Chapter 6: Q4E (page 374)
Using the correction for continuity, determine the probability required in Exercise 2 of Sec. 6.3.
Probability that the number of people from the suburbs attending the concert will be fewer than 270 is \(0.0210\).
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
Suppose that we model the occurrence of defects on a fabric manufacturing line as a Poisson process with rate 0.01 per square foot. Use the central limit theorem (both with and without the correction for continuity) to approximate the probability that one would find at least 15 defects in 2000 square feet of fabric.
Let X have the binomial distribution with parameters n and p. Let Y have the binomial distribution with parameters n and p/k with k > 1. Let \(Z = kY\).
a. Show that X and Z have the same mean.
b. Find the variances of X and Z. Show that, if p is small, then the variance of Z is approximately k times as large as the variance of X.
c. Show why the results above explain the higher variability in the bar heights in Fig. 6.2 compared to Fig. 6.1.
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 ½, the next three have length 1/3, etc.
Show that, for every \(x \in \left[ {0,1} \right),\;{h_n}\left( x \right) = 1\) for one and only one n among \(1 + \frac{{\left( {{k_n} - 1} \right){k_n}}}{2},...,\frac{{{k_n}\left( {{k_n} + 1} \right)}}{2}\)
Suppose that three girls, A, B, and C throw snowballs at a target. Suppose also that girl A throws 10 times, and the probability that she will hit the target on any given throw is 0.3; girl B throws 15 times, and the probability that she will hit the target on any given throw is 0.2, and girl C throws 20 times, and the probability that she will hit the target on any given throw is 0.1. Determine the probability that the target will be hit at least 12 times
In this exercise, we construct an example of a sequence of random variables\({{\bf{Z}}_{\bf{n}}}\)that\({{\bf{Z}}_{\bf{n}}}\)converges to 0 with probability 1 but\({{\bf{Z}}_{\bf{n}}}\)fails to converge to 0 in a quadratic mean. Let X be a random variable with a uniform interval distribution [0, 1]. Define the sequence\({{\bf{Z}}_{\bf{n}}}\)by\({{\bf{Z}}_{\bf{n}}}{\bf{ = }}{{\bf{n}}^{\bf{2}}}\)if\({\bf{0 < X < }}{\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/{\vphantom {{\bf{1}} {\bf{n}}}}\right.}\!\lower0.7ex\hbox{\({\bf{n}}\)}}\) and\({{\bf{Z}}_{\bf{n}}}{\bf{ = 0}}\)otherwise.
a. Prove that\({{\bf{Z}}_{\bf{n}}}\)converges to 0 with probability 1.
b. Prove that\({{\bf{Z}}_{\bf{n}}}\)it does not converge to 0 in quadratic mean.
What do you think about this solution?
We value your feedback to improve our textbook solutions.
