Mathematics, 19.03.2021 18:40, britt729

According to the U. S. Census, the proportion of adults in a certain county who owned their own home was 0.71. An SRS of 100 adults in a certain section of the county found that 65 owned their home. What represents the approximate probability of obtaining a sample of 100 adults in which 65 or fewer own their home, assuming that this section of the county has the same overall proportion of adults who own their home as does the entire county?

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answered: uticabadgirl

0.0764 = 7.64% approximate probability of obtaining a sample of 100 adults in which 65 or fewer own their home.

Step-by-step explanation:

To solve this question, we use the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

According to the U.S. Census, the proportion of adults in a certain county who owned their own home was 0.71.

This means that p = 0.71

Sample of 100:

This means that n = 100

Mean

$\mu = E(X) = np = 100*0.71 = 71$

Standard deviation:

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.71*0.29} = 4.5376$

What represents the approximate probability of obtaining a sample of 100 adults in which 65 or fewer own their home, assuming that this section of the county has the same overall proportion of adults who own their home as does the entire county?

This is, using continuity correction, $P(X \leq 65 - 0.5) = P(X \leq 64.5)$ , which is the pvalue of Z when X = 64.5. So