Question

Over 500 million tweets are sent per day (Digital Marketing Ramblings website, December 15, 2014). Assume that the number of tweets per hour follows a Poisson distribution and that Bob receives on average 7 tweets during his lunch hour.

a. What is the probability that Bob receives no tweets during his lunch hour (to 4 decimals)?

b. What is the probability that Bob receives at least 4 tweets during his lunch hour (to 4 decimals)?

c. What is the expected number of tweets Bob receives during the first 30 minutes of his lunch hour (to 1 decimal)?

d. What is the probability that Bob receives no tweets during the first 30 minutes of his lunch hour (to 4 decimals)?

Answer 1

Concepts and reason

Poisson distribution: The Poisson distribution is a discrete probability distribution. This distribution is used when the number of occurrences for the given event is within a fixed interval of space or time.

Requirements for the Poisson probability distribution:

• The occurrences should be random.

• The occurrences should be independent to each other.

• For the given interval, the occurrences must be uniformly distributed.

Fundamentals

The Poisson probability distribution is,

$P\left( {X = x|\lambda } \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\,\,\,;\,\,X = 0,1....$

Where,

$\lambda$ : Parameter of the distribution.

x: Number of successes.

(a)

The probability that Bob receives no tweets during his lunch hour is obtained below:

Let X be a random variable that denote the number of tweets per hour which follows the Poisson distribution. From the information, the average tweet per hour is 7. That is $\lambda = 7$ .

The required probability is,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 7}}{{\left( 7 \right)}^0}}}{{0!}}\\\\ = \frac{{0.0009 \times 1}}{1}\\\\ = 0.0009\\\end{array}$

(b)

The probability that Bob receives at least 4 tweets during his lunch hour is obtained below:

Let X be a random variable that denote the number of tweets per hour which follows the Poisson distribution. From the information, the average tweet per hour is 7. That is $\lambda = 7$ .

The required probability is,

$\begin{array}{c}\\P\left( {X \ge 4} \right) = 1 - P\left( {X < 4} \right)\\\\ = 1 - \left[ {P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right)} \right]\\\\ = 1 - \left[ {\frac{{{e^{ - 7}}{{\left( 7 \right)}^0}}}{{0!}} + \frac{{{e^{ - 7}}{{\left( 7 \right)}^1}}}{{1!}} + \frac{{{e^{ - 7}}{{\left( 7 \right)}^2}}}{{2!}} + \frac{{{e^{ - 7}}{{\left( 7 \right)}^3}}}{{3!}}} \right]\\\\ = 1 - \left[ {0.0009 + 0.0064 + 0.0223 + 0.0521} \right]\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.0817\\\\ = 0.9183\\\end{array}$

(c)

The expected number of tweets Bob receives during the first 30 minutes of his lunch hour is obtained below:

Let X be a random variable that denote the number of tweets Bob receives during the first 30 minutes of his lunch hour which follows the Poisson distribution. From the information, the average tweet per hour is 7. That is $\lambda = 7$ .

The required expected number is,

$\begin{array}{c}\\E\left( X \right) = \frac{\lambda }{n}\\\\ = \frac{7}{2}\\\\ = 3.5\\\end{array}$

(d)

The probability that Bob receives no tweets during the first 30 minutes of his lunch hour is obtained below:

Let X be a random variable that denote the number of tweets per 30 minutes which follows the Poisson distribution. From the information, the average tweet per 30 minutes is 3.5. That is $\lambda = 3.5$ .

The required probability is,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 3.5}}{{\left( {3.5} \right)}^0}}}{{0!}}\\\\ = \frac{{0.0302 \times 1}}{1}\\\\ = 0.0302\\\end{array}$

Ans: Part a

Thus, the probability that Bob receives no tweets during his lunch hour is 0.0009.

Part b

Thus, the probability that Bob receives at least 4 tweets during his lunch hour is 0.9183.

Part c

Thus, the expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

Part d

Thus, the probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.