document.write( "Question 1171843: At a small community library, the number of visitors per hour during the day has the distribution shown. Find the variance and standard deviation given that:
\n" ); document.write( "∑XP(X)= 9.88
\n" ); document.write( "∑X^2P(X)= 99.12
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #850836 by CPhill(1987)\"\" \"About 
You can put this solution on YOUR website!
Absolutely! Let's break down how to find the variance and standard deviation using the provided information.\r
\n" ); document.write( "\n" ); document.write( "**Understanding the Formulas**\r
\n" ); document.write( "\n" ); document.write( "* **Mean (μ or E[X]):** The average number of visitors. In this case, it's given as ∑XP(X) = 9.88.
\n" ); document.write( "* **Variance (σ² or Var(X)):** A measure of how spread out the data is. The formula is:
\n" ); document.write( " * Var(X) = ∑X²P(X) - (∑XP(X))²
\n" ); document.write( "* **Standard Deviation (σ or SD(X)):** The square root of the variance. It tells you the typical deviation of values from the mean.
\n" ); document.write( " * SD(X) = √Var(X)\r
\n" ); document.write( "\n" ); document.write( "**Calculations**\r
\n" ); document.write( "\n" ); document.write( "1. **Mean (μ):**
\n" ); document.write( " * μ = ∑XP(X) = 9.88\r
\n" ); document.write( "\n" ); document.write( "2. **Variance (σ²):**
\n" ); document.write( " * σ² = ∑X²P(X) - (∑XP(X))²
\n" ); document.write( " * σ² = 99.12 - (9.88)²
\n" ); document.write( " * σ² = 99.12 - 97.6144
\n" ); document.write( " * σ² = 1.5056\r
\n" ); document.write( "\n" ); document.write( "3. **Standard Deviation (σ):**
\n" ); document.write( " * σ = √σ²
\n" ); document.write( " * σ = √1.5056
\n" ); document.write( " * σ ≈ 1.2270\r
\n" ); document.write( "\n" ); document.write( "**Results**\r
\n" ); document.write( "\n" ); document.write( "* Variance: 1.5056
\n" ); document.write( "* Standard Deviation: approximately 1.2270
\n" ); document.write( " \n" ); document.write( "
\n" );