Question

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of 2014, Google’s Chrome browser exceeded a 20% market share for the first time, with a 20.37% share of the browser market (Forbes website). For a randomly selected group of 20 Internet browser users, answer the following questions.

(a) Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.
(b) Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.
(c) For the sample of 20 Internet browser users, compute the expected number of Chrome users.
(d) For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

Answer 1

Answer:

a) 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) 80.5% probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

c) 4.074

d) Variance 3.24, standard deviation 1.8

Step-by-step explanation:

For each internet browser user, there are only two possible outcomes. Either they use chrome, or they do not. They are chosen at random, which means that the probability of an user using chrome is independent from other users. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

20.37% share of the browser market

This means that $p = 0.2037$

Group of 20 Internet browser users

This means that $n = 20$

(a) Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is $P(X = 8)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243$

2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

(b) Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

Either less than 3 users use Chrome, or at least 3 do. The sum of the probabilities of these events is decimal 1. So

$P(X < 3) + P(X \geq 3) = 1$

So

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105$

$P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538$

$P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.195$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.195 = 0.805$

80.5% probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

(c) For the sample of 20 Internet browser users, compute the expected number of Chrome users.

$E(X) = np = 20*0.2037 = 4.074$

(d) For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

$V(X) = np(1-p)$

$V(X) = 20*0.2037*0.7963 = 3.24$

The standard deviation is the square root of the variance. SO

$\sqrt{V(X)} = \sqrt{3.24} = 1.8$