document.write( "Question 1196344: Let Z~N(0,1) and z_αthe value such that P(Z≥z_α )=α. Express each of the probabilities below in terms of α. Show the steps used to get to the answers.
\n" ); document.write( "3.1 P(Z\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #848416 by ElectricPavlov(122)\"\" \"About 
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**1. P(Z >= 0)**\r
\n" ); document.write( "\n" ); document.write( "* **Understanding z_α:**
\n" ); document.write( " * z_α represents the z-score that corresponds to the right-tail probability of α in a standard normal distribution.
\n" ); document.write( " * In other words, P(Z >= z_α) = α\r
\n" ); document.write( "\n" ); document.write( "* **Symmetry of Standard Normal Distribution:**
\n" ); document.write( " * The standard normal distribution is symmetric around 0.
\n" ); document.write( " * Therefore, P(Z >= 0) = 0.5\r
\n" ); document.write( "\n" ); document.write( "**2. P(Z <= 5)**\r
\n" ); document.write( "\n" ); document.write( "* **Large Z-values:** For very large values of z (like 5), the probability of Z being less than that value is extremely close to 1.
\n" ); document.write( " * This is because the standard normal distribution extends indefinitely to the right, but with diminishing probability density.\r
\n" ); document.write( "\n" ); document.write( "* **In terms of α:** Since P(Z >= z_α) = α represents a right-tail probability, and P(Z <= 5) covers almost the entire distribution, we can approximate:
\n" ); document.write( " * P(Z <= 5) ≈ 1 - α \r
\n" ); document.write( "\n" ); document.write( "**In summary:**\r
\n" ); document.write( "\n" ); document.write( "* P(Z >= 0) = 0.5
\n" ); document.write( "* P(Z <= 5) ≈ 1 - α
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