Question

1) Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number X has a Poisson distribution with parameter ? = 0.14.

(a) What is the probability that a disk has exactly one missing pulse? (Round to four decimal places)

(b) What is the probability that a disk has at least two missing pulses? (Round to four decimal places)

(c) If two disks are independently selected, what is the probability that neither contains a missing pulse?(Round to four decimal places)

The National Center for Health Statistics estimates that 16.9 percent of U.S. children and adolescents aged 2 to 19 were obese in 2009-2010. Assume that this obesity rate holds true today. If you were to randomly select 16 U.S. children or adolescents, find each of the following:
Note1: carry at least 6 digit precision for any intermediate calculations then round your answer to 4 decimal places
Note2: Any value that is less than 0.0001, (1 x 10^-4) , can be rounded to 0

The probability that at least 2 people in the sample will be obese is: _______________

Answer 1

Concepts and reason

The concept of Poisson and Binomial distribution are used to solve this problem.

Poisson distribution is a distribution which calculates the probability of the occurrence of a particular number of events in a particular number of trials.

The binomial distribution describes the probability of the number of successes and failures.

Fundamentals

The binomial distribution is the distribution that calculates the number of successes. The probability of obtaining x success in n trials in a binomial experiment is formulated by,

$P\left( {X = x} \right) = \left( {_x^n} \right){p^x}{\left( {1 - p} \right)^{n - x}}$

Here, p is the probability of successes.

The probability mass function of a Poisson distribution is defined as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}},{\rm{ }}x = 0,1,2, \cdot \cdot \cdot$

Here, x is a Poisson variate, and$\lambda$is a Poisson parameter. The mean and variance of Poisson distribution are equal to Poisson parameter$\lambda$.

Poisson distribution is the limiting case of Binomial distribution.

(1.a)

Let X be the number of missing pulses and follows Poisson distribution with $\lambda =$0.14. The probability that a disk has exactly one missing pulse is calculated as,

$\begin{array}{c}\\P\left( {X = 1} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\\\\ = \frac{{{e^{ - 0.14}} \times {{\left( {0.14} \right)}^1}}}{{1!}}\\\\ = 0.1217\\\end{array}$

(1.b)

Let X be the number of missing pulses and follows Poisson distribution with $\lambda =$0.14. The probability that a disk has at least two missing pulse is calculated as,

$\begin{array}{c}\\P\left( {X \ge 2} \right) = 1 - P\left( {X < 2} \right)\\\\ = 1 - \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right)} \right)\\\\ = 1 - \left( {\frac{{{e^{ - 0.14}}{{\left( {0.14} \right)}^0}}}{{0!}} + \frac{{{e^{ - 0.14}}{{\left( {0.14} \right)}^1}}}{{1!}}} \right)\\\\ = 1 - \left( {0.8693 + 0.1217} \right)\\\end{array}$

$= 0.009$

(1.c)

Let X be the number of missing pulses and follows Poisson distribution with $\lambda =$0.14. The probability that two independently selected disks do not contains missing pulse is calculated as,

$\begin{array}{c}\\P\left( {{\rm{two disks with no missing pulse}}} \right) = \left( {P\left( {X = 0} \right) \times P\left( {X = 0} \right)} \right)\\\\ = \left( {\frac{{{e^{ - 0.14}}{{\left( {0.14} \right)}^0}}}{{0!}} \times \frac{{{e^{ - 0.14}}{{\left( {0.14} \right)}^0}}}{{0!}}} \right)\\\\ = \left( {0.8693 \times 0.8693} \right)\\\\ = 0.7556\\\end{array}$

Let X denotes the number of obese people in 2009-2010 that follows binomial distribution. The probability of success$\left( p \right)$ that is 16.9% of the U.S. children aged 2 to 19 are obese. If 16 U.S. children$\left( n \right)$ are randomly selected, the probability that at least 2 people in the sample will be obese is calculated as,

$\begin{array}{c}\\P\left( {X \ge 2} \right) = 1 - P\left( {X < 2} \right)\\\\ = 1 - P\left( {X = 0} \right) + P\left( {X = 1} \right)\\\\ = 1 - \left( {\left( {_0^{16}} \right) \times 0.{{\left( {169} \right)}^0} \times {{\left( {0.831} \right)}^{16}}} \right) + \left( {\left( {_1^{16}} \right) \times {{\left( {0.169} \right)}^1} \times {{\left( {0.831} \right)}^{15}}} \right)\\\\ = 1 - \left( {0.0517 + 0.1682} \right)\\\end{array}$

$= 0.7801$

Ans: Part 1.a

The probability that a disk has exactly one missing pulse is calculated as 0.1217.

Part 1.b

The probability that a disk has at least two missing pulse is calculated as 0.009.

Part 1.c

The probability that two selected disks have no missing pulses is calculated as 0.7556.

The probability that at least 2 people in the sample will be obese is 0.7801.