Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of , Google's Chrome browser exceeded a market share for the first time, with a share of the browser market (Forbes website, December ). For a randomly selected group of Internet browser users, answer the following questions. a. Compute the probability that exactly of the Internet browser users use Chrome as their Internet browser (to 4 decimals). For this question, if you compute the probability manually, make sure to carry at least six decimal digits in your calculations. b. Compute the probability that at least of the Internet browser users use Chrome as their Internet browser (to 4 decimals).c. For the sample of Internet browser users, compute the expected number of Chrome users (to 3 decimals).d. For the sample of Internet browser users, compute the variance and standard deviation for the number of Chrome users (to 3 decimals).VarianceStandard deviation

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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, December ). 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 (to 4 decimals). For this question, if you compute the probability manually, make sure to carry at least six decimal digits in your calculations.

b. Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals).

c. For the sample of Internet browser users, compute the expected number of Chrome users (to 3 decimals).

d. For the sample of Internet browser users, compute the variance and standard deviation for the number of Chrome users (to 3 decimals).

a. P(x = 8) = 0.024273

b. P(x ≥ 3) = 0.8050

c. E(x) = 4.074

d. var(x) = 3.244 and SD = 1.801

Step-by-step explanation:

The given problem can be solved using the binomial distribution

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ  

Where n is the number of trials, x is the variable of interest and p is the probability of success.  

For the given problem,

Probability of success = p = 20.37% = 0.2037

Probability of failure = q = 1 - p = 1- 0.2037 = 0.7963

Number of trials  = n = 20

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

In this case, we have x = 8

P(x = 8) = ²⁰C₈×0.2037⁸×(1 - 0.2037)²⁰⁻⁸  

P(x = 8) = 125970×0.2037⁸×0.7963¹²

P(x = 8) = 0.024273

b. Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals).

P(x ≥ 3) = 1 - P(x < 2)

But we know that

P(x < 2) = P(x = 0) + P(x = 1) + P(x = 2)

So,

P(x ≥ 3) = 1 - [ P(x = 0) + P(x = 1) + P(x = 2) ]

For P(x = 0):

Here we have x = 0, n = 20 and p = 0.2037

P(x = 0) = ²⁰C₀×0.2037⁰×(1 - 0.2037)²⁰⁻⁰

P(x = 0) = 0.0105

For P(x = 1):

Here we have x = 1, n = 20 and p = 0.2037

P(x = 1) = ²⁰C₁×0.2037¹×(1 - 0.2037)²⁰⁻¹

P(x = 1) = 0.0538

For P(x = 2):

Here we have x = 2, n = 20 and p = 0.2037

P(x = 2) = ²⁰C₂×0.2037²×(1 - 0.2037)²⁰⁻²

P(x = 2) = 0.1307

Finally,

P(x ≥ 3) = 1 - [P(x = 0) + P(x = 1) + P(x = 2)]  

P(x ≥ 3) = 1 - [ 0.0105 + 0.0538  + 0.1307]  

P(x ≥ 3) = 1 - [0.1950]

P(x ≥ 3) = 0.8050

c. For the sample of Internet browser users, compute the expected number of Chrome users (to 3 decimals).

The expected number of Chrome users is given by

E(x) = n×p

Where n is the number of trials and p is the probability of success

E(x) = 20×0.2037

E(x) = 4.074

d. For the sample of Internet browser users, compute the variance and standard deviation for the number of Chrome users (to 3 decimals).

The variance for the number of Chrome users is given by

var(x) = n×p×q

Where n is the number of trials and p is the probability of success and q is the probability of failure.

var(x) = 20×0.2037×0.7963

var(x) = 3.244

The standard deviation for the number of Chrome users is given by

SD = √(n×p×q)

SD = √var(x)

SD = √(3.244)

SD = 1.801

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step-by-step explanation:

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