In 2014, 85 percent of households in the United States had a computer. For a randomly selected sample of 200 households in 2014,

Question

3 years ago

7

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 ?

2 answers:

DochEvi [55] · Answer 1 · 2022-04-12T01:29:22+03:00

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)

pickupchik [31] · Answer 2 · 2022-04-12T03:27:14+03:00

5 0

Answer:

The mean is 170 households, and the standard deviation is 5.05 households.

Step-by-step explanation: