Question

1. college prep a survey claims that 32% of u.s. adults think that more emphasis should be placed on producing domestic energy from nuclear power. you simulate taking a large number of samples of size 50, and the distribution has a standard deviation of about 0.02. which interval is likely to contain 95% of the data? 28% - 36% 30% - 34% 46% - 54% 48% - 52% 23. structure a developer claims that the percent of city residents who favor building a splash pad at a local park is likely between 72.3% and 81.7%. about how many residents were surveyed? round your answer to the nearest whole number. about residents were surveyed. correct answers:

Explanation:

Step1: Recall the empirical rule for normal distribution

For a normal - distributed data, about 95% of the data lies within 2 standard deviations of the mean.

Step2: Calculate the lower and upper bounds for the first problem

The mean proportion is $p = 0.32$ and the standard deviation $\sigma=0.02$.
The lower bound is $p - 2\sigma=0.32-2\times0.02 = 0.32 - 0.04=0.28$ or 28%.
The upper bound is $p + 2\sigma=0.32 + 2\times0.02=0.32 + 0.04 = 0.36$ or 36%.

Step3: For the second problem, recall the formula for margin of error

The margin of error $E$ for a proportion in a confidence interval is given by $E = z\sqrt{\frac{p(1 - p)}{n}}$. For a 95% confidence interval, $z = 1.96\approx2$. The mid - point of the interval $72.3\%$ and $81.7\%$ is $\frac{0.723 + 0.817}{2}=0.77$. The margin of error $E=\frac{0.817 - 0.723}{2}=0.047$.

Step4: Rearrange the margin of error formula to solve for $n$

We know that $E = z\sqrt{\frac{p(1 - p)}{n}}$. Squaring both sides gives $E^{2}=z^{2}\frac{p(1 - p)}{n}$. Then $n=\frac{z^{2}p(1 - p)}{E^{2}}$. Substituting $z = 2$, $p = 0.77$, and $E=0.047$, we have $n=\frac{2^{2}\times0.77\times(1 - 0.77)}{0.047^{2}}=\frac{4\times0.77\times0.23}{0.047^{2}}=\frac{4\times0.1771}{0.002209}=\frac{0.7084}{0.002209}\approx321$.

Answer:

1. 28% - 36%
2. 321