Your car needs new brakes, which cost $285. The

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

DochEvi [55]

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)

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A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .