we are evaluating the probability that a randomly chosen ball will have a diameter between 1 and 3 standard deviations above the mean. The graphic below shows the portion of the normal distribution included between 1 and 3 SD’s:
  ::由于“0.5”和“9”,我们正在评估随机选择的球的直径在1至3标准差高于平均值的概率。下图显示1至3SD的正常分布部分:

  

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  The percentage of the data spanning the 2 ^nd and 3 ^rd SD’s is  $\begin{array}{r}13.5\mathrm{\%}+2.35\mathrm{\%}=15.85\mathrm{\%}\end{array}$
  ::第2和第3SD的数据百分比为13.52.3515.85%。

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  The probability that a randomly chosen basketball will have a diameter between 9.5 and 10.5 inches is 15.85%.
  ::随机选择的篮球直径在9.5至10.5英寸之间的概率为15.85%。

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  2. If the depth of the snow in my yard is normally distributed, with  $\begin{array}{r}\mu ={2.5}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$ and $\begin{array}{r}\sigma ={.25}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$ , what is the probability that a randomly chosen location will have a snow depth between 2.25 and 2.75 inches?
  ::2. 如果我家院子里的雪的深度通常分布在2.5英寸和2.5英寸之间,那么随机选择的地点在2.25英寸至2.75英寸之间,其下雪深度的概率是多少?

  

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  2.25 inches is $\begin{array}{r}\mu -1\sigma \end{array}$ , and 2.75 inches is $\begin{array}{r}\mu +1\sigma \end{array}$ , so the area encompassed approximately represents $\begin{array}{r}34\mathrm{\%}+34\mathrm{\%}=68\mathrm{\%}\end{array}$ .
  ::2.25英寸是1-13, 2.75英寸是1-12, 所以覆盖面积大约为34-34-68%。

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  The probability that a randomly chosen location will have a depth between 2.25 and 2.75 inches is 68%.
  ::随机选择位置的深度在2.25至2.75英寸之间的概率为68%。

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  3. If the height of women in the U.S. is normally distributed with  $\begin{array}{r}\mu =5^{\mathrm{\prime }}{8}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$ and $\begin{array}{r}\sigma ={1.5}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$ , what is the probability that a randomly chosen woman in the U.S. is shorter than 5′5″?
  ::3. 如果美国妇女的身高通常与5855555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555,

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  This one is slightly different, since we aren’t looking for the probability of a limited range of values. We want to evaluate the probability of a value occurring anywhere below 5′5″. Since the domain of a normal distribution is infinite, we can’t actually state the probability of the portion of the distribution on ‘that end’ because it has no ‘end’! What we need to do is add up the probabilities that we do know and subtract them from 100% to get the remainder.
  ::这一点略有不同, 因为我们并不在寻找有限值范围的概率。 我们想要评估一个值发生于5'55555°的概率。 由于正常分布的域是无限的, 我们实际上无法说明`此端 ' 分配部分的概率, 因为它没有` 结束 ' ! 我们需要做的是将我们所知道的概率加在一起, 从100%中减去, 以获得剩余值 。

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  Here is that normal distribution graphic again, with the height data inserted:
  ::这里再次显示正常的分布图形, 并插入高度数据 :

  

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  Recall that a normal distribution always has 50% of the data on each side of the mean. That indicates that 50% of U.S. females are taller than 5′8″, and gives us a solid starting point to calculate from. There is another 34% between 5′6.5″ and 5′8″ and a final 13.5% between 5′5″ and 5′6.5″. Ultimately that totals: $\begin{array}{r}50\mathrm{\%}+34\mathrm{\%}+13.5\mathrm{\%}=97.5\mathrm{\%}\end{array}$ .
  ::回顾正常分布总是有50%的平均值数据。 这表明50%的美国女性高于5885, 并给了我们一个坚实的起点来计算。 还有34%在56.55至585之间,最后的13.5%在555至56.5之间。 最终的总数是:503413.5/97.5%。

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  Since 87.5% of U.S. females are 5′5″ or taller, that leaves 2.5% that are less than 5′5″ tall.
  ::由于87.5%的美国女性为5'5或更高,因此2.5%的美国女性不到5'555。

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  Earlier Problem Revisited
  ::重审先前的问题

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  If the price per pound of USDA Choice Beef is normally distributed with a mean of $4.85/lb and a standard deviation of $0.35/lb, what is the estimated probability that a randomly chosen sample (from a randomly chosen market) will be between $5.20 and $5.55 per pound?
  ::如果USDA选择牛肉每磅的价格通常平均分配为4.85美元/磅,标准差为0.35/磅/磅,则随机选择的样本(从随机选择的市场上)每磅在5.20美元至5.55美元之间的估计概率是多少?

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  $5.20 is $\begin{array}{r}\mu +1\sigma \end{array}$ , and $5.55 is $\begin{array}{r}\mu +2\sigma \end{array}$ , so the probability of a value occurring in that range is approximately 13.5%.
  ::5.20是1-12美元,5.55是2-12美元,因此该范围内值的概率约为13.5%。

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  Examples
  ::实例

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  Example 1
  ::例1

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  A normally distributed data set has  $\begin{array}{r}\mu =10\end{array}$  and  $\begin{array}{r}\sigma =2.5\end{array}$ , what is the probability of randomly selecting a value greater than 17.5 from the set?
  ::通常分布的数据集有 10 和 2.5, 从集中随机选择值大于17.5的概率是多少?

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  If  $\begin{array}{r}\mu =10\end{array}$ and $\begin{array}{r}\sigma =2.5\end{array}$ , then $\begin{array}{r}17.5=\mu +3\sigma \end{array}$ . Since we are looking for all data above that point, we need to subtract the probability that a value will occur below that value from 100%:
  ::===============================================================================================================================================)=============================================================================================================================================================================================================================================================================================================================================================================

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  The probability that a value will be less than 10 is 50%, since 10 is the mean. There is another 34% between 10 and 12.5, another 13.5% between 12.5 and 15, and a final 2.35% between 15 and 17.5.
  ::数值小于 10 的概率为 50% , 因为平均值为 10 。 在 10 和 12.5 之间还有 34% , 在 12.5 和 15 之间还有 13.5% , 最后为 2.35% 在 15 和 17.5 之间。

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  $\begin{array}{r}100\mathrm{\%}-50\mathrm{\%}-34\mathrm{\%}-13.5\mathrm{\%}-2.35\mathrm{\%}=0.15\mathrm{\%}\end{array}$  probability of a value greater than 17.5
  ::数值大于17.5的概率为100.50 34 13.5 2.35 0.15%

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  Example 2
  ::例2

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  A normally distributed set has  $\begin{array}{r}\mu =.05\end{array}$  and  $\begin{array}{r}\sigma =.01\end{array}$ , what is the probability of randomly choosing a value between .05 and .07 from the set?
  ::通常的分布集为 .05 和 .01, 随机选择 . 05 和 . 07 之间值的概率是多少?

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  0.05 is the mean, and 0.07 is 2 standard deviations above the mean, so the probability of a value in that range is  $\begin{array}{r}34\mathrm{\%}+13.5\mathrm{\%}=47.5\mathrm{\%}\end{array}$
  ::0.05是平均值,0.07是高于平均值的2个标准偏差,因此该范围内值的概率为3413.547.5%。

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  Example 3
  ::例3

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  A normally distributed data set has  $\begin{array}{r}\mu =514\end{array}$  and an unknown standard deviation, what is the probability that a randomly selected value will be less than 514?
  ::通常分布的数据集有514和未知的标准偏差,随机选择值低于514的概率有多大?

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  514 is the mean, so the probability of a value less than that is 50%.
  ::514是平均值,所以值低于该值的概率为50%。

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  Review 
  ::回顾

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  For questions 1-15, assume all distributions to be normal or approximately normal, and calculate percentages using the  $\begin{array}{r}68-95-99.7\end{array}$ rule.
  ::对于问题1-15, 假定所有分布均正常或大致正常,并使用68-95-99.7细则计算百分比。

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  1. Given mean 63 and standard deviation of 168, find the approximate percentage of the distribution that lies between -105 and 567.
     ::考虑到平均值63,标准差168,找到105至567之间分配的大致百分比。
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  2. Approximately what percent of a normal distribution is between 2 standard deviations and 3 standard deviations from the mean?
     ::正常分配的大约多少百分比是在2个标准差和3个标准差之间?
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  3. Given standard deviation of 74 and mean of 124, approximately what percentage of the values are greater than 198?
     ::鉴于标准偏差为74,平均值为124,值中的百分比大约大于198?
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  4. Given  $\begin{array}{r}\sigma =39\end{array}$ and $\begin{array}{r}\mu =101\end{array}$ , approximately what percentage of the values are less than 23?
     ::鉴于 39 和101, 大约多少百分比的值小于23?
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  5. Given mean 92 and standard deviation 189, find the approximate percentage of the distribution that lies between -286 and 470.
     ::考虑到平均值92和标准差189,找到 -286至470之间的分配的大致百分比。
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  6. Approximately what percent of a normal distribution lies between  $\begin{array}{r}\mu +1\sigma \end{array}$ and $\begin{array}{r}\mu +2\sigma \end{array}$ ?
     ::正常分配的大约百分之几在1+1和2+2之间?
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  7. Given standard deviation of 113 and mean 81, approximately what percentage of the values are less than -145?
     ::鉴于标准偏差113和平均值81,价值中大约多少百分比低于-145?
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  8. Given mean 23 and standard deviation 157, find the approximate percentage of the distribution that lies between 23 and 337.
     ::参照平均值23和标准偏差157,找到23至337之间分布的大致百分比。
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  9. Given  $\begin{array}{r}\sigma =3\end{array}$ and $\begin{array}{r}\mu =84\end{array}$ , approximately what percentage of the values are greater than 90?
     ::鉴于3和84, 数值中大约多少百分率大于90?
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  10. Approximately what percent of a normal distribution is between  $\begin{array}{r}\mu \end{array}$ and $\begin{array}{r}\mu +1\sigma \end{array}$ ?
      ::通常分配量的大约百分之几在 μ 和 1- 12 之间?
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  11. Given mean 118 and standard deviation 145, find the approximate percentage of the distribution that lies between -27 and 118.
      ::考虑到平均值118和标准差145,找到27至118之间的分配的大致百分比。
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  12. Given standard deviation of 81 and mean 67, approximately what percentage of values are greater than 310?
      ::鉴于标准偏差81和平均值67,价值中大约比310的百分比大于310?
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  13. Approximately what percent of a normal distribution is less than 2 standard deviations from the mean?
      ::正常分配的大约多少比例低于平均值的2个标准差?
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  14. Given  $\begin{array}{r}\mu +1\sigma =247\end{array}$ and $\begin{array}{r}\mu +2\sigma =428\end{array}$ , find the approximate percentage of the distribution that lies between 66 and 428.
      ::考虑到1247和2428,找到66至428之间分布的大致百分比。
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  15. Given  $\begin{array}{r}\mu -1\sigma =-131\end{array}$ and $\begin{array}{r}\mu +1\sigma =233\end{array}$ , approximately what percentage of the values are greater than -495?
      ::考虑到1131和1233, 数值中有多大比例大于-495?

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。