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evaluate $\\sigma=\\sqrt{np(1 - p)}$ for $n = 41$, $p = 0.5$. $\\sigma=…

Question

evaluate $\sigma=\sqrt{np(1 - p)}$ for $n = 41$, $p = 0.5$.
$\sigma=\square$
(simplify your answer. type an integer or decimal rounded to one decimal place as needed.)

Explanation:

Step1: Substitute values into formula

Given \(n = 41\), \(p=0.5\), then \(1 - p=1 - 0.5 = 0.5\). Substitute into \(\sigma=\sqrt{np(1 - p)}\), we get \(\sigma=\sqrt{41\times0.5\times0.5}\).

Step2: Calculate the product inside the square - root

First calculate \(41\times0.5\times0.5\). \(41\times0.5 = 20.5\), then \(20.5\times0.5=10.25\). So \(\sigma=\sqrt{10.25}\).

Step3: Find the square - root

Since \(\sqrt{10.25}=3.2\) (because \(3.2\times3.2 = 10.24\approx10.25\) when rounded to one decimal place).

Answer:

\(3.2\)