Key Concepts and Formulas

To accurately determine the mean and variance of transformed observations, we rely on the fundamental definitions of these statistical measures and their properties under linear transformations.

1. Mean ($\bar{z}$): The arithmetic average of a set of $N$ observations, $z_1, z_2, \ldots, z_N$. $\bar{z} = \frac{1}{N} \sum_{i=1}^{N} z_i$

2. Variance ($\sigma_z^2$): A measure of the spread or dispersion of data points around their mean. It is the average of the squared differences from the mean. $\sigma_z^2 = \frac{1}{N} \sum_{i=1}^{N} (z_i - \bar{z})^2$ An alternative, often more convenient, formula for variance is: $\sigma_z^2 = \frac{1}{N} \sum_{i=1}^{N} z_i^2 - (\bar{z})^2$

3. Properties of Mean and Variance under Linear Transformation: If a new set of observations $y_i$ is obtained from an original set $x_i$ by a linear transformation $y_i = a x_i + b$ (where $a$ and $b$ are constants), then:

   • The new mean $\bar{y}$ is related to the original mean $\bar{x}$ by: $\bar{y} = a \bar{x} + b$ Reasoning: Adding a constant $b$ shifts all data points, and consequently the mean, by $b$. Multiplying by a constant $a$ scales all data points, and thus scales the mean by $a$.
   • The new variance $\sigma_y^2$ is related to the original variance $\sigma_x^2$ by: $\sigma_y^2 = a^2 \sigma_x^2$ Reasoning: Adding a constant $b$ to each observation shifts the entire distribution but does not change its spread, so it does not affect the variance. Multiplying each observation by a constant $a$ scales the deviations from the mean by $a$, and therefore scales the squared deviations (and thus the variance) by $a^2$. The standard deviation $\sigma_y$ would be $|a|\sigma_x$.

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Step-by-Step Solution

Let the given observations be $x_1, x_2, \ldots, x_{10}$. There are $N=10$ observations.

Step 1: Simplify the given information using a substitution.

• What we are doing: We introduce a temporary variable to simplify the expressions given in the problem statement. This makes the initial calculations more straightforward.
• Why we are doing it: The given sums involve the term $(x_i - 5)$. By substituting this term with a new variable, say $d_i$, we can directly work with the sums of $d_i$ and $d_i^2$, which are in a standard format for calculating mean and variance.
• The math: Let $d_i = x_i - 5$. The first given equation is $\sum_{i=1}^{10} (x_i - 5) = 10$. Substituting $d_i$, this becomes: $\sum_{i=1}^{10} d_i = 10$ The second given equation is $\sum_{i=1}^{10} (x_i - 5)^2 = 40$. Substituting $d_i$, this becomes: $\sum_{i=1}^{10} d_i^2 = 40$

Step 2: Calculate the mean and variance of the substituted variable $d_i$.

• What we are doing: We use the fundamental definitions of mean and variance to calculate these measures for the intermediate variable $d_i$, for which we now have direct sums.
• Why we are doing it: These values ($\bar{d}$ and $\sigma_d^2$) are the building blocks from which we will derive the mean and variance of $x_i$ and subsequently the final desired observations.
• The math: The number of observations is $N=10$.
  • Calculate the mean of $d_i$ ($\bar{d}$): Using the definition $\bar{d} = \frac{1}{N} \sum_{i=1}^{N} d_i$: $\bar{d} = \frac{10}{10} = 1$
  • Calculate the variance of $d_i$ ($\sigma_d^2$): Using the computational formula $\sigma_d^2 = \frac{1}{N} \sum_{i=1}^{N} d_i^2 - (\bar{d})^2$: $\sigma_d^2 = \frac{40}{10} - (1)^2$ $\sigma_d^2 = 4 - 1 = 3$

Step 3: Calculate the mean and variance of the original observations $x_i$.

• What we are doing: We use the properties of linear transformations to find the mean and variance of the original observations $x_i$ from the mean and variance of $d_i$.
• Why we are doing it: The final observations are based on $x_i$, so we need to first understand the statistical properties of $x_i$.
• The math: We know that $d_i = x_i - 5$. Rearranging this, we get $x_i = d_i + 5$. This is a linear transformation of the form $x_i = a d_i + b$, where $a=1$ and $b=5$.
  • Calculate the mean of $x_i$ ($\bar{x}$): Using the property $\bar{x} = a\bar{d} + b$: $\bar{x} = 1 \cdot \bar{d} + 5$ Substitute $\bar{d} = 1$: $\bar{x} = 1 + 5 = 6$ Reasoning: Adding a constant (5) to each observation shifts the mean by that same constant.
  • Calculate the variance of $x_i$ ($\sigma_x^2$): Using the property $\sigma_x^2 = a^2 \sigma_d^2$: $\sigma_x^2 = (1)^2 \cdot \sigma_d^2$ Substitute $\sigma_d^2 = 3$: $\sigma_x^2 = 1 \cdot 3 = 3$ Reasoning: Adding a constant (5) to each observation does not change the spread of the data, and therefore does not affect the variance.

Step 4: Calculate the mean ($\mu$) and variance ($\lambda$) of the final observations.

• What we are doing: We apply the properties of linear transformations one final time to find the mean ($\mu$) and variance ($\lambda$) of the observations $y_i = x_i - 3$.
• Why we are doing it: This is the ultimate goal of the problem, to find the ordered pair $(\mu, \lambda)$.
• The math: The new observations are $y_i = x_i - 3$. This is a linear transformation of the form $y_i = a x_i + b$, where $a=1$ and $b=-3$.
  • Calculate the mean of $y_i$ ($\mu$): Using the property $\mu = a\bar{x} + b$: $\mu = 1 \cdot \bar{x} - 3$ Substitute $\bar{x} = 6$: $\mu = 6 - 3 = 3$ Reasoning: Subtracting a constant (3) from each observation shifts the mean by that same constant.
  • Calculate the variance of $y_i$ ($\lambda$): Using the property $\lambda = a^2 \sigma_x^2$: $\lambda = (1)^2 \cdot \sigma_x^2$ Substitute $\sigma_x^2 = 3$: $\lambda = 1 \cdot 3 = 3$ Reasoning: Subtracting a constant (3) from each observation does not change the spread of the data, and therefore does not affect the variance.

Step 5: Form the ordered pair.

• What we are doing: We combine the calculated mean ($\mu$) and variance ($\lambda$) into the ordered pair requested by the problem.
• Why we are doing it: This is the final answer format specified in the question.
• The math: We found $\mu = 3$ and $\lambda = 3$. Therefore, the ordered pair is $(\mu, \lambda) = (3, 3)$.

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Common Mistakes & Tips

• Misapplying Variance Properties: A frequent error is to assume that adding or subtracting a constant affects the variance. Remember, $\sigma_{x+b}^2 = \sigma_x^2$ and $\sigma_{x-b}^2 = \sigma_x^2$. Only multiplication by $a$ changes variance, by $a^2$.
• Incorrectly Squaring 'a': When applying the variance transformation $\sigma_y^2 = a^2 \sigma_x^2$, ensure you square the constant $a$.
• Ignoring Substitution Benefits: Failing to use a substitution like $d_i = x_i - 5$ can lead to more complex algebraic manipulations, increasing the chance of errors. Embrace simplification!
• Confusing Sum of Squares with Squared Sum: Be careful with the variance formula: $\frac{1}{N} \sum z_i^2 - (\bar{z})^2$. Ensure you're not mistakenly calculating $\frac{1}{N} (\sum z_i)^2$.

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Summary

This problem efficiently tests your understanding of mean and variance calculations, particularly how these measures transform under linear operations. By first simplifying the given sums using a substitution ($d_i = x_i - 5$), we calculated the mean and variance of $d_i$. Then, using the properties of linear transformations, we sequentially found the mean and variance of $x_i$, and finally the mean ($\mu$) and variance ($\lambda$) of the desired observations $x_i - 3$. The final calculated ordered pair was $(3, 3)$.

The final answer is $\boxed{(3,3)}$, which corresponds to option (B).