Q.The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations.

Concept and Intuition

The mean deviation from the mean measures how spread out the data is, on average, from the center (the mean). Unlike variance or standard deviation, it uses absolute values, so it treats all deviations equally without squaring them. This makes it more intuitive: it answers, "If I pick a random data point, how far from the average should I expect it to be, in the original units?"

Why does this work? Because we first find the balance point (the mean), then measure each point's distance from that balance, and finally average those distances. The result is a single number summarizing typical deviation.

Step-by-Step Solution

1. Find the mean of the data. The data set is: $2, 3, 5, 7, 11, 13, 17, 19, 22$. Sum of values:

Number of values: $n = 9$.

Mean ($\bar{x}$):

2. Compute each deviation from the mean (absolute value).

   Deviation = $|x_i - \bar{x}|$:

   • $|2 - 11| = 9$
   • $|3 - 11| = 8$
   • $|5 - 11| = 6$
   • $|7 - 11| = 4$
   • $|11 - 11| = 0$
   • $|13 - 11| = 2$
   • $|17 - 11| = 6$
   • $|19 - 11| = 8$
   • $|22 - 11| = 11$

3. Sum the absolute deviations.

4. Divide by the number of data points to get the mean deviation.

Concept & Intuition

Variance measures spread — how far data points are from the mean.

• If you add a constant to every observation, the whole distribution shifts left/right, but the distances between points stay the same. So variance doesn’t change.
• If you multiply every observation by a constant, all distances stretch by that factor. Since variance is an average of squared distances, it multiplies by the square of that factor.

Here: new value = $6 \times \text{old} - 5$. The “-5” is just a shift (no effect), and the “×6” is a stretch. So new variance = $6^2 \times \text{old variance}$.

Step-by-step reasoning

1. Original variance

   Let the original observations be $x_1, x_2, \dots, x_{50}$.

   Given: $\text{Var}(x) = 7$.

2. Transformation

   Each observation becomes $y_i = 6x_i - 5$.

3. Effect on variance

   Variance is invariant under translation:

because subtracting 5 doesn’t change spread.

4. Scaling property For any constant $c$, $\text{Var}(c x) = c^2 \,\text{Var}(x)$. Here $c = 6$, so

5. Conclusion The new variance is 252, which corresponds to option (D).

The mean deviation from the mean is a measure of dispersion, telling us, on average, how much the data points deviate from the central value (the mean). It helps us understand the spread or variability within a dataset.

To calculate it, we follow these steps:

1. Find the mean ($\bar{x}$) of the data. The mean is the sum of all observations divided by the number of observations.
2. Calculate the absolute deviation of each observation from the mean. For each data point $x_i$, we find $|x_i - \bar{x}|$. We use absolute values because we are interested in the magnitude of the deviation, regardless of whether the data point is above or below the mean. If we didn't use absolute values, the sum of deviations from the mean would always be zero, making it useless as a measure of spread.
3. Sum all the absolute deviations.
4. Divide the sum of absolute deviations by the total number of observations ($n$). This gives us the average absolute deviation, which is the mean deviation from the mean.

Let's apply these steps to the given data: 1, 3, 4, 7, 11, 18, 29, 47, 78.

The number of observations, $n = 9$.

1. Calculate the mean ($\bar{x}$):

   First, sum all the data points:

   $\sum x_i = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 78 = 198$

   Now, divide by the number of observations:

   $\bar{x} = \frac{\sum x_i}{n} = \frac{198}{9} = 22$

   So, the mean of the data is 22.

2. Calculate the absolute deviations from the mean ($|x_i - \bar{x}|$):

   We will find the absolute difference between each data point and the mean, 22.

   | $x_i$ | $|x_i - \bar{x}|$ = $|x_i - 22|$ |

   | :---- | :-------------------------------- |

   | 1 | $|1 - 22| = |-21| = 21$ |

   | 3 | $|3 - 22| = |-19| = 19$ |

   | 4 | $|4 - 22| = |-18| = 18$ |

   | 7 | $|7 - 22| = |-15| = 15$ |

   | 11 | $|11 - 22| = |-11| = 11$ |

   | 18 | $|18 - 22| = |-4| = 4$ |

   | 29 | $|29 - 22| = |7| = 7$ |

   | 47 | $|47 - 22| = |25| = 25$ |

   | 78 | $|78 - 22| = |56| = 56$ |

3. Sum all the absolute deviations ($\sum |x_i - \bar{x}|$):

   Sum of absolute deviations $= 21 + 19 + 18 + 15 + 11 + 4 + 7 + 25 + 56 = 176$

4. Calculate the mean deviation from the mean ($\text{MD}_{\bar{x}}$):

   $\text{MD}_{\bar{x}} = \frac{\sum |x_i - \bar{x}|}{n} = \frac{176}{9}$

Comparing this result with the given options:

(A) 22

(B) 24

(C) $\dfrac{176}{9}$

(D) $\dfrac{182}{9}$

The calculated mean deviation matches option (C).

Find $k$. The probabilities must sum to $1$:

Solving, $k=\dfrac{-6\pm\sqrt{36+28}}{14}=\dfrac{-6\pm 8}{14}$. The valid (non-negative) root is $k=\dfrac{2}{14}=\dfrac{1}{7}.$

Compute the mean.

Substitute $k=\tfrac{1}{7}$: