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Chapter 6: Q3SE (page 375)

Suppose that X has the Poisson distribution with mean 10. Use the central limit theorem, both without and with the correction for continuity, to determine an approximate value for\({\rm P}\left( {8 \le X \le 12} \right)\)Use the table of Poisson probabilities given in the back of this book to assess the quality of these approximations.

Short Answer

Expert verified

Approximate value of \(P\left( {8 \le X \le 12} \right)\)is 0.571

Step by step solution

01

Given information

A random variable X follows Poisson distribution with mean 10.

02

Calculate approximate value of \({\rm P}\left( {8 \le X \le 12} \right)\)

X has approximately a normal distribution with mean 10 and standard deviation\({\left( {10} \right)^{\frac{1}{2}}} = 3.162\).

Thus by central limit theorem without correction for continuity:

\(\begin{array}{c}P\left( {8 \le X \le 12} \right) = P\left( {\frac{{8 - 10}}{{3.162}} \le Z \le \frac{{12 - 10}}{{3.162}}} \right)\\ = P\left( {\frac{{ - 2}}{{3.162}} \le Z \le \frac{2}{{3.162}}} \right)\end{array}\)

\(\begin{array}{c}P\left( {8 \le X \le 12} \right) = \phi \left( { - 0.6325} \right) - \phi \left( {0.6325} \right)\\ = 0.473\end{array}\)

By central limit theorem with correction for continuity:

\(\begin{array}{c}P\left( {7.5 \le X \le 12.5} \right) = P\left( {\frac{{7.5 - 10}}{{3.162}} \le Z \le \frac{{12.5 - 10}}{{3.162}}} \right)\\ = P\left( {\frac{{ - 2.5}}{{3.162}} \le Z \le \frac{{2.5}}{{3.162}}} \right)\end{array}\)

\(\begin{array}{c}{\rm P}\left( {7.5 \le X \le 12.5} \right) = \phi \left( {0.7906} \right) - \phi \left( { - 0.7906} \right)\\ = 0.571\end{array}\)

The exact probability is found from the poisson table to be

\(\left( {0.1126} \right) + \left( {0.1251} \right) + \left( {0.1251} \right) + \left( {0.1137} \right) + \left( {0.0948} \right) = 0.571\)

And the approximate value of\({\rm P}\left( {8 \le X \le 12} \right)\)is 0.571

Thus the approximation with the correction for continuity is almost perfect.

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Most popular questions from this chapter

Suppose that X is a random variable for which E(X) = 渭 and \({\bf{E}}\left[ {{{\left( {{\bf{X - \mu }}} \right)}^{\bf{4}}}} \right]{\bf{ = }}{{\bf{\beta }}^{\bf{4}}}\) Prove that

\({\bf{P}}\left( {\left| {{\bf{X - \mu }}} \right| \ge {\bf{t}}} \right) \le \frac{{{{\bf{\beta }}_{\bf{4}}}}}{{{{\bf{t}}^{\bf{4}}}}}\)

Suppose that, on average, 1/3 of the graduating seniors at a certain college have two parents attend the graduation ceremony, another third of these seniors have one parent attend the ceremony, and the remaining third of these seniors have no parents attend. If there are 600 graduating seniors in a particular class, what is the probability that not more than 650 parents will attend the graduation ceremony?

Suppose people attending a party pour drinks from a bottle containing 63 ounces of a particular liquid. Suppose also that the expected size of each drink is 2 ounces, the standard deviation of each drink is 1/2 ounce, and all drinks are poured independently. Determine the probability that the bottle will not be empty after 36 drinks have been poured.

Suppose that\({X_1},{X_2}....\)is a sequence of positive integer-valued random variables. Suppose that there is a function\(f\)such that for every\(m = 1,2...\),\(\mathop {{\bf{lim}}}\limits_{{\bf{\delta x}} \in {\bf{0}}} {\bf{{\rm P}}}\left( {{{\bf{X}}_{\bf{n}}}{\bf{ = m}}} \right){\bf{ = f}}\left( {\bf{m}} \right)\),\(\sum\limits_{{\bf{m = 1}}}^{\bf{\ currency}} {{\bf{f}}\left( {\bf{m}} \right){\bf{ = 1}}} \), and\(f\left( x \right) = 0\)for every\(x\)thatis not a positive integer. Let\(F\)be the discrete c.d.f. whose p.f. is\(f\).

Prove that\({X_n}\)converges in distribution to\(F\)

In this exercise, we construct an example of a sequence of random variables \({Z_n}\) such that but \(\Pr \left( {\mathop {\lim }\limits_{n \to \infty } \;{Z_n} = 0} \right) = 0\).That is, \({Z_n}\)converges in probability to 0, but \({Z_n}\)does not converge to 0 with probability 1. Indeed, \({Z_n}\)converges to 0 with probability 0.

Let Xbe a random variable having the uniform distribution on the interval\(\left[ {0,1} \right]\). We will construct a sequence of functions \({h_n}\left( x \right)\;for\;n = 1,2,...\)and define\({Z_n} = {h_n}\left( X \right)\). Each function \({h_n}\) will take only two values, 0 and 1. The set of x where \({h_n}\left( x \right) = 1\) is determined by dividing the interval \(\left[ {0,1} \right]\)into k non-overlapping subintervals of length\(\frac{1}{k}\;for\;k = 1,2,...\)arranging these intervals in sequence, and letting \({h_n}\left( x \right) = 1\) on the nth interval in the sequence for \(n = 1,2,...\)For each k, there are k non overlapping subintervals, so the number of subintervals with lengths \(1,\frac{1}{2},\frac{1}{3},...,\frac{1}{k}\) is \(1 + 2 + 3 + ... + k = \frac{{k\left( {k + 1} \right)}}{2}\)

The remainder of the construction is based on this formula. The first interval in the sequence has length 1, the next two have length 陆, the next three have length 1/3, etc.

  1. For each\(n = 1,2,...\), proved that there is a unique positive integer \({k_n}\) such that \(\frac{{\left( {{k_n} - 1} \right){k_n}}}{2} < n \le \frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)
  2. For each\(n = 1,2,..\;\), let\({j_n} = n - \frac{{\left( {{k_n} - 1} \right){k_n}}}{2}\). Show that \({j_n}\) takes the values \(1,...,{k_n}\;\)as n runs through\(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\).
  3. Define \({h_n}\left( x \right) = \left\{ \begin{array}{l}1\;if\;\frac{{\left( {{j_n} - 1} \right)}}{{{k_n}}} \le x < \frac{{{j_n}}}{{{k_n}}},\\0\;if\;not\end{array} \right.\)

Show that, for every \(x \in \left[ {0,1} \right),\;{h_n}\left( x \right) = 1\) for one and only one n among \(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)

  1. Show that \({Z_n} = {h_n}\left( X \right)\)takes the value 1 infinitely often with probability 1 \(\)
  1. Show that(6.2.18) holds.
  1. Show that\(\Pr \left( {{Z_n} = 0} \right) = 1 - \frac{1}{{{K_n}}}\)and \(\mathop {\lim }\limits_{n \to \infty } \;{k_n} = \infty \;\).
  2. Show that
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