Math Problem Statement

Suppose that a line has been observed 10 times using the same equipment and procedures. The results in ft. are shown below. It is assumed that no mistakes exist and that the observations have already been corrected the error having 50% probability. 538.57,538.39,538.37,538.39,538.48,538.49,538.33,538.46,538.47,538.55

Solution

To analyze the given data of line observations, let's calculate some common statistics that help summarize the data and assess its variability. These statistics include:

  1. Mean (Average): It gives us a central value for the observed data.
  2. Variance: It shows the spread or dispersion of the observations from the mean.
  3. Standard Deviation: It provides the average amount by which observations deviate from the mean.
  4. Median: This is the middle value when the data is arranged in increasing order.
  5. Range: It shows the difference between the maximum and minimum observations.

The observed line lengths (in feet) are:
538.57,538.39,538.37,538.39,538.48,538.49,538.33,538.46,538.47,538.55538.57, 538.39, 538.37, 538.39, 538.48, 538.49, 538.33, 538.46, 538.47, 538.55

Let's compute these values.

Step 1: Mean

The mean μ\mu is given by: μ=i=1nxin\mu = \frac{\sum_{i=1}^{n} x_i}{n} Where xix_i are the individual observations, and nn is the number of observations (here n=10n = 10).

Step 2: Variance

Variance σ2\sigma^2 is calculated using: σ2=i=1n(xiμ)2n\sigma^2 = \frac{\sum_{i=1}^{n} (x_i - \mu)^2}{n} This helps in determining how spread out the values are around the mean.

Step 3: Standard Deviation

The standard deviation is the square root of the variance: σ=σ2\sigma = \sqrt{\sigma^2}

Step 4: Median

The median is the middle value when the data set is ordered from least to greatest.

Step 5: Range

The range is the difference between the maximum and minimum values: Range=max(xi)min(xi)\text{Range} = \max(x_i) - \min(x_i)

I'll calculate these values now.Here are the computed values based on the given dataHere are the computed values based on the given

  1. Mean (Average): 538.45ft538.45 \, \text{ft}
  2. Variance: 0.00554ft20.00554 \, \text{ft}^2
  3. Standard Deviation: 0.0744ft0.0744 \, \text{ft}
  4. Median: 538.465ft538.465 \, \text{ft}
  5. Range: 0.24ft0.24 \, \text{ft}

These statistics provide an overview of the data's central tendency and spread. The small variance and standard deviation suggest that the observations are closely clustered around the mean.

Do you need more details or have any questions?

Here are 5 related questions to explore:

  1. How does variance help assess the accuracy of repeated measurements?
  2. What is the significance of the median in a dataset?
  3. How would the presence of an outlier affect the mean and standard deviation?
  4. In what practical situations is the range used as a measure of variability?
  5. How do these statistics help in detecting errors in measurement procedures?

Tip: For a large dataset, the standard deviation gives a better idea of data spread than the range.

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Statistics

Formulas

Mean: μ = (Σx_i) / n
Variance: σ² = (Σ(x_i - μ)²) / n
Standard Deviation: σ = √σ²
Range: max(x_i) - min(x_i)
Median: middle value of ordered data

Theorems

None

Suitable Grade Level

Grades 9-12