20 Normal Approximation to Poisson Assume that the number of asbestos particles in a squared

Example 4-20 Normal Approximation to Poisson Assume that the number of asbestos particles in a squared

  meter of dust on a surface follows a Poisson distribution with a mean of 1000. If a squared meter

  of dust is analyzed, what is the probability that 950 or fewer particles are found?

  This probability can be expressed exactly as

  PX ( ≤ 950 ) = ∑

  e x − 1000

  = 0 x !

  The computational diffi culty is clear. The probability can be approximated as

  ( ⎞

  ≤ 950 ) = ( ≤ 950 5 . ) ≈ P Z ≤

  ⎝⎜

  1000 ⎠⎟

  Practical Interpretation: Poisson probabilities that are diffi cult to compute exactly can be approximated with easy- to-compute probabilities based on the normal distribution.

  FIGURE 4-21 Conditions for approximating hypergeometric and binomial probabilities.

  Chapter 4Continuous Random Variables and Probability Distributions

  EXERCISES

  FOR SECTION 4-7

  Problem available in WileyPLUS at instructor’s discretion.

  Tutoring problem available in WileyPLUS at instructor’s discretion.

  4-95.

  Suppose that X is a binomial random variable with

  4-103.

  Suppose that the number of asbestos particles in

  n = 200 and p = . . 0 4 Approximate the following probabilities:

  a sample of 1 squared centimeter of dust is a Poisson random

  (a) PX ( ≤ 70 (b) P(70 < X < 90) (c) P(X = 80) )

  variable with a mean of 1000. What is the probability that 10

  4-96.

  Suppose that X is a Poisson random variable

  squared centimeters of dust contains more than 10,000 particles?

  with λ = 6.

  4-104.

  A high-volume printer produces minor print-quality

  (a) Compute the exact probability that

  X is less than four.

  errors on a test pattern of 1000 pages of text according to a Pois-

  (b) Approximate the probability that

  X is less than four and

  son distribution with a mean of 0.4 per page.

  compare to the result in part (a).

  (a) Why are the numbers of errors on each page independent

  (c) Approximate the probability that 8 < X 12 .

  random variables?

  4-97.

  Suppose that X has a Poisson distribution with a

  (b) What is the mean number of pages with errors (one or more)?

  mean of 64. Approximate the following probabilities:

  (c) Approximate the probability that more than 350 pages con-

  (a) P X > 72 ( ) (b) P X < 64 ( ) (c) P ( 60 < X ≤ 68 ) tain errors (one or more).

  4-105. Hits to a high-volume Web site are assumed to follow

  4-98.

  The manufacturing of semiconductor chips pro-

  a Poisson distribution with a mean of 10,000 per day. Approxi-

  duces 2 defective chips. Assume that the chips are inde-

  mate each of the following:

  pendent and that a lot contains 1000 chips. Approximate the

  (a) Probability of more than 20,000 hits in a day

  following probabilities:

  (b) Probability of less than 90 hits in a day

  (a) More than 25 chips are defective.

  (c) Value such that the probability that the number of hits in a

  (b) Between 20 and 30 chips are defective.

  day exceeds the value is 0.01

  4-99.

  There were 49.7 million people with some type of

  (d) Expected number of days in a year (365 days) that exceed

  long-lasting condition or disability living in the United States

  10,200 hits.

  in 2000. This represented 19.3 percent of the majority of civil-

  (e) Probability that over a year (365 days), each of the more

  ians aged five and over (http:factfinder.census.gov). A sample

  than 15 days has more than 10,200 hits.

  of 1000 persons is selected at random.

  4-106. An acticle in Biometrics [“Integrative Analysis of

  (a) Approximate the probability that more than 200 persons in

  Transcriptomic and Proteomic Data of Desulfovibrio Vulgaris:

  the sample have a disability.

  A Nonlinear Model to Predict Abundance of Undetected Pro-

  (b) Approximate the probability that between 180 and 300

  teins” (2009)] reported that protein abundance from an operon

  people in the sample have a disability. 4-100. (a set of biologically related genes) was less dispersed than

  Phoenix water is provided to approximately 1.4 million

  from randomly selected genes. In the research, 1000 sets of

  people who are served through more than 362,000 accounts (http:

  genes were randomly constructed, and of these sets, 75 were

  phoenix.govWATERwtrfacts.html). All accounts are metered and

  more disperse than a specific opteron. If the probability that

  billed monthly. The probability that an account has an error in a

  a random set is more disperse than this opteron is truly 0.5,

  month is 0.001, and accounts can be assumed to be independent.

  approximate the probability that 750 or more random sets

  (a) What are the mean and standard deviation of the number of

  exceed the opteron. From this result, what do you conclude

  account errors each month?

  about the dispersion in the opteron versus random genes?

  (b) Approximate the probability of fewer than 350 errors in a month.

  4-107. An article in Atmospheric Chemistry and Physics

  (c) Approximate a value so that the probability that the number

  [“Relationship Between Particulate Matter and Childhood Asthma

  of errors exceeds this value is 0.05.

  – Basis of a Future Warning System for Central Phoenix,” 2012,

  (d) Approximate the probability of more than 400 errors per

  Vol. 12, pp. 2479-2490] linked air quality to childhood asthma

  month in the next two months. Assume that results between

  incidents. The study region in central Phoenix, Arizona recorded

  months are independent.

  10,500 asthma incidents in children in a 21-month period. Assume

  4-101.

  An electronic office product contains 5000 elec-

  that the number of asthma incidents follows a Poisson distribution.

  tronic components. Assume that the probability that each compo-

  (a) Approximate the probability of more than 550 asthma inci-

  nent operates without failure during the useful life of the product

  dents in a month.

  is 0.999, and assume that the components fail independently.

  (b) Approximate the probability of 450 to 550 asthma inci-

  Approximate the probability that 10 or more of the original 5000

  dents in a month.

  components fail during the useful life of the product.

  (c) Approximate the number of asthma incidents exceeded

  4-102.

  A corporate Web site contains errors on 50 of 1000

  with probability 5.

  pages. If 100 pages are sampled randomly without replace-

  (d) If the number of asthma incidents was greater during the

  ment, approximate the probability that at least one of the pages

  winter than the summer, what would this imply about the

  in error is in the sample.

  Poisson distribution assumption?

  Section 4-8Exponential Distribution

  4-108.

  A set of 200 independent patients take antiacid medi-

  (b) Approximate the probability that more than 65 cabs pass

  cation at the start of symptoms, and 80 experience moderate

  within a 10-hour day.

  to substantial relief within 90 minutes. Historically, 30 of

  (c) Approximate the probability that between 50 and 65 cabs

  patients experience relief within 90 minutes with no medi-

  pass in a 10-hour day.

  cation. If the medication has no effect, approximate the

  (d) Determine the mean hourly rate so that the probability

  probability that 80 or more patients experience relief of symp-

  is approximately 0.95 that 100 or more cabs pass in a

  toms. What can you conclude about the effectiveness of this

  10-hour data.

  medication?

  4-111. The number of (large) inclusions in cast iron follows

  4-109. a Poisson distribution with a mean of 2.5 per cubic millimeter.

  Among homeowners in a metropolitan area, 75 recy-

  Approximate the following probabilities:

  cle plastic bottles each week. A waste management company

  (a) Determine the mean and standard deviation of the number

  services 1500 homeowners (assumed independent). Approxi-

  of inclusions in a cubic centimeter (cc).

  mate the following probabilities:

  (b) Approximate the probability that fewer than 2600 inclu-

  (a) At least 1150 recycle plastic bottles in a week

  sions occur in a cc.

  (b) Between 1075 and 1175 recycle plastic bottles in a week

  (c) Approximate the probability that more than 2400 inclu-

  4-110. Cabs pass your workplace according to a Poisson pro-

  sions occur in a cc.

  cess with a mean of five cabs per hour.

  (d) Determine the mean number of inclusions per cubic mil-

  (a) Determine the mean and standard deviation of the number

  limeter such that the probability is approximately 0.9 that

  of cabs per 10-hour day.

  500 or fewer inclusions occur in a cc.

  4-8 Exponential Distribution

  The discussion of the Poisson distribution defined a random variable to be the number of flaws along a length of copper wire. The distance between flaws is another random variable that is often of interest. Let the random variable X denote the length from any starting point on the wire until

  a flaw is detected. As you might expect, the distribution of X can be obtained from knowledge of the distribution of the number of flaws. The key to the relationship is the following concept. The distance to the first flaw exceeds three millimeters if and only if there are no flaws within a length of three millimeters—simple but sufficient for an analysis of the distribution of X.

  In general, let the random variable N denote the number of flaws in x millimeters of wire. If the mean number of flaws is λ per millimeter, N has a Poisson distribution with mean λx. We assume that the wire is longer than the value of x. Now

  −λ e x

  () λ x

  PX>x = PN 0 =

  Fx () PX

  = −λ ( ≤ x ) =− 1 e , x ≥ 0

  is the cumulative distribution function of X. By differentiating F x () , the probability density function of X is calculated to be

  fx x () =λ

  e −λ ,x ≥ 0

  The derivation of the distribution of X depends only on the assumption that the flaws in the wire follow a Poisson process . Also, the starting point for measuring X does not matter because the probability of the number of flaws in an interval of a Poisson process depends only on the length of the interval, not on the location. For any Poisson process, the following general result applies.

  Exponential Distribution

  The random variable X that equals the distance between successive events from a

  Poisson process with mean number of events λ > 0 per unit interval is an exponential

  random variable with parameter λ. The probability density function of X is

  fx

  e −λ () x =λ for 0 ≤ x< ∞ (4-14)

  The exponential distribution obtains its name from the exponential function in the prob- ability density function. See plots of the exponential distribution for selected values of λ in

  Chapter 4Continuous Random Variables and Probability Distributions

  FIGURE 4-22 Probability density function of exponential

  FIGURE 4-23 Probability for the exponential

  random variables for selected values of λ.

  distribution in Example 4-21.

  Fig. 4-22. For any value of λ, the exponential distribution is quite skewed. The following results are easily obtained and are left as an exercise.

  Mean and Variance

  If the random variable X has an exponential distribution with parameter λ,

  μ= EX () 2 = and σ= VX () = (4-15)

  λ

  λ

  It is important to use consistent units to express intervals, X, and λ. The following exam- ple illustrates unit conversions.