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Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of 2014, Googles Chrome browser exceeded a 20% market share for the first time, with a 20.37% share of the browser market (Forbes website, December 15, 2014). 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 c) For the sample of 20 Internet browser users, compute the expected number of d) For the sample of 20 Internet browser users, compute the variance and standard Chrome as their Internet browser. Chrome users. deviation for the number of Chrome users.

math   Statistics-And-Probability   
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The probability that a randomly selected  internet browser users uses Chrome as their internet browser is 0.2037

Let X be the number of Chrome users among a randomly selected sample of 20 internet browser users. X has a Binomial distribution with parameters, number of trials, n=20 and success probability p=0.2037.

the probability that X=x users use Chrome in a sample of 20 internet browser users is given by

egin{align*} P(X=x)&=inom{n}{x}p^x(1-p)^{n-x} &=inom{20}{x}0.2037^x(1-0.2037)^{20-x} &=rac{20!}{x!(20-x)!}0.2037^x(1-0.2037)^{20-x} end{align*}

a) The probability that exactly 8 out of the 20 internet browser users use Chrome as their internet browser is

egin{align*} P(X=8)&=rac{20!}{8!(20-8)!}0.2037^8(1-0.2037)^{20-8} &=rac{20 imes 19 imes ... imes 1}{(8 imes 7 imes ... imes 1)(12 imes 11 imes ... imes 1)}0.2037^8(0.7963)^{12} &=0.0243 end{align*}

b) The probability that at least 3 out of the 20 internet browser users use Chrome as their internet browser is

P(X > 3) = 1-P(X < 3) = 1-(P(X = 0) + P(x = 1) + P(X = 2)) -1-( 0!(22-W + İ (20-1) 0 20371 (1-02037)2 201 -- -020370(1-0.2037)20-0 0.20371 (1-0.2037)20-1 + 0.20372(1 -0.2037)20-2 2) + 0!(20 0)! 2!(20 - 2)! / 21 20 1 (0.01050.05380.1307) 1-0, 1949 0.8051

c) The expected number of X is (using the standard results for Binomial distribution)

E(X) = np= 20 × 0.2037= 4.074

ans: For the sample of 20 internet browser users, the expected number of Chrome users is 4.074

d) The Variance of X is (using the standard results for Binomial distribution)

Var(X) = np(1-p) 20 × 0.2037 × (1-0.2037) 3.2441

The standard deviation of X is

SD(X) = V/Var(X) = V3.2441 = 1.8011

ans: For the sample of 20 internet browser users, the Variance for number of Chrome users is 3.2441 and standard deviation is 1.8011

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