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b:head_first_statistics:using_the_normal_distribution

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b:head_first_statistics:using_the_normal_distribution [2025/10/08 03:10] – [Apply a continuity correction] hkimscilb:head_first_statistics:using_the_normal_distribution [2025/10/29 02:12] (current) – [All aboard the Love Train] hkimscil
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 [1] 0.9406201 [1] 0.9406201
 > pnorm(-1.56, lower.tail = FALSE) > pnorm(-1.56, lower.tail = FALSE)
 +[1] 0.9406201
 +> pnorm(-1.56, 0, 1, lower.tail = F)
 [1] 0.9406201 [1] 0.9406201
 > pnorm(64, 71, sqrt(20.25), lower.tail = FALSE) > pnorm(64, 71, sqrt(20.25), lower.tail = FALSE)
Line 151: Line 153:
 >  > 
 </code> </code>
 +
 +Note: 이제는 x축이 discrete 하지 않으므로 dnorm()과 같은 펑션을 써서 더할 수 없다 (할 수 있기는 하지만 간단하지 않다). 
 +
  
 ==== exercise ==== ==== exercise ====
 <WRAP box> <WRAP box>
-1. N(10, 4), value 6  +  - N(10, 4), value 6  
-2. N(6.3, 9), value 0.3 +  - N(6.3, 9), value 0.3 
-3. N(2, 4). If the standard score is 0.5, what’s the value?  +  - N(2, 4). If the standard score is 0.5, what’s the value?  
-4. The standard score of value 20 is 2. If the variance is 16, what’s the mean?+  - The standard score of value 20 is 2. If the variance is 16, what’s the mean? 
 +</WRAP> 
 +<WRAP box> 
 +<code> 
 +  * 1 
 +pnorm(6, 10, sqrt(4), lower.tail = F) 
 +  * 2 
 +pnorm(0.3, 6.3, sqrt(9), lower.tail = F) 
 +  * 3 
 +0.5 = (v - 2)/sqrt(4) 
 +v-2 = 1 
 +v = 3 
 +  * 4 
 +z = (v - mean) / sd  
 +2 = (20 - mean) / sqrt(16) 
 +mean = 12 
 +</code>
 </WRAP> </WRAP>
  
Line 202: Line 223:
 \end{eqnarray*} \end{eqnarray*}
  
-<code>> 1-pnorm(-0.44)+<code> 
 +> 1-pnorm(-0.44)
 [1] 0.6700314 [1] 0.6700314
 >  > 
 +> pnorm(69, 71, sqrt(20.25), lower.tail = F)
 +[1] 0.6716394
 +> 
 +> z <- (69 - 71)/ sqrt(20.25)
 +> z
 +[1] -0.4444444
 +> pnorm(z, lower.tail = F)
 +[1] 0.6716394
 +> 
 +
 </code> </code>
  
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 rnorm(n, mean = 0, sd = 1) rnorm(n, mean = 0, sd = 1)
 </code> </code>
-</WRAP> 
-{{  :b:head_first_statistics:pasted:20201204-175705.png?500}} 
- 
- 
  
 +{{:b:head_first_statistics:pasted:20201204-175705.png?500}}
 따라서  따라서 
 $$P(X + Y < 380) = 0.9082409 $$ $$P(X + Y < 380) = 0.9082409 $$
 +</WRAP>
  
 ===== exercise ===== ===== exercise =====
-<WRAP info>+<WRAP box>
 Julie’s matchmaker is at it again. What's the **probability that a man will be at least 5 inches taller than a woman**? In Statsville, the height of men in inches is distributed as N(71, 20.25), and the height of women in inches is distributed as N(64, 16). Julie’s matchmaker is at it again. What's the **probability that a man will be at least 5 inches taller than a woman**? In Statsville, the height of men in inches is distributed as N(71, 20.25), and the height of women in inches is distributed as N(64, 16).
 </WRAP> </WRAP>
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 **probability that a man will be at least 5 inches taller than a woman**? = "probability that a man will be at least 5 inches taller than (an average) woman" 이므로 $P(X > F + 5)$ 을 구하라는 문제.  **probability that a man will be at least 5 inches taller than a woman**? = "probability that a man will be at least 5 inches taller than (an average) woman" 이므로 $P(X > F + 5)$ 을 구하라는 문제. 
 +
 \begin{align*} \begin{align*}
 P(X > F + 5) & = P(X - F > 5) P(X > F + 5) & = P(X - F > 5)
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 {{:b:head_first_statistics:pasted:20191114-072427.png}} {{:b:head_first_statistics:pasted:20191114-072427.png}}
  
 +기억: 
 +E(ax + b) = a E(x) + b
 +V(ax + b) = a^2 V(x) + 0
 +
 + 
 ===== Independent Observation  ===== ===== Independent Observation  =====
 Rather than transforming the weight of each adult, what we really need to figure out is <fc #ff0000>the probability distribution for the combined weight of four separate adults</fc>. In other words, we need to work out <fc #ff0000>the probability distribution of four independent observations of X</fc>. Rather than transforming the weight of each adult, what we really need to figure out is <fc #ff0000>the probability distribution for the combined weight of four separate adults</fc>. In other words, we need to work out <fc #ff0000>the probability distribution of four independent observations of X</fc>.
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 > pbinom(29,40, 1/4, lower.tail = F) > pbinom(29,40, 1/4, lower.tail = F)
 [1] 4.630881e-11 [1] 4.630881e-11
 +> dbinom(30:40,  40, 1/4)
 + [1] 4.140329e-11 4.451967e-12 4.173719e-13 3.372702e-14 2.314599e-15
 + [6] 1.322628e-16 6.123279e-18 2.206587e-19 5.806808e-21 9.926167e-23
 +[11] 8.271806e-25
 +> 1 - dbinom(0:29, 40, 1/4)
 + [1] 0.9999899 0.9998659 0.9991284 0.9963200 0.9886534 0.9727683
 + [7] 0.9470494 0.9142704 0.8821219 0.8602926 0.8556357 0.8687597
 +[13] 0.8942786 0.9240975 0.9512055 0.9718076 0.9853165 0.9930901
 +[19] 0.9970569 0.9988641 0.9996024 0.9998738 0.9999637 0.9999905
 +[25] 0.9999978 0.9999995 0.9999999 1.0000000 1.0000000 1.0000000
 +> sum(dbinom(30:40,  40, 1/4))
 +[1] 4.630881e-11
 +> 1 - sum(dbinom(0:29, 40, 1/4))
 +[1] 4.630896e-11
 +> 
 +
 </code> </code>
  
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 <WRAP box> <WRAP box>
-이를 R을 이용하여 구하면, +위를 R에서 해보면 
 <code> <code>
-pbinom(5, 12, 1/2)+> dbinom(0, 12, 1/2) + dbinom(1, 12, 1/2) + dbinom(2, 12, 1/2)  
 +>  + dbinom(3, 12, 1/2) + dbinom(4, 12, 1/2) + dbinom(5, 12, 1/2) 
 +[1] 0.387207
 </code> </code>
 +그러나, R에서는 더 간단한 방법으로 
 <code> <code>
 > pbinom(5, 12, 1/2) > pbinom(5, 12, 1/2)
Line 788: Line 842:
 </code> </code>
  
-위는 아래와 같음을 이해해야 한다+그리고, 위의 dbinom으로 하나씩 계산한다고 하더라도 아래처럼 하게 된다
 <code> <code>
 > sum(dbinom(c(0:5),12,1/2)) > sum(dbinom(c(0:5),12,1/2))
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 ===== Pool Puzzle ===== ===== Pool Puzzle =====
 <wrap #continuity_correction_egs /> <wrap #continuity_correction_egs />
-<WRAP help> +<WRAP box> 
-X < 3  ----  <wrap spoiler> X < 2.5 </wrap> +X < 3   <wrap spoiler> X < 2.5 </wrap> 
-X > 3  ----  <wrap spoiler> X > 3.5 </wrap> +X > 3   <wrap spoiler> X > 3.5 </wrap> 
-X <_ 3  ----  <wrap spoiler> X < 3.5 </wrap> +X <_ 3   <wrap spoiler> X < 3.5 </wrap> 
-X >_ 3  ----  <wrap spoiler> X > 2.5 </wrap> +X >_ 3   <wrap spoiler> X > 2.5 </wrap> 
-3 <_ X < 10   ----  <wrap spoiler> 2.5 < X < 9.5 </wrap> +3 <_ X < 10    <wrap spoiler> 2.5 < X < 9.5 </wrap> 
-X = 0  ----  <wrap spoiler> -0.5 < X < 0.5 </wrap> +X = 0   <wrap spoiler> -0.5 < X < 0.5 </wrap> 
-3 <_ X <_ 10  ----  <wrap spoiler> 2.5 < X < 10.5 </wrap> +3 <_ X <_ 10   <wrap spoiler> 2.5 < X < 10.5 </wrap> 
-3 < X <_ 10  ----  <wrap spoiler> 3.5 < X < 10.5 </wrap> +3 < X <_ 10   <wrap spoiler> 3.5 < X < 10.5 </wrap> 
-X > 0  ----  <wrap spoiler> X > 0.5 </wrap> +X > 0   <wrap spoiler> X > 0.5 </wrap> 
-3 < X < 10  ----  <wrap spoiler> 3.5 < X < 9.5 </wrap>+3 < X < 10   <wrap spoiler> 3.5 < X < 9.5 </wrap>
 </WRAP> </WRAP>
  
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 {{:b:head_first_statistics:pasted:20191118-113020.png}} {{:b:head_first_statistics:pasted:20191118-113020.png}}
  
-$\lambda > 15$ 일 때, Poisson distribution, $X \sim Po(\lambda)$는 $X \sim N(\lambda, \lambda)$ 의 성격을 취한다.+<fc #ff0000>$\lambda > 15$ 일 때,</fc> Poisson distribution, $X \sim Po(\lambda)$는 $X \sim N(\lambda, \lambda)$ 의 성격을 취한다.
  
 예)  예) 
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 $0.9654916 \sim 0.9656205$ $0.9654916 \sim 0.9656205$
 +
 +R에서 ppois을 이용하면
 +<code>
 +> ppois(51, 40)
 +[1] 0.9612598
 +> 
 +
 +</code>
  
 ===== Check up ===== ===== Check up =====
b/head_first_statistics/using_the_normal_distribution.1759893052.txt.gz · Last modified: by hkimscil
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