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Statistics & Probability
Statistics
Hard

Question

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then a+3b5\mathrm{a+3 b-5} is equal to ___________.

Answer: 7

Solution

Key Concepts and Formulas

  1. Mean (xˉ\bar{x}): The arithmetic average of a set of observations. For nn observations x1,x2,,xnx_1, x_2, \ldots, x_n, the mean is calculated as: xˉ=i=1nxin\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} This formula can be rearranged to find the sum of observations: xi=nxˉ\sum x_i = n \cdot \bar{x}.

  2. Variance (σ2\sigma^2): A measure of how much the observations in a dataset deviate from their mean. The most practical formula for calculation, especially when dealing with changes in data, is: σ2=i=1nxi2n(xˉ)2\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - \left( \bar{x} \right)^2 This formula can be rearranged to find the sum of squares of observations: xi2=n(σ2+xˉ2)\sum x_i^2 = n \left( \sigma^2 + \bar{x}^2 \right).

  3. Impact of Adding/Removing Observations: When an observation is added or removed, the total sum of observations (xi\sum x_i) and the total sum of squares of observations (xi2\sum x_i^2) are updated by adding or subtracting the value (and its square) of the respective observation. The number of observations (nn) also changes accordingly.


Step-by-Step Solution

This problem involves adjusting the mean and variance of a dataset after one observation is removed. Our strategy will be to first determine the initial sums of observations and squares of observations, then adjust these sums for the removed observation, and finally calculate the new mean and variance.

Step 1: Calculate Initial Sums for 7 Observations

We are given the following information for the initial set of 7 observations:

  • Number of observations (n1n_1) = 7
  • Mean (xˉ1\bar{x}_1) = 8
  • Variance (σ12\sigma_1^2) = 16

The goal of this step is to find the initial sum of observations (xi\sum x_i) and the initial sum of squares of observations (xi2\sum x_i^2). These are the fundamental quantities that will be adjusted when an observation is removed.

  • Step 1.1: Calculate the Sum of the Initial Observations (xi\sum x_i)

    • Why: The definition of the mean allows us to directly calculate the sum of all observations.
    • Formula: xi=n1xˉ1\sum x_i = n_1 \cdot \bar{x}_1
    • Calculation: xi=7×8=56\sum x_i = 7 \times 8 = 56
    • Reasoning: The sum of all 7 original observations is 56.
  • Step 1.2: Calculate the Sum of the Squares of the Initial Observations (xi2\sum x_i^2)

    • Why: The variance formula involves the sum of squares. By using the given variance, mean, and number of observations, we can determine the initial sum of squares. This is crucial for updating the variance later.
    • Formula: σ12=xi2n1(xˉ1)2\sigma_1^2 = \frac{\sum x_i^2}{n_1} - (\bar{x}_1)^2
    • Calculation: Substitute the given values into the variance formula: 16=xi27(8)216 = \frac{\sum x_i^2}{7} - (8)^2 16=xi276416 = \frac{\sum x_i^2}{7} - 64 Add 64 to both sides to isolate the term with xi2\sum x_i^2: 16+64=xi2716 + 64 = \frac{\sum x_i^2}{7} 80=xi2780 = \frac{\sum x_i^2}{7} Multiply by 7 to solve for xi2\sum x_i^2: xi2=80×7=560\sum x_i^2 = 80 \times 7 = 560
    • Reasoning: The sum of the squares of all 7 original observations is 560. Note that this is different from the square of the sum (562=313656^2 = 3136).

Step 2: Adjust Sums for the Omitted Observation

One observation, with a value of 14, is omitted from the dataset. This changes the number of observations, the total sum, and the total sum of squares.

  • The new number of observations (n2n_2) will be 71=67 - 1 = 6.

  • Step 2.1: Calculate the New Sum of Observations for 6 Observations

    • Why: When a specific observation is removed, its value is subtracted from the total sum of all observations.
    • Calculation: Let the new sum of observations be xi\sum x'_i. xi=(Old xi)(Omitted observation) \sum x'_i = (\text{Old } \sum x_i) - (\text{Omitted observation}) xi=5614=42\sum x'_i = 56 - 14 = 42
    • Reasoning: The sum of the remaining 6 observations is now 42.
  • Step 2.2: Calculate the New Sum of Squares of Observations for 6 Observations

    • Why: Similarly, when an observation is removed, its square is subtracted from the total sum of the squares of all observations.
    • Calculation: Let the new sum of squares be (xi)2\sum (x'_i)^2. (xi)2=(Old xi2)(Square of omitted observation) \sum (x'_i)^2 = (\text{Old } \sum x_i^2) - (\text{Square of omitted observation}) (xi)2=560(14)2\sum (x'_i)^2 = 560 - (14)^2 (xi)2=560196=364\sum (x'_i)^2 = 560 - 196 = 364
    • Reasoning: The sum of the squares of the remaining 6 observations is 364.

Step 3: Calculate the New Mean (a) and Variance (b) of the Remaining 6 Observations

Now we have all the necessary components for the modified dataset:

  • New number of observations (n2n_2) = 6

  • New sum of observations (xi\sum x'_i) = 42

  • New sum of squares of observations ((xi)2\sum (x'_i)^2) = 364

  • Step 3.1: Calculate the New Mean (a)

    • Why: Apply the mean formula using the new sum of observations and the new number of observations.
    • Formula: a=xin2a = \frac{\sum x'_i}{n_2}
    • Calculation: a=426=7a = \frac{42}{6} = 7
    • Reasoning: The mean of the remaining 6 observations is 7. This is the value of 'a'.
  • Step 3.2: Calculate the New Variance (b)

    • Why: Apply the variance formula using the new sum of squares, the new mean (a), and the new number of observations.
    • Formula: b=(xi)2n2(a)2b = \frac{\sum (x'_i)^2}{n_2} - (a)^2
    • Calculation: Substitute the values we've calculated: b=3646(7)2b = \frac{364}{6} - (7)^2 Simplify the fraction 3646\frac{364}{6}: 3646=1823\frac{364}{6} = \frac{182}{3} Now, substitute this back into the equation for bb: b=182349b = \frac{182}{3} - 49 To combine these terms, find a common denominator: b=182349×33=1821473=353b = \frac{182}{3} - \frac{49 \times 3}{3} = \frac{182 - 147}{3} = \frac{35}{3}
    • Reasoning: The variance of the remaining 6 observations is 353\frac{35}{3}. This is the value of 'b'.

Step 4: Calculate the Final Expression (a + 3b - 5)

We have found the values of aa and bb:

  • a=7a = 7
  • b=353b = \frac{35}{3}

Now, substitute these values into the expression a+3b5a + 3b - 5: a+3b5=7+3(353)5a + 3b - 5 = 7 + 3\left(\frac{35}{3}\right) - 5 a+3b5=7+355a + 3b - 5 = 7 + 35 - 5 a+3b5=425a + 3b - 5 = 42 - 5 a+3b5=37a + 3b - 5 = 37 Reasoning: Substituting the derived values of 'a' and 'b' into the given expression yields the final answer.


Common Mistakes & Tips

  • Confusing Sum of Squares with Square of Sum: Ensure you correctly calculate xi2\sum x_i^2 and do not confuse it with (xi)2(\sum x_i)^2. The variance formula explicitly uses xi2\sum x_i^2.
  • Arithmetic Errors: This problem involves several arithmetic steps (multiplication, division, subtraction). Double-check each calculation, especially when dealing with fractions and squares.
  • Updating All Parameters: Remember to update not only the sum of observations and sum of squares but also the number of observations (nn) when an observation is added or removed.

Summary

We began by utilizing the initial mean and variance of 7 observations to calculate their total sum and sum of squares. Then, we adjusted these sums by subtracting the value and square of the omitted observation (14). With the new sums and the reduced number of observations (6), we calculated the new mean (a) and variance (b). Finally, we substituted these values into the given expression a+3b5a + 3b - 5 to arrive at the solution.

The final answer is 37\boxed{37}.

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