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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).
Answer:
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
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