In 2014, 85 percent of households in the United States had a computer. For a randomly selected sample of 200 households in 2014, let the random variable C represent the number of households in the sample that had a computer. What are the mean and standard deviation of C ?

Answers

Answer 1

Answer:

The mean of C is 170 households

The standard deviation of C, is approximately 5 households

Step-by-step explanation:

The given parameters are;

The percentage of households in the United States that had a computer in 2014 = 85%

The size of the randomly selected sample in 2014, n = 200

The random variable representing the number of households that had a computer = C

Therefore, we have;

The probability of a household having a computer P = 85/100 = 0.85

Let

Therefore;

The mean (expected) number in the sample, μₓ, = E(x) = n × P is given as follows;

μₓ = 200 × 0.85 = 170

The mean of C = μₓ = 170

The variance, σ² = n × P × (1 - P) = 200 × 0.85 × (1 - 0.85) = 25.5

Therefore;

The standard deviation, σ = √(σ²) = √(25.5) ≈ 5.05

The standard deviation of C, σ ≈ 5 households (we round (down) to the nearest whole number)

Answer 2

The mean and the standard deviation of C are 170 and 5.05 respectively

The given parameters are:

[tex]\mathbf{n = 200}[/tex] -- the sample size

[tex]\mathbf{p = 85\%}[/tex] -- the proportion of household that had a computer

(a) The mean

This is calculated as:

[tex]\mathbf{\bar x = np}[/tex]

So, we have:

[tex]\mathbf{\bar x = 200 \times 85\%}[/tex]

[tex]\mathbf{\bar x = 170}[/tex]

(b) The standard deviation

This is calculated as:

[tex]\mathbf{\sigma = \sqrt{np(1 - p)}}[/tex]

So, we have:

[tex]\mathbf{\sigma = \sqrt{170 \times (1 - 85\%)}}[/tex]

[tex]\mathbf{\sigma = \sqrt{170 \times 15\%}}[/tex]

[tex]\mathbf{\sigma = \sqrt{25.5}}[/tex]

Take square roots

[tex]\mathbf{\sigma = 5.05}[/tex]

Hence, the mean and the standard deviation of C are 170 and 5.05 respectively

Read more about mean and standard deviation at:

https://brainly.com/question/10729938


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since we cant see what info is given maybe these tips can help...

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