exercise:3780563cb4: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Let <math>S_n</math> be the number of successes in <math>n</math> independent trials. Use the program ''' BinomialProbabilities''' (Section \ref{sec 3.2}) to compute, for given <math>n</math>, <math>p</math>, and <math>j</math>, the probability...") |
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Let <math>S_n</math> be the number of successes in <math>n</math> independent trials. Use the program ''' BinomialProbabilities''' ([[guide:E54e650503|Combinations]]) to compute, for given <math>n</math>, <math>p</math>, and <math>j</math>, the probability | |||
trials. Use the program ''' BinomialProbabilities''' ( | |||
for given <math>n</math>, <math>p</math>, and <math>j</math>, the probability | |||
<math display="block"> | <math display="block"> | ||
P(-j\sqrt{npq} < S_n - np < j\sqrt{npq})\ . | P(-j\sqrt{npq} < S_n - np < j\sqrt{npq})\ . | ||
</math> | </math> | ||
<ul><li> Let <math>p = .5</math>, and compute this probability for <math>j = 1</math>, 2, 3 and <math>n = | <ul style="list-style-type:lower-alpha"><li> Let <math>p = .5</math>, and compute this probability for <math>j = 1</math>, 2, 3 and <math>n = | ||
10</math>, 30, 50. Do the same for <math>p = .2</math>. | 10</math>, 30, 50. Do the same for <math>p = .2</math>. | ||
</li> | </li> | ||
Latest revision as of 22:09, 14 June 2024
Let [math]S_n[/math] be the number of successes in [math]n[/math] independent trials. Use the program BinomialProbabilities (Combinations) to compute, for given [math]n[/math], [math]p[/math], and [math]j[/math], the probability
[[math]]
P(-j\sqrt{npq} \lt S_n - np \lt j\sqrt{npq})\ .
[[/math]]
- Let [math]p = .5[/math], and compute this probability for [math]j = 1[/math], 2, 3 and [math]n = 10[/math], 30, 50. Do the same for [math]p = .2[/math].
- Show that the standardized random variable [math]S_n^* = (S_n - np)/\sqrt{npq}[/math] has expected value 0 and variance 1. What do your results from (a) tell you about this standardized quantity [math]S_n^*[/math]?