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wolverine [178]
3 years ago
10

What is the value of 15/124×230/30÷230/124

Mathematics
1 answer:
KonstantinChe [14]3 years ago
5 0
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Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in
arsen [322]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

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

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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Answer:

B

Step-by-step explanation:

A, C, and D can be a possibility but they are not exactly at the middle. B is the only one that shows a slightly triangular shape, which you can see the middle that shows the best measure of the center.

Hope this helped!

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Mr. Grant's class is working on a recycling project. The students are collecting aluminum cans. Over 3 days, they collected 88 c
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Eric took a test with 15 questions. he answered 2/5 of the questions correctly, how many did he get wrong?
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