SOLUTION: 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: ∑XP(X)= 9.88 ∑X^

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:
∑XP(X)= 9.88
∑X^2P(X)= 99.12

Answer by CPhill(1987) (Show Source): 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.
**Understanding the Formulas**
* **Mean (μ or E[X]):** The average number of visitors. In this case, it's given as ∑XP(X) = 9.88.
* **Variance (σ² or Var(X)):** A measure of how spread out the data is. The formula is:
* Var(X) = ∑X²P(X) - (∑XP(X))²
* **Standard Deviation (σ or SD(X)):** The square root of the variance. It tells you the typical deviation of values from the mean.
* SD(X) = √Var(X)
**Calculations**
1. **Mean (μ):**
* μ = ∑XP(X) = 9.88
2. **Variance (σ²):**
* σ² = ∑X²P(X) - (∑XP(X))²
* σ² = 99.12 - (9.88)²
* σ² = 99.12 - 97.6144
* σ² = 1.5056
3. **Standard Deviation (σ):**
* σ = √σ²
* σ = √1.5056
* σ ≈ 1.2270
**Results**
* Variance: 1.5056
* Standard Deviation: approximately 1.2270

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