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--- b:head_first_statistics:using_the_normal_distribution [2025/10/08 03:08] – [Swivel chair again] hkimscil
+++ b:head_first_statistics:using_the_normal_distribution [2025/10/29 02:12] (current) – [All aboard the Love Train] hkimscil
@@ Line 146: / Line 146: @@
 [1] 0.9406201
 > pnorm(-1.56, lower.tail = FALSE)
+[1] 0.9406201
+> pnorm(-1.56, 0, 1, lower.tail = F)
 [1] 0.9406201
 > pnorm(64, 71, sqrt(20.25), lower.tail = FALSE)
@@ Line 151: / Line 153: @@
 >
 </code>
+Note: 이제는 x축이 discrete 하지 않으므로 dnorm()과 같은 펑션을 써서 더할 수 없다 (할 수 있기는 하지만 간단하지 않다).
 ==== exercise ====
 <WRAP box>
-. N(10, 4), value 6
+  - N(10, 4), value 6
-. N(6.3, 9), value 0.3
+  - N(6.3, 9), value 0.3
-. 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?
-. 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
+.5 = (v - 2)/sqrt(4)
+v-2 = 1
+v = 3
+  * 4
+z = (v - mean) / sd
+= (20 - mean) / sqrt(16)
+mean = 12
+</code>
 </WRAP>
@@ Line 202: / Line 223: @@
 \end{eqnarray*}
-<code>> 1-pnorm(-0.44)
+<code>
+> 1-pnorm(-0.44)
 [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>
@@ Line 513: / Line 545: @@
 rnorm(n, mean = 0, sd = 1)
 </code>
-</WRAP>
-{{  :b:head_first_statistics:pasted:20201204-175705.png?500}}
+{{:b:head_first_statistics:pasted:20201204-175705.png?500}}
 따라서
 $$P(X + Y < 380) = 0.9082409 $$
+</WRAP>
 ===== 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).
 </WRAP>
@@ Line 532: / Line 562: @@
 **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*}
 P(X > F + 5) & = P(X - F > 5)
@@ Line 581: / Line 612: @@
 {{: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  =====
 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>.
@@ Line 656: / Line 692: @@
 > pbinom(29,40, 1/4, lower.tail = F)
 [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>
@@ Line 777: / Line 829: @@
 {{:b:head_first_statistics:pasted:20191118-095652.png}}
-<WRAP info>
+<WRAP box>
-이를 R을 이용하여 구하면,
+위를 R에서 해보면
 <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>
+그러나, R에서는 더 간단한 방법으로
 <code>
 > pbinom(5, 12, 1/2)
@@ Line 788: / Line 842: @@
 </code>
-위는 아래와 같음을 이해해야 한다
+그리고, 위의 dbinom으로 하나씩 계산한다고 하더라도 아래처럼 하게 된다
 <code>
 > sum(dbinom(c(0:5),12,1/2))
@@ Line 844: / Line 898: @@
 이 값은 위의 0.387에 근사하다.
-<WRAP info>
+<WRAP box>
   * In particular circumstances you can **use the normal distribution to approximate the binomial**. If X ~ B(n, p) and np > 5 and nq > 5 then you can approximate X using X ~ N(np, npq)
   * If you’re approximating the binomial distribution with the normal distribution, then you need to **<fc #ff0000>apply a continuity correction</fc>** to make sure your results are accurate.
@@ Line 851: / Line 905: @@
 {{:b:head_first_statistics:pasted:20191118-103328.png}}
-<WRAP info>
+<WRAP box>
 Q:Does it really save time to approximate the binomial distribution with the normal?
@@ Line 874: / Line 928: @@
 ===== Pool Puzzle =====
 <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>
-<_ X < 10   ----  <wrap spoiler> 2.5 < X < 9.5 </wrap>
+<_ 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>
-<_ X <_ 10  ----  <wrap spoiler> 2.5 < X < 10.5 </wrap>
+<_ X <_ 10   <wrap spoiler> 2.5 < X < 10.5 </wrap>
-< X <_ 10  ----  <wrap spoiler> 3.5 < X < 10.5 </wrap>
+< 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>
-< X < 10  ----  <wrap spoiler> 3.5 < X < 9.5 </wrap>
+< X < 10   <wrap spoiler> 3.5 < X < 9.5 </wrap>
 </WRAP>
@@ Line 921: / Line 975: @@
 {{: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)
@@ Line 970: / Line 1024: @@
 $0.9654916 \sim 0.9656205$
+R에서 ppois을 이용하면
+<code>
+> ppois(51, 40)
+[1] 0.9612598
+>
+</code>
 ===== Check up =====