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Chapter 4: Discrete Random Variables

4.6 Poisson Distribution

Learning Objectives

By the end of this section, the student should be able to:

  • Identify the components of a Poisson experiment
  • Use the formulas for a Poisson random variable to compute the mean, variance, and standard deviation

There are two main characteristics of a Poisson experiment.

  1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages.
  2. The number of events occurring has a constant average and the events are independent of the “time” since the last event. Continuing the book example, the average number of misspelled words is 5, and finding a word that is misspelled does not influence the page on which the next misspelled word lies. Additionally, two events shouldn’t occur simultaneously.

The Poisson distribution is frequently used for experiments where there are a large number of possible but rare events. For instance, the Poisson distribution may be used to approximate the binomial if the probability of success is “small” (such as 0.01) and the number of trials is “large” (such as 1,000). You will verify the relationship in the homework exercises.

The random variable [latex]X =[/latex]the number of occurrences in the interval of interest. Because the average per interval is constant, we can find the average for any size interval by scaling to the size of the interval.

Example

The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes, selected randomly. The time interval of interest is five minutes. What is the probability that the number of loaves put on the shelf in five minutes is three?

Let [latex]X =[/latex] the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is [latex]\left(\frac{5}{30}\right)[/latex](12) = 2 loaves of bread.

The probability question asks you to find [latex]P(X = 3)[/latex].

Your Turn!

You notice that a news reporter says “uh,” on average, two times per broadcast. What is the probability that the news reporter says “uh” more than two times per broadcast.

This is a Poisson problem because you are interested in knowing the number of times the news reporter says “uh” during a broadcast.

a. What is the interval of interest?

Solution

one broadcast

 

b. What is the average number of times the news reporter says “uh” during one broadcast?

Solution

2

 

c. Let [latex]X = \underline{\hspace{2cm}}[/latex]. What values does X take on?

Solution

Let [latex]X =[/latex] the number of times the news reporter says “uh” during one broadcast.

[latex]x = 0, 1, 2, 3, ...[/latex]

 

d. The probability question is [latex]P( \underline{\hspace{2cm}} )[/latex].

Solution

[latex]P(x > 2)[/latex]

Your Turn!

An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reasons why this would be a Poisson distribution.

Solution

This problem wants to find the probability of events occurring in a fixed interval of time with a known average rate (5 per hour). The events are independent.

Your Turn!

The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. What is the average number of fish caught in 15 minutes?

Solution

[latex]\left( \frac{15}{60} \right) (8) = 2 \text{ fish}[/latex]

Notation and Mean

The notation for the Poisson Probability Distribution function is [latex]P[/latex], and we denote “[latex]X[/latex] is a random variable with a Poisson probability distribution as [latex]X \sim P(\lambda)[/latex]. The parameter is [latex]\lambda[/latex], which is also the mean [latex]\mu[/latex]. So, we may also write [latex]X \sim P(\mu)[/latex].

The mean is given as part of a Poisson experiment, [latex]\mu = \lambda[/latex].

The formula for the variance is [latex]\sigma^2 = \lambda = \mu[/latex], that is, the average is also the variance of a Poisson random variable.

The formula for the standard deviation is [latex]\sigma = \sqrt{\lambda}[/latex].

Note

To compute Poisson probabilities on a graphing calculator, go into 2nd DISTR. The syntax for the instructions are as follows. The TI calculators use [latex]\lambda[/latex] (lambda) for the mean.

To calculate [latex]P(X = x)[/latex]: poissonpdf([latex]\lambda[/latex], x).

To calculate [latex]P(X \leq x)[/latex]: poissoncdf([latex]\lambda[/latex], x).

If you wanted to instead find [latex]P(X>x)[/latex], use 1 – poissoncdf([latex]\lambda,x[/latex]).

In Excel, both Poisson probabilities are computed using POISSON.DIST(x, [latex]\lambda[/latex], True/False), where False computes [latex]P(X=x)[/latex] and is equivalent to poissonpdf, and True computes [latex]P(X \leq x)[/latex] and is equivalent to poissoncdf.

Example

Leah’s school email receives an average of six emails between 8 a.m. and 10 a.m. each day. What is the probability that Leah receives more than one email in the next 15 minutes?

Let [latex]X =[/latex] the number of emails Leah receives in 15 minutes. (The interval of interest is 15 minutes or [latex]\frac{1}{4}[/latex] hour.)

[latex]x = 0, 1, 2, 3, ...[/latex]

If Leah receives, on the average, six emails in two hours, and there are eight 15 minute intervals in two hours, then Leah receives

[latex]\left(\frac{1}{8}\right)(6) = 0.75[/latex] emails in 15 minutes, on average. So, [latex]\lambda = \mu = 0.75[/latex] for this problem, and [latex]X \sim P(0.75)[/latex].

Find [latex]P(X > 1)[/latex]: (calculator or computer)

  • Press 1 – and then press 2nd DISTR.
  • Arrow down to poissoncdf(. Press ENTER.
  • Enter .75,1).
  • The result is [latex]P(X > 1) = 0.1734.[/latex]

The probability that Leah receives more than one email in the next 15 minutes is about 0.1734:

[latex]P(X > 1) = 1 - poissoncdf(0.75, 1) = 0.1734[/latex].

The graph of [latex]X \sim P(0.75)[/latex] is:

This graph shows a Poisson probability distribution. It has 5 bars that decrease in height from left to right. The x-axis shows values in increments of 1 starting with 0, representing the number of calls Leah receives within 15 minutes. The y-axis ranges from 0 to 0.5 in increments of 0.1.

The y-axis contains the probability of x where X = the number of emails in 15 minutes.

Your Turn!

Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,500 arrivals and departures each day.

  1. How many airplanes arrive and depart the airport per hour?
  2. What is the probability that there are exactly 100 arrivals and departures in one hour?
  3. What is the probability that there are at most 100 arrivals and departures in one hour?
Solution
  1. Let [latex]X =[/latex] the number of airplanes arriving and departing from Hartsfield-Jackson in one hour. The average number of arrivals and departures per hour is [latex]\frac{2,500}{24} \approx 104.1667[/latex].
  2. [latex]X \sim P(104.1667) , \text{ so } P(X=100)=\text{poissonpdf}(104.1667,100)≈0.0366[/latex]
  3. [latex]P(X \le 100)= \text{poissoncdf} (104.1667,100) \approx 0.3651[/latex]

 

Example

According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let [latex]X=[/latex] the number of emails an email user receives per day. The discrete random variable [latex]X[/latex] takes on the values [latex]x= 0, 1, 2, \ldots[/latex]. The random variable [latex]X[/latex] has a Poisson distribution: [latex]X \sim P(147)[/latex]. The mean is 147 emails.

  1. What is the probability that an email user receives exactly 160 emails per day?
  2. What is the probability that an email user receives at most 160 emails per day?
  3. What is the standard deviation?
Solution
  1. [latex]P(x = 160)[/latex] = poissonpdf(147, 160) ≈ 0.0180
  2. [latex]P(x \leq 160)[/latex] = poissoncdf(147, 160) ≈ 0.8666
  3. Standard Deviation = [latex]\sigma =\sqrt{\mu }=\sqrt{147}\approx 12.1244[/latex]

Your Turn!

According to a recent poll by the Pew Internet Project, girls between the ages of 14 and 17 send an average of 187 text messages each day. Let [latex]X =[/latex]the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable [latex]X[/latex] takes on the values [latex]x = 0, 1, 2 …[/latex]. The random variable [latex]X[/latex] has a Poisson distribution: [latex]X \sim P(187)[/latex]. The mean is 187 text messages.

  1. What is the probability that a teen girl sends exactly 175 texts per day?
  2. What is the probability that a teen girl sends at most 150 texts per day?
  3. What is the standard deviation?
Solution
  1. [latex]P(X=175)=\text{poissonpdf}(187,175) \approx 0.0203[/latex]
  2. [latex]P(X\le150)= \text{poissoncdf} (187,150) \approx 0.0030[/latex]
  3. [latex]\text{Standard Deviation} = \sigma=\sqrt{\mu} = \sqrt{187} \approx 13.6748[/latex]

    The Poisson distribution can be used to approximate probabilities for a binomial distribution. The Poisson approximation to a binomial distribution was commonly used in the days before technology made both values very easy to calculate.

    Let [latex]n[/latex] represent the number of binomial trials and let [latex]p[/latex] represent the probability of a success for each trial. If [latex]n[/latex] is large enough and [latex]p[/latex] is small enough then the Poisson approximates the binomial very well. In general, [latex]n[/latex] is considered “large enough” if it is greater than or equal to 20. The probability [latex]p[/latex] from the binomial distribution should be less than or equal to 0.05. When the Poisson is used to approximate the binomial, we use the binomial mean [latex]\mu = np[/latex].

    Example

    On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they similar?

    Your Turn!

    On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

    Section 4.6 Review

    A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is “small” (less than or equal to 0.05) and the number of trials is “large” (greater than or equal to 20).

    Formula Review

    [latex]X \sim P(\mu)[/latex] means that [latex]X[/latex] has a Poisson probability distribution where [latex]X= \text{the number of occurrences in the interval of interest}[/latex].

    [latex]X[/latex] takes on the values [latex]x = 0, 1, 2, 3, \ldots[/latex]

    The mean [latex]\mu[/latex] is typically given.

    The variance is [latex]\sigma^2 = \mu[/latex], and the standard deviation is

    [latex]\sigma \text{ = }\sqrt{\mu }[/latex].

    When [latex]P(\mu)[/latex] is used to approximate a binomial distribution, [latex]\mu = np[/latex] where [latex]n[/latex] represents the number of independent trials and [latex]p[/latex] represents the probability of success in a single trial.

    Section 4.6 Practice

    Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day.

    1. Assume the event occurs independently on any given day. Define the random variable [latex]X[/latex].

    Solution

    [latex]X \sim P(120)[/latex]

     

    2. What values does [latex]X[/latex] take on?

    Solution

    [latex]0, 1, 2, 3, 4, …[/latex]

     

    3. What is the probability of getting 150 customers in one day?

    Solution

    0.0010

     

    4. What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.

    Solution

    0.0485

     

    5. What is the probability that the store will have more than 12 customers in the first hour?

    Solution

    0.2084

     

    6. What is the probability that the store will have fewer than 12 customers in the first two hours?

    Solution

    0.0214

    Which type of distribution can the Poisson model be used to approximate? When would you do this?

    Solution

    The Poisson distribution can approximate a binomial distribution, which you would do if the probability of success is small and the number of trials is large.

    Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age.

    1. Assume the event occurs independently on any given day. In words, define the random variable [latex]X[/latex].

    Solution

    [latex]X =[/latex] the number of U.S. teens who die from motor vehicle injuries per day.

     

    2. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]

    Solution

    [latex]P(8)[/latex]

     

    3. What values does [latex]X[/latex] take on?

    Solution

    [latex]0, 1, 2, 3, 4, ...[/latex]

     

    4. For the given values of the random variable X, fill in the corresponding probabilities.

     

    5. Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.

    Solution

    No

     

    6. Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.

    Solution

    No

    The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let [latex]X =[/latex] the number of calls received at noon.

    1. Find the mean and standard deviation of [latex]X[/latex].
    2. What is the probability that the office receives at most six calls at noon on Monday?
    3. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon?
    4. What is the probability that the office receives more than eight calls at noon?
    Solution
    1. [latex]X \sim P(5.5); \mu = 5.5; \sigma = \sqrt{5.5} \approx 2.3452[/latex]
    2. [latex]P(x \le 6) = \text{poissoncdf} (5.5, 6) \approx 0.6860[/latex]
    3. There is a 15.7% probability that the law staff will receive more calls than they can handle.
    4. [latex]P(x > 8) = 1 – P(x \le 8) = 1 – \text{poissoncdf} (5.5, 8) \approx 1 – 0.8944 = 0.1056[/latex]

    The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let [latex]X =[/latex] the number of births in an hour.

    1. Find the mean and standard deviation of [latex]X[/latex].
    2. Sketch a graph of the probability distribution of [latex]X[/latex].
    3. What is the probability that the maternity ward will deliver three babies in one hour?
    4. What is the probability that the maternity ward will deliver at most three babies in one hour?
    5. What is the probability that the maternity ward will deliver more than five babies in one hour?

    A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.

    Solution

    Let [latex]X =[/latex]the number of defective bulbs in a string.

    Using the Poisson distribution:

    [latex]\mu = np = 100(0.03) = 3[/latex]
    [latex]X \sim P(3)[/latex]
    [latex]P(x \le 4) = \text{poissoncdf} (3, 4) \approx 0.8153[/latex]

     

    Using the binomial distribution:

    [latex]X \sim B(100, 0.03)[/latex]
    [latex]P(x \le 4) = \text{binomcdf} (100, 0.03, 4) \approx 0.8179[/latex]

    The Poisson approximation is very good—the difference between the probabilities is only 0.0026.

    The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. Find the probability that she has no children.
    5. Find the probability that she has fewer children than the Japanese average.
    6. Find the probability that she has more children than the Japanese average.

    The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. Find the probability that she has no children.
    5. Find the probability that she has fewer children than the Spanish average.
    6. Find the probability that she has more children than the Spanish average.
    Solution
    1. [latex]X =[/latex] the number of children for a Spanish woman
    2. [latex]0, 1, 2, 3,...[/latex]
    3. [latex]X \sim P(1.47)[/latex]
    4. 0.2299
    5. 0.5679
    6. 0.4321

    Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces:

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}}[/latex]
    4. Find the probability that she has no litters in one year.
    5. Find the probability that she has at least two litters in one year.
    6. Find the probability that she has exactly three litters in one year.
    Solution
    1. [latex]X =[/latex] the number of litters a fertile, female cat produces per year
    2. [latex]0, 1, 2, 3, ...[/latex]
    3. [latex]X \sim P(3)[/latex]
    4. 0.0498
    5. 0.8009
    6. 0.2240

    The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. How many cookies do we expect to have an extra fortune?
    5. Find the probability that none of the cookies have an extra fortune.
    6. Find the probability that more than three have an extra fortune.
    7. As [latex]n[/latex] increases, what happens involving the probabilities using the two distributions? Explain in complete sentences.
    Solution
    1. [latex]X =[/latex] the number of fortune cookies that have an extra fortune
    2. [latex]0, 1, 2, 3,... 144[/latex]
    3. [latex]X \sim B(144, 0.03)[/latex] or [latex]P(4.32)[/latex]
    4. 4.32
    5. 0.0124 or 0.0133
    6. 0.6300 or 0.6264
    7. As [latex]n[/latex] gets larger, the probabilities get closer together.

    According to the South Carolina Department of Mental Health website, for every 200 U.S. women, the average number who suffer from anorexia is one. Out of a randomly chosen group of 600 U.S. women determine the following.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. How many are expected to suffer from anorexia?
    5. Find the probability that no one suffers from anorexia.
    6. Find the probability that more than four suffer from anorexia.
    Solution
    1. [latex]X =[/latex] the number of women that suffer from anorexia
    2. [latex]0, 1, 2, 3, ..., 600[/latex]
    3. [latex]X \sim P(3)[/latex]
    4. 3
    5. 0.0498
    6. 0.1847

    The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. Suppose that 100 people with tax returns over $25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to answer the following questions.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. How many are expected to be audited?
    5. Find the probability that no one was audited.
    6. Find the probability that at least three were audited.
    Solution
    1. [latex]X =[/latex] the number of people audited in one year
    2. [latex]0, 1, 2, ..., 100[/latex]
    3. [latex]X \sim P(2)[/latex]
    4. 2
    5. 0.1353
    6. 0.3233

    Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. How many seniors are expected to have participated in after-school sports all four years of high school?
    5. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
    6. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
    Solution
    1. [latex]X =[/latex] the number of seniors who participated in after-school sports all four years of high school
    2. [latex]0, 1, 2, 3, ... 60[/latex]
    3. [latex]X \sim P(4.8)[/latex]
    4. 4.8
    5. Yes
    6. 4

    On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes.

    1. In words, define the random variable [latex]X[/latex].
    2. List the values that [latex]X[/latex] may take on.
    3. Give the distribution of [latex]X[/latex]. [latex]X \sim \underline{\hspace{2cm}} ( \underline{\hspace{2cm}} , \underline{\hspace{2cm}} )[/latex]
    4. On average, how many pieces of egg shell do you expect to be in the cake?
    5. What is the probability that there will not be any pieces of egg shell in the cake?
    6. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes?
    7. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why?
    Solution
    1. [latex]X =[/latex] the number of shell pieces in one cake
    2. [latex]0, 1, 2, 3,...[/latex]
    3. [latex]X \sim P(1.5)[/latex]
    4. 1.5
    5. 0.2231
    6. 0.0001
    7. Yes

    Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week.

    1. In words, the random variable [latex]X = \underline{\hspace{2cm}}[/latex]

    a. the number of times Mrs. Plum’s cats wake her up each week.
    b. the number of times Mrs. Plum’s cats wake her up each hour.
    c. the number of times Mrs. Plum’s cats wake her up each night.
    d. the number of times Mrs. Plum’s cats wake her up.

    Solution

    a

     

    2. Find the probability that her cats will wake her up no more than five times next week.

    a. 0.5000
    b. 0.9329
    c. 0.0378
    d. 0.0671

    Solution

    d

    References

    “ATL Fact Sheet,” Department of Aviation at the Hartsfield-Jackson Atlanta International Airport, 2013. Available online at http://www.atlanta-airport.com/Airport/ATL/ATL_FactSheet.aspx (accessed May 15, 2013).

    Center for Disease Control and Prevention. “Teen Drivers: Fact Sheet,” Injury Prevention & Control: Motor Vehicle Safety, October 2, 2012. Available online at http://www.cdc.gov/Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html (accessed May 15, 2013).

    “Children and Childrearing,” Ministry of Health, Labour, and Welfare. Available online at http://www.mhlw.go.jp/english/policy/children/children-childrearing/index.html (accessed May 15, 2013).

    “Eating Disorder Statistics,” South Carolina Department of Mental Health, 2006. Available online at http://www.state.sc.us/dmh/anorexia/statistics.htm (accessed May 15, 2013).

    “Giving Birth in Manila: The maternity ward at the Dr Jose Fabella Memorial Hospital in Manila, the busiest in the Philippines, where there is an average of 60 births a day,” theguardian, 2013. Available online at http://www.theguardian.com/world/gallery/2011/jun/08/philippines-health#/?picture=375471900&index=2 (accessed May 15, 2013).

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    Lenhart, Amanda. “Teens, Smartphones & Testing: Texting volume is up while the frequency of voice calling is down. About one in four teens say they own smartphones,” Pew Internet, 2012. Available online at http://www.pewinternet.org/~/media/Files/Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf (accessed May 15, 2013).

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