Mathematics, 11.05.2021 02:20, lily3934

Suppose that there is a school bond referendum in Greensboro, and 73% of voters support it. You randomly ask 20 Greensboro voters whether they support the bond referendum. The standard error of the sample proportion is . The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is . Group of answer choices 0.0993; 0.6000 0.0993; 0.0951 0.1095; 0.1170 0.1095; 0.8830

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answered: jaelynnm

The standard error of the sample proportion is 0.0993.

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is 0.0951.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

73% of voters support it.

This means that p = 0.73

Sample of 20 voters

This means that n = 20

Mean and standard deviation:

$\mu = p = 0.73$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.73*0.27}{20}} = 0.0993$

The standard error of the sample proportion is 0.0993.

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is

12/20 = 0.6, so this is the p-value of Z when X = 0.6.

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

$Z = \frac{0.6 - 0.73}{0.0993}$

Z = -1.31

Z = -1.31 has a p-value of 0.0951.

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is 0.0951.

answered: Guest

so, converting firstly the mixed fractions to improper fractions and then "adding".

$\bf \stackrel{mixed}{2\frac{9}{10}}\implies \cfrac{2\cdot 10+9}{10}\implies \stackrel{improper}{\cfrac{29}{10}}~\hfill \stackrel{mixed}{5\frac{4}{15}}\implies \cfrac{5\cdot 15+4}{15}\implies \stackrel{improper}{\cfrac{79}{15}} [-0.35em] ~\dotfill$

$\bf -\cfrac{29}{10}+\left( -\cfrac{79}{15} \right)\implies -\cfrac{29}{10}-\cfrac{79}{15}\implies \stackrel{\textit{using the lcd of 30}}{\cfrac{-(3)29-(2)79}{30}}\implies \cfrac{-87-158}{30} \cfrac{-245}{30}\implies -\cfrac{49}{6}\implies -8\frac{1}{6}$