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

Answer:

(a) The probability that Bob receives no tweets during his lunch hour is 0.0002.

(b) The probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c) The expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

(d) The probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.

Step-by-step explanation:

Let X = number of tweets.

The random variable X follows a Poisson distribution with parameter λ = 7.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-7}(7)^{x}}{x!};\ x=0, 1, 2, 3...$

(a)

Compute the probability that Bob receives no tweets during his lunch hour as follows:

$P(X=0)=\frac{e^{-7}(7)^{0}}{0!}\\=\frac{0.000192\times1}{1} \\=0.000192\\\approx0.0002$

Thus, the probability that Bob receives no tweets during his lunch hour is 0.0002.

(b)

Compute the probability that Bob receives at least 4 tweets during his lunch hour as follows:

P (X ≥ 4) = 1 - P (X < 4)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

              $=1-\frac{e^{-7}(7)^{0}}{0!}-\frac{e^{-7}(7)^{1}}{1!}-\frac{e^{-7}(7)^{2}}{2!} - \frac{e^{-7}(7)^{3}}{3!}\\=1-0.0002-0.0064-0.0223-0.0521\\=0.9190$

Thus, the probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c)

The average number of tweets in 60 minutes is 7.

Then the average number of tweets in 1 minute is, $\frac{7}{60}$.

Hence, the average number of tweets during 30 minutes is, $\frac{7}{60}\times30=3.5$

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

(d)

Compute the probability that Bob receives no tweets during the first 30 minutes of his lunch hour as follows:

$P(No\ tweets) = \frac{e^{-3.5}(3.5)^{0}}{0!}=\frac{0.0302\times1}{1} =0.0302$

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