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The UCLA School of Law Williams Institute estimates that \( 0.5 \% \) of U.S. identify as transgender. If the researchers at the Williams Institute want to use a normal distribution to estimate probabilities for the proportion of adults that identify as transgender, which of the following sample sizes are sufficiently large? Select all that apply 2600 1800 4500 3700 1900

Ask by Schwartz Howell. in the United States
Mar 14,2025

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The sample sizes of 2600, 4500, and 3700 are sufficiently large for using a normal distribution to estimate the proportion of adults identifying as transgender.

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We need to check the condition for using the normal approximation for the distribution of the sample proportion. A common requirement is that both \[ np \ge 10 \quad \text{and} \quad n(1-p) \ge 10, \] where \( p \) is the population proportion. Here, \( p = 0.005 \). ### Step 1: Checking \( np \ge 10 \) The condition becomes: \[ n \cdot 0.005 \ge 10 \quad \Rightarrow \quad n \ge \frac{10}{0.005} = 2000. \] So, the sample size must be at least \( 2000 \) for \( np \) to be at least \( 10 \). ### Step 2: Checking \( n(1-p) \ge 10 \) Since \( 1-p = 0.995 \), for any \( n \ge 2000 \): \[ n \cdot 0.995 \ge 2000 \cdot 0.995 = 1990, \] which is much greater than \( 10 \). Therefore, the condition \( n(1-p) \ge 10 \) is automatically satisfied for these choices of \( n \). ### Step 3: Evaluating Given Sample Sizes We now check each sample size: - **2600:** Since \( 2600 \ge 2000 \), it is sufficient. - **1800:** Since \( 1800 < 2000 \), it is not sufficient. - **4500:** Since \( 4500 \ge 2000 \), it is sufficient. - **3700:** Since \( 3700 \ge 2000 \), it is sufficient. - **1900:** Since \( 1900 < 2000 \), it is not sufficient. ### Final Answer The sample sizes that are sufficiently large are: - \(2600\) - \(4500\) - \(3700\)

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To determine if a sample size is sufficiently large for estimating proportions using a normal distribution, the rule of thumb is that both \( np \) and \( n(1-p) \) should be greater than or equal to 10, where \( n \) is the sample size and \( p \) is the population proportion. In this case, \( p = 0.005 \) (0.5%) and \( 1-p = 0.995 \). Calculating for each of the sample sizes: - For \( n = 2600 \): - \( np = 2600 \times 0.005 = 13 \) (sufficient) - \( n(1-p) = 2600 \times 0.995 = 2587 \) (sufficient) - For \( n = 1800 \): - \( np = 1800 \times 0.005 = 9 \) (not sufficient) - \( n(1-p) = 1800 \times 0.995 = 1791 \) (sufficient) - For \( n = 4500 \): - \( np = 4500 \times 0.005 = 22.5 \) (sufficient) - \( n(1-p) = 4500 \times 0.995 = 4477.5 \) (sufficient) - For \( n = 3700 \): - \( np = 3700 \times 0.005 = 18.5 \) (sufficient) - \( n(1-p) = 3700 \times 0.995 = 3685.5 \) (sufficient) - For \( n = 1900 \): - \( np = 1900 \times 0.005 = 9.5 \) (not sufficient) - \( n(1-p) = 1900 \times 0.995 = 1890.5 \) (sufficient) Thus, the sample sizes that are sufficiently large for this situation are **2600**, **4500**, and **3700**.

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