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n=((1.96 sigma)/(E))^(2) Elon is planning a study of hours of internet use per month per household. He uses the given equation to determine the minimum number of randomly selected households, n, that he needs to include in his sample given a standard deviation of sigma hours and a maximum allowable error of E hours. Which of the following equations correctly gives the standard deviation in terms of the maximum allowable error and the minimum number of randomly selected households? Choose 1 answer: (A) sigma=sqrt((En)/(1.96)) (B) sigma=(Esqrtn)/(1.96) (c) sigma=sqrt((1.96 n)/(n))

n=(1.96σE)2 n=\left(\frac{1.96 \sigma}{E}\right)^{2} \newlineElon is planning a study of hours of internet use per month per household. He uses the given equation to determine the minimum number of randomly selected households, n n , that he needs to include in his sample given a standard deviation of σ \sigma hours and a maximum allowable error of E E hours. Which of the following equations correctly gives the standard deviation in terms of the maximum allowable error and the minimum number of randomly selected households?\newlineChoose 11 answer:\newline(A) σ=En1.96 \sigma=\sqrt{\frac{E n}{1.96}} \newline(B) σ=En1.96 \sigma=\frac{E \sqrt{n}}{1.96} \newline(C) σ=1.96nπ \sigma=\sqrt{\frac{1.96 n}{\pi}}

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Q. n=(1.96σE)2 n=\left(\frac{1.96 \sigma}{E}\right)^{2} \newlineElon is planning a study of hours of internet use per month per household. He uses the given equation to determine the minimum number of randomly selected households, n n , that he needs to include in his sample given a standard deviation of σ \sigma hours and a maximum allowable error of E E hours. Which of the following equations correctly gives the standard deviation in terms of the maximum allowable error and the minimum number of randomly selected households?\newlineChoose 11 answer:\newline(A) σ=En1.96 \sigma=\sqrt{\frac{E n}{1.96}} \newline(B) σ=En1.96 \sigma=\frac{E \sqrt{n}}{1.96} \newline(C) σ=1.96nπ \sigma=\sqrt{\frac{1.96 n}{\pi}}
  1. Given Equation: We start with the given equation:\newlinen=(1.96×σE)2n = \left(\frac{1.96 \times \sigma}{E}\right)^2\newlineWe want to solve for σ\sigma, the standard deviation.
  2. Square Root Elimination: First, we take the square root of both sides to eliminate the square on the right-hand side: n=(1.96×σE)2\sqrt{n} = \sqrt{\left(\frac{1.96 \times \sigma}{E}\right)^2}
  3. Simplify Right Hand Side: Simplifying the square root of the square on the right-hand side gives us: n=1.96×σE\sqrt{n} = \frac{1.96 \times \sigma}{E}
  4. Isolate Sigma Term: Next, we multiply both sides by EE to isolate the term with σ\sigma on the right-hand side:\newlineE⋅n=1.96⋅σE \cdot \sqrt{n} = 1.96 \cdot \sigma
  5. Solve for Sigma: Finally, we divide both sides by 1.961.96 to solve for sigma:\newlineσ=E⋅n1.96\sigma = \frac{E \cdot \sqrt{n}}{1.96}
  6. Check Options: Now we check the options given to see which one matches our derived equation for sigma:\newline(A) σ=(E⋅n)/1.96\sigma = \sqrt{(E \cdot n) / 1.96} does not match because there is no square root over nn.\newline(B) σ=(E⋅n)/1.96\sigma = (E \cdot \sqrt{n}) / 1.96 matches our derived equation.\newline(C) σ=(1.96⋅n)/E\sigma = \sqrt{(1.96 \cdot n) / E} does not match because the numerator is incorrect.

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