Math Problem Statement

The user is asked to calculate the average density, standard deviation of density, 90% confidence limit, and relative error based on trials of water density measurements.

Solution

To solve the given tasks, I'll first extract the relevant data from the table and proceed with the required calculations.

Step 1: Extract Data

  • Volume of water delivered (mL): 50 mL for all trials.
  • Mass of beaker + water delivered (g):
    • Trial 1: 96.190 g
    • Trial 2: 93.964 g
    • Trial 3: 92.850 g
  • Mass of water delivered (g):
    • Trial 1: 48.445 g
    • Trial 2: 46.219 g
    • Trial 3: 45.105 g
  • Density of water (g/mL):
    • Trial 1: 0.9689 g/mL
    • Trial 2: 0.9244 g/mL
    • Trial 3: 0.9021 g/mL

Step 2: Calculating the Average Density

To find the average density: Average density=0.9689+0.9244+0.90213=2.79543=0.9318 g/mL\text{Average density} = \frac{0.9689 + 0.9244 + 0.9021}{3} = \frac{2.7954}{3} = 0.9318 \text{ g/mL}

Step 3: Calculating the Standard Deviation

The standard deviation (σ) of the densities is calculated as: σ=1N1i=1N(xixˉ)2\sigma = \sqrt{\frac{1}{N-1} \sum_{i=1}^{N} (x_i - \bar{x})^2} Where xix_i are the individual densities, xˉ\bar{x} is the mean density, and NN is the number of trials.

Plugging in the values: σ=131[(0.96890.9318)2+(0.92440.9318)2+(0.90210.9318)2]\sigma = \sqrt{\frac{1}{3-1} \left[(0.9689 - 0.9318)^2 + (0.9244 - 0.9318)^2 + (0.9021 - 0.9318)^2\right]} σ=12[0.001372+0.000055+0.000876]\sigma = \sqrt{\frac{1}{2} \left[0.001372 + 0.000055 + 0.000876\right]} σ=12×0.002303=0.00115150.0339\sigma = \sqrt{\frac{1}{2} \times 0.002303} = \sqrt{0.0011515} \approx 0.0339

Step 4: Calculating the 90% Confidence Limit for the Average Density

The 90% confidence interval can be calculated using: Confidence limit=xˉ±tσN\text{Confidence limit} = \bar{x} \pm t \frac{\sigma}{\sqrt{N}} For a 90% confidence level and N=3N = 3, the t-value (from the t-distribution table) is approximately 1.886.

Thus, the confidence limit is: 0.9318±1.886×0.033930.9318 \pm 1.886 \times \frac{0.0339}{\sqrt{3}} 0.9318±1.886×0.0196=0.9318±0.03700.9318 \pm 1.886 \times 0.0196 = 0.9318 \pm 0.0370 So, the 90% confidence interval is approximately 0.89480.8948 to 0.96880.9688 g/mL.

Step 5: Calculating the Relative Error

The relative error can be calculated as: Relative Error=Standard deviationAverage density×100\text{Relative Error} = \frac{\text{Standard deviation}}{\text{Average density}} \times 100 Relative Error=0.03390.9318×1003.64%\text{Relative Error} = \frac{0.0339}{0.9318} \times 100 \approx 3.64\%

Summary

  1. Average Density: 0.9318 g/mL
  2. Standard Deviation: 0.0339 g/mL
  3. 90% Confidence Interval: 0.8948 to 0.9688 g/mL
  4. Relative Error: 3.64%

Would you like further details or explanations on any of these steps?

Related Questions:

  1. How does increasing the number of trials affect the accuracy of the average density?
  2. What is the significance of using a t-value in calculating the confidence interval?
  3. How would the results change if the volume of water delivered was different for each trial?
  4. Why is it important to use the standard deviation when analyzing experimental data?
  5. How can outliers affect the calculation of standard deviation and mean in this experiment?

Tip:

When calculating confidence intervals, always ensure you have the correct t-value based on your sample size and desired confidence level.

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Math Problem Analysis

Mathematical Concepts

Statistics
Measurement
Standard Deviation
Confidence Interval
Relative Error

Formulas

Average density formula: (sum of densities) / (number of trials)
Standard deviation formula: sqrt((1/(N-1)) * sum((xi - x̄)^2))
Confidence limit formula: x̄ ± t(σ/√N)
Relative error formula: (Standard deviation / Average density) * 100

Theorems

t-distribution for confidence intervals
Error propagation in measurements

Suitable Grade Level

Grades 11-12 (Advanced High School or Introductory College Level)