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b:head_first_statistics:using_the_normal_distribution [2025/10/08 12:07] – [Swivel chair again] hkimscilb:head_first_statistics:using_the_normal_distribution [2025/10/09 06:26] (current) – [X - Y] hkimscil
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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}} 
- 
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 +{{: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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-<Wrap box>+<WRAP box>
 There are 40 questions, which means there are 40 trials.  There are 40 questions, which means there are 40 trials. 
  
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 </WRAP> </WRAP>
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 <code> <code>
 > pbinom(29,40, 1/4, lower.tail = F) > pbinom(29,40, 1/4, lower.tail = F)
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 {{:b:head_first_statistics:pasted:20191118-095652.png}} {{:b:head_first_statistics:pasted:20191118-095652.png}}
  
-<WRAP info>+<WRAP box>
 이를 R을 이용하여 구하면,  이를 R을 이용하여 구하면, 
 <code> <code>
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 이 값은 위의 0.387에 근사하다.  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)   * 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.   * 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.
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 {{:b:head_first_statistics:pasted:20191118-103328.png}} {{:b:head_first_statistics:pasted:20191118-103328.png}}
  
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 Q:Does it really save time to approximate the binomial distribution with the normal? Q:Does it really save time to approximate the binomial distribution with the normal?
  
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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>
b/head_first_statistics/using_the_normal_distribution.1759892859.txt.gz · Last modified: by hkimscil

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