Class 11 Mathematics Chapter 13 Statistics

Key Concept: Basic Definition

b) The mean of absolute values of deviations from ‘a’

[Solution Description] Mean deviation about a central value ‘a’ is defined as the mean of the absolute values of the deviations of the observations from ‘a’.

Key Concept: Conceptual Analysis, Critical Thinking, Statistical Significance

d) Its inability to highlight outliers makes it statistically insignificant.

[Solution Description] In a dataset with significant outliers, mean deviation fails to account properly for extreme values because it treats all deviations equally by taking their absolute value. This limits its sensitivity to variations and often leads to underestimation of dispersion, making it less statistically significant compared to robust measures like standard deviation that square deviations, thereby highlighting variances caused by outliers.

Key Concept: Real-World Application, Advanced Calculation

a) Product A is more consistent with less variability.

[Solution Description]

To determine which product’s revenue is more consistent, we calculate the standard deviation for both products. For Product A, the standard deviation is $$\sqrt{16} = 4$$. Product B already has a standard deviation of 5.

Comparing the standard deviations:

– Product A’s standard deviation is 4.

– Product B’s standard deviation is 5.

Since Product A has a lower standard deviation compared to Product B, its revenue is more consistent, suggesting less variability.

Key Concept: Basic Formula Understanding

b) $$h^2 \left(\frac{1}{N} \sum f_i y_i^2 – \left(\frac{1}{N} \sum f_i y_i \right)^2 \right)$$

[Solution Description]

In the shortcut method, the formula for variance ($$\sigma^2$$) is given by:

$$\sigma^2 = h^2 \left(\frac{1}{N} \sum f_i y_i^2 – \left(\frac{1}{N} \sum f_i y_i \right)^2 \right)$$

where $$h$$ is the class interval width, $$f_i$$ is frequency, and $$y_i = \frac{x_i – A}{h}$$.

Key Concept: Complex Data Analysis, Historical Impact

b) Region Y

[Solution Description] Consistent sales performance means less variability in data. We use variance or standard deviation to measure this consistency.

Steps:

– Calculate the mean for both regions.

– Determine the variance for each region by finding the mean of squared deviations from the mean.

Calculation for Region X:

Mean for X ($$\bar{x}_X$$) = $$\frac{120 + 125 + 130 + 135 + 140 + 145 + 150 + 155 + 160 + 165}{10} = 142.5$$

Variance for X ($$\sigma_X^2$$):

\begin{align*}

\sigma_X^2 &= \frac{(120-142.5)^2 + (125-142.5)^2 + \ldots +(165-142.5)^2 }{10}

&= \frac{5062.5}{10} = 506.25

\end{align*}

Calculation for Region Y:

Mean for Y ($$\bar{x}_Y$$) = $$\frac{100 + 100 + 105 + 110 + 115 + 120 + 125 + 130 + 135 + 140}{10} = 118$$

Variance for Y ($$\sigma_Y^2$$):

\begin{align*}

\sigma_Y^2 &= \frac{(100-118)^2 + (100-118)^2 + \ldots + (140-118)^2}{10}

&= \frac{1225}{10} = 122.5

\end{align*}

Since Region Y has a smaller variance than Region X, Region Y shows more consistent sales performance.

Key Concept: Complex Variance Calculation, Real-world Application

c) 47.34

[Solution Description]

To calculate the variance $$\sigma^2 = \frac{1}{N} \sum f_i (x_i – \bar{x})^2$$, we first need to find the mid-point for each class interval:

$$x_1 = \frac{10 + 20}{2} = 15$$, $$x_2 = \frac{20 + 30}{2} = 25$$, $$x_3 = \frac{30 + 40}{2} = 35$$

Now calculate the total frequency ($$N$$):

$$N = 5 + 15 + 10 = 30$$

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

$$\bar{x} = \frac{\sum f_ix_i}{N} = \frac{(5 \times 15) + (15 \times 25) + (10 \times 35)}{30} = \frac{75 + 375 + 350}{30} = 26.67$$

Using these, we calculate the variance:

$$\sigma^2 = \frac{1}{30} [(5 \times (15-26.67)^2) + (15 \times (25-26.67)^2) + (10 \times (35-26.67)^2)]$$

$$= \frac{1}{30} [(5 \times 136.89) + (15 \times 2.78) + (10 \times 69.44)] = \frac{1}{30} [684.45 + 41.7 + 694.4] = 47.34$$

Hence, the variance is approximately $$47.34$$.

Key Concept: Variance and Standard Deviation Relationship, Dispersion in Data Sets

b) Portfolio B

[Solution Description] Standard deviation is the square root of variance. We need to calculate the standard deviation for each portfolio.

For Portfolio A: $$\sigma = \sqrt{49} = 7$$,

For Portfolio B: $$\sigma = \sqrt{64} = 8$$,

For Portfolio C: $$\sigma = \sqrt{36} = 6$$.

Thus, Portfolio B, with a standard deviation of 8, has the highest dispersion among the portfolios.

Key Concept: Basic Concept

c) When the degree of variability is very high

[Solution Description] In cases where there is high variability in the series, the median fails to represent the central tendency accurately because it does not reflect the spread or scale of the data.

Key Concept: Real-world Application, Complex Grouped Data

b) 9.69

[Solution Description]

To find the mean deviation about the median:

Step 1: Calculate the cumulative frequency for each class interval and find the median class. The cumulative frequencies are: 5, 17, 32, 42, 50. Since there are 50 employees, the median position is $$\frac{50}{2} = 25$$. Therefore, the median class is 30-40 minutes.

Step 2: Use the median formula for grouped data:

$$\text{Median} = L + \left( \frac{\frac{N}{2} – CF}{f} \right) \times h$$

where $$L = 30$$, $$N = 50$$, $$CF = 17$$ (cumulative frequency before median class), $$f = 15$$, and $$h = 10$$ (class width).

$$\text{Median} = 30 + \left( \frac{25 – 17}{15} \right) \times 10$$

$$\text{Median} = 30 + \left(\frac{8}{15}\right) \times 10 = 30 + 5.33 = 35.33 \text{ minutes}$$

Step 3: Calculate deviations from the median and their absolute values, then multiply by frequencies:

For 10-20: $$(\mid 15 – 35.33 \mid \times 5) = (20.33 \times 5) = 101.65$$

For 20-30: $$(\mid 25 – 35.33 \mid \times 12) = (10.33 \times 12) = 123.96$$

For 30-40: $$(\mid 35 – 35.33 \mid \times 15) = (0.33 \times 15) = 4.95$$

For 40-50: $$(\mid 45 – 35.33 \mid \times 10) = (9.67 \times 10) = 96.7$$

For 50-60: $$(\mid 55 – 35.33 \mid \times 8) = (19.67 \times 8) = 157.36$$

Total sum = $$484.62$$

Step 4: Mean Deviation = $$\frac{\text{Total sum}}{N} = \frac{484.62}{50}= 9.69$$

Hence, the mean deviation about the median is approximately $$9.69$$ minutes.

Key Concept: Real-World Application, Advanced Deviation Calculation

b) There is significant variation in the product weights.

[Solution Description] The given mean deviation indicates that on average, the weights of the products deviate from the mean weight (55 kg) by 4 kg. This suggests that there is some degree of variability in the production process. Since mean deviation uses absolute values, it provides an unbiased measure of dispersion around the mean, allowing the company to understand the spread of product weights.

Key Concept: Complex Grouped Data, Advanced Calculation

a) 9.29

[Solution Description]

To calculate the mean deviation about the mean using the step-deviation method, follow these steps:

Step 1: Calculate the midpoints ($$x_i$$) for each class interval:

$$\text{Midpoints: } 10, 20, 30, 40, 50$$

Step 2: Choose an assumed mean $$A = 30$$, and calculate deviations from the assumed mean:

$$d_i = \frac{x_i – A}{c}$$

where $$c = 10$$.

Deviations:

$$d_i = -2, -1, 0, 1, 2$$

Step 3: Multiply each deviation by its corresponding frequency to find $$f_id_i$$:

$$f_id_i = -6, -7, 0, 8, 10$$

Step 4: Sum up all $$f_id_i$$:

$$\sum f_id_i = -6 + (-7) + 0 + 8 + 10 = 5$$

Step 5: Calculate the mean using the formula:

$$\bar{x} = A + c \left(\frac{\sum f_id_i}{\sum f_i}\right)$$

$$\bar{x} = 30 + 10 \left(\frac{5}{35}\right) = 31.43$$

Step 6: Find absolute deviations from the calculated mean:

$$|x_i – \bar{x}| = |10 – 31.43|, |20 – 31.43|, |30 – 31.43|, |40 – 31.43|, |50 – 31.43|$$

$$= 21.43, 11.43, 1.43, 8.57, 18.57$$

Step 7: Multiply the absolute deviations by their frequencies and sum them:

$$\sum f_i|x_i – \bar{x}| = 3 \times 21.43 + 7 \times 11.43 + 12 \times 1.43 + 8 \times 8.57 + 5 \times 18.57 = 325.29$$

Step 8: Divide by total frequency to find the mean deviation:

$$M.D.(\bar{x}) = \frac{325.29}{35} = 9.29$$

Therefore, the mean deviation about the mean is approximately 9.29.

Key Concept: Ungrouped Data Application

b) 3.84

[Solution Description]

To find the mean deviation about the mean, we first calculate the mean of the data:

Mean $$x = \frac{6 + 8 + 10 + 14 + 18}{5} = \frac{56}{5} = 11.2$$.

Now, we find the deviations from the mean and their absolute values:

\begin{align*}

|6 – 11.2| &= 5.2

|8 – 11.2| &= 3.2

|10 – 11.2| &= 1.2

|14 – 11.2| &= 2.8

|18 – 11.2| &= 6.8

\end{align*}

The sum of these absolute deviations is:

$$5.2 + 3.2 + 1.2 + 2.8 + 6.8 = 19.2$$

Thus, the mean deviation about the mean is:

$$M.D. (x) = \frac{19.2}{5} = 3.84$$

Key Concept: Comparative Analysis, Advanced Reasoning

a) 5.66

[Solution Description]

Given that the variance of data set C is twice that of D, we know:

Let $$Var(D) = \sigma_D^2 = 16$$ (since $$\sigma_D = 4$$).

So, $$Var(C) = 2 \times Var(D) = 2 \times 16 = 32$$.

The standard deviation of data set C is the square root of its variance:

$$\sigma_C = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \approx 5.66$$

Key Concept: Central Tendency

a) Mean

[Solution Description] The measure of central tendency used when calculating mean deviation about the mean is the “mean” itself.

Key Concept: Variance Calculation

c) 61

[Solution Description]

To calculate the variance, first find the midpoints of each class interval and then use the variance formula.

Midpoints are calculated as follows:

– For 20-30: $$\frac{20 + 30}{2} = 25$$

– For 30-40: $$\frac{30 + 40}{2} = 35$$

– For 40-50: $$\frac{40 + 50}{2} = 45$$

Let $$x_i$$ be the midpoints. Now, calculate the mean ($$\bar{x}$$).

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{(4 \times 25) + (6 \times 35) + (10 \times 45)}{4 + 6 + 10}$$

$$\bar{x} = \frac{100 + 210 + 450}{20} = \frac{760}{20} = 38$$

Now, calculate the variance:

$$\sigma^2 = \frac{1}{N} \sum_{i=1}^{3} f_i (x_i – \bar{x})^2$$

$$= \frac{1}{20} [(4 \times (25 – 38)^2) + (6 \times (35 – 38)^2) + (10 \times (45 – 38)^2)]$$

$$= \frac{1}{20} [(4 \times 169) + (6 \times 9) + (10 \times 49)]$$

$$= \frac{1}{20} [676 + 54 + 490] = \frac{1220}{20} = 61$$

The variance is 61.

Key Concept: Complex Mean Deviation, Real-world Application

d) 45.34

[Solution Description]

The first step is to find the cumulative frequencies for each class interval:

\begin{align*}

&F_1 = 8

&F_2 = 8 + 12 = 20

&F_3 = 20 + 20 = 40

&F_4 = 40 + 10 = 50

&F_5 = 50 + 5 = 55

\end{align*}

Total frequency $$N$$ is 55. The median class is identified where $$\frac{N}{2} = 27.5$$, which lies within the third class (mid-point 150).

Using interpolation,

$$L = 125,\ n = 20,\ f_m = 20$$

Median ($$M$$) calculation involves

$$M = L + \left(\frac{\frac{N}{2} – F_2}{f_m}\right) \times c$$

Substituting values,

\begin{align*}

M &= 125 + \left(\frac{27.5 – 20}{20}\right) \times 50

&= 125 + 18.75

&= 143.75

\end{align*}

Now calculate the mean deviation using

$$M.D.(M) = \frac{\sum f_i |x_i – M|}{N}$$

Calculation of absolute deviations:

\begin{align*}

&|50-143.75| = 93.75

&|100-143.75| = 43.75

&|150-143.75| = 6.25

&|200-143.75| = 56.25

&|250-143.75| = 106.25

\end{align*}

Frequency weighted deviations sum is

$$\sum f_i |x_i – M| = (8 \times 93.75) + (12 \times 43.75) + (20 \times 6.25) + (10 \times 56.25) + (5 \times 106.25)$$

Evaluating this, we get

\begin{align*}

&750 + 525 + 125 + 562.5 + 531.25 = 2493.75

\end{align*}

Finally, divide by total frequency:

$$M.D.(M) = \frac{2493.75}{55} = 45.34$$

Key Concept: Concept Understanding

b) Absolute deviations avoid cancellation of positive and negative deviations.

Absolute deviations make it easier to calculate a standard deviation.

Key Concept: Real-World Application, Complex Data Sets

a) Portfolio A

[Solution Description] To find which portfolio has a lower mean deviation about the median, we first calculate the median for both data sets.

**For Portfolio A:**

Ordered: -2, -1, 3, 4, 5, 6, 7, 8, 9, 10.

Median = average of 5th and 6th values = $$\frac{5 + 6}{2} = 5.5$$.

Absolute deviations: $$|4 – 5.5|, |8 – 5.5|, |3 – 5.5|, |7 – 5.5|, |-2 – 5.5|, |5 – 5.5|, |6 – 5.5|, |10 – 5.5|, |-1 – 5.5|, |9 – 5.5|$$.

Mean deviation: $$\frac{1.5 + 2.5 + 2.5 + 1.5 + 7.5 + 0.5 + 0.5 + 4.5 + 6.5 + 3.5}{10} = \frac{31.5}{10} = 3.15$$.

**For Portfolio B:**

Ordered: -25, -20, -15, -10, -5, 15, 20, 25, 30, 35.

Median = average of 5th and 6th values = $$\frac{-5 + 15}{2} = 5$$.

Absolute deviations: $$|15 – 5|, |-5 – 5|, |20 – 5|, |-10 – 5|, |25 – 5|, |-15 – 5|, |30 – 5|, |-20 – 5|, |35 – 5|, |-25 – 5|$$.

Mean deviation: $$\frac{10 + 10 + 15 + 15 + 20 + 20 + 25 + 25 + 30 + 30}{10} = \frac{190}{10} = 19.0$$.

Comparing these results, Portfolio A has a lower mean deviation about the median, indicating more stable returns.

Key Concept: Basic Calculation

a) 3

[Solution Description] To calculate the mean deviation about the mean:

– Calculate the mean of the data set.

$$\text{Mean} = \frac{3 + 6 + 9 + 12}{4} = \frac{30}{4} = 7.5$$

– Find the absolute deviations from the mean.

$$|3 – 7.5| = 4.5, \quad |6 – 7.5| = 1.5, \quad |9 – 7.5| = 1.5, \quad |12 – 7.5| = 4.5$$

– Mean deviation about the mean is calculated as:

$$M.D.(x) = \frac{1}{4} (4.5 + 1.5 + 1.5 + 4.5) = \frac{12}{4} = 3$$

The mean deviation about the mean is 3.

Key Concept: Theoretical Implications, Advanced Calculation

b) 3.67

[Solution Description]

The mean deviation cannot be algebraically treated further due to its reliance on absolute deviations rather than squared deviations, unlike variance or standard deviation which handle algebraic expressions better.

To find the mean deviation for the given data:

– Calculate the mean ($$\bar{x}$$):

$$\bar{x} = \frac{\sum x_i}{n} = \frac{7 + 9 + 10 + 13 + 17 + 20}{6} = 12.67$$

– Calculate absolute deviations from the mean:

$$|x_i – \bar{x}|: \{|7-12.67|, |9-12.67|, |10-12.67|,

|13-12.67|, |17-12.67|, |20-12.67|\}$$

$$= \{5.67, 3.67, 2.67, 0.33, 4.33, 7.33\}$$

– Calculate the mean deviation:

$$M.D.(\bar{x}) = \frac{1}{6} \left(5.67 + 3.67 + 2.67 + 0.33 + 4.33 + 7.33\right) = 3.67$$

Therefore, the mean deviation for the given dataset is 3.67.

Key Concept: Theoretical Development, Real-World Application

c) 17th century

[Solution Description]

The field of statistics started developing theoretically during the mid-seventeenth century. This period marks the beginning of systematic collection and analysis of data, laying the foundation for modern statistical methods. Over time, these developments led to numerous applications in biology, agriculture, and other areas by figures such as Sir Ronald A. Fisher in the early 20th century.

Therefore, it was the seventeenth century that marked the beginning of theoretical development in statistics.

Key Concept: Deriving Formulas, Advanced Reasoning

c) 12

[Solution Description]

The formula to derive standard deviation from variance is:

$$\sigma = \sqrt{Variance}$$

Substituting the given variance:

$$\sigma = \sqrt{144} = 12$$

Therefore, the correct representation of standard deviation is 12.

Key Concept: Flexibility in Products

b) General stores

[Solution Description] General stores offer flexibility because they stock various everyday products, allowing them to adapt and vary their product lines as needed for daily consumer demands.

Key Concept: Median Calculation

c) 22

[Solution Description] To find the median of an ordered data set, for an odd number of observations, we use the formula for the median position: $$\left(\frac{n+1}{2}\right)^{th}$$ observation. Here, the number of observations $$n=7$$, which is odd. The median is at position $$\left(\frac{7+1}{2}\right)=4$$. Thus, the median value is the 4th observation in the ordered list, which is 22.

Key Concept: Real-World Application, Probability Distribution

c) $$\frac{47}{66}$$

[Solution Description]

Total number of ways to draw two balls from 12 is given by combination:

$$n(S) = \binom{12}{2} = \frac{12 \times 11}{2} = 66$$

For the drawing of two balls of the same color, consider each color:

– Two red: $$\binom{5}{2} = 10$$

– Two green: $$\binom{4}{2} = 6$$

– Two blue: $$\binom{3}{2} = 3$$

Total for same color:

$$n(A) = 10 + 6 + 3 = 19$$

So, the probability of drawing two balls of different colors:

$$P(\text{different color}) = 1 – \frac{n(A)}{n(S)} = 1 – \frac{19}{66} = \frac{47}{66}$$

Key Concept: Complex Calculation, Problem Solving

c) 3.238

[Solution Description] To find the standard deviation, follow these steps:

– Calculate the mean (\$ \bar{x} \$):

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{(3 \times 4) + (6 \times 5) + (9 \times 7) + (12 \times 8)}{4 + 5 + 7 + 8}

= \frac{12 + 30 + 63 + 96}{24} = \frac{201}{24} = 8.375$$

– Use the variance formula:

$$\sigma^2 = \frac{\sum f_i (x_i – \bar{x})^2}{N}

= \frac{4(3 – 8.375)^2 + 5(6 – 8.375)^2 + 7(9 – 8.375)^2 + 8(12 – 8.375)^2}{24}

= \frac{4(28.890625) + 5(5.640625) + 7(0.390625) + 8(13.140625)}{24}

= \frac{115.5625 + 28.203125 + 2.734375 + 105.125}{24} = \frac{251.625}{24} = 10.484375$$

– Calculate the standard deviation:

$$\sigma = \sqrt{10.484375} = 3.238$$

Thus, the standard deviation is approximately 3.238.

Key Concept: Mean Deviation Calculation

b) 4.8

[Solution Description]

To find the mean deviation about the median:

– First, order the data: $$3, 7, 11, 15, 19$$

– The median is the middle value: $$11$$

– Calculate the absolute deviations from the median:

$$|3 – 11| = 8$$, $$|7 – 11| = 4$$, $$|11 – 11| = 0$$, $$|15 – 11| = 4$$, $$|19 – 11| = 8$$

– Find the mean of these absolute deviations:

$$\text{Mean Deviation} = \frac{(8 + 4 + 0 + 4 + 8)}{5} = \frac{24}{5} = 4.8$$

Key Concept: Complex Mean Deviation, Standard Deviation Application

c) 6

[Solution Description]

First, we calculate the median of the dataset: 5, 10, 15, 20, 25.

Since there are five numbers in ascending order, the median is the middle number, which is 15.

Next, calculate the deviations from the median:

$$|5 – 15| = 10, |10 – 15| = 5, |15 – 15| = 0, |20 – 15| = 5, |25 – 15| = 10$$

Now, find the average of these absolute deviations:

$$\frac{(10 + 5 + 0 + 5 + 10)}{5} = \frac{30}{5} = 6$$

Thus, the mean deviation about the median is 6. This value indicates the average distance of each score from the median. Compared to the standard deviation (approximately 7.07), the mean deviation shows a relatively similar spread around the median, considering they measure centrality differently.

Key Concept: Real-World Application, Theoretical Application

c) 140.36

[Solution Description] To compute the variance using the shortcut method:

– Assume the mean $$A$$ as the midpoint of the second interval: $$71$$.

– Calculate midpoints: 60, 71, 82, 93.

– Deviations $$d_i$$: For 60, $$d_1 = -11$$, for 71, $$d_2 = 0$$, for 82, $$d_3 = 11$$, for 93, $$d_4 = 22$$.

– Frequencies $$f_i$$ are already known.

– Step deviations $$u_i$$: $$u_1 = -1$$, $$u_2 = 0$$, $$u_3 = 1$$, $$u_4 = 2$$.

– Calculate product of frequency and step deviation squared: $$f_iu_i^2$$ gives $$9(1)$$, $$13(0)$$, $$21(1)$$, $$7(4)$$ resulting in $$9 + 0 + 21 + 28 = 58$$.

– Sum frequencies $$\sum f_i = 50$$.

– Use formula $$\sigma_x^2 = h^2 \frac{\sum f_i u_i^2}{\sum f_i}$$.

– Substitute values: $$\sigma_x^2 = 11^2 \times \frac{58}{50} = 121 \times 1.16 = 140.36$$.

Key Concept: Application of Formula

c) 42

[Solution Description]

To calculate the mean, we use the formula $$x = a + h \sum \frac{f_i d_i}{N}$$. Here, $$a = 30$$, $$h = 4$$, $$\sum f_i d_i = 36$$, and $$N = 12$$.

We substitute these values into the formula:

$$x = 30 + 4 \cdot \left(\frac{36}{12}\right)$$

Simplifying inside the parentheses first:

$$x = 30 + 4 \cdot 3$$

Then multiply:

$$x = 30 + 12 = 42$$

Thus, the calculated mean is 42.

Key Concept: Real-World Application, Advanced Formula Application

c) \$8,400.5

[Solution Description]

To calculate the mean deviation about the median salary, we follow these steps:

1. Determine the median interval using cumulative frequency:

The total number of employees is $$5 + 10 + 15 + 20 = 50$$.

The median position is $$\frac{50+1}{2} = 25.5$$.

2. Identify the median class:

Accumulating the frequencies, we get:

– Up to 40,000, there are 5 employees.

– Up to 50,000, there are 5 + 10 = 15 employees.

– Up to 60,000, there are 15 + 15 = 30 employees.

Therefore, the median class is 50,001-60,000.

3. Use the formula for the median in a grouped data set:

$$\text{Median} = L + \left( \frac{\frac{N}{2} – F}{f} \right) \times h$$

where $$L$$ is the lower boundary of the median class, $$F$$ is the cumulative frequency before the median class, $$f$$ is the frequency of the median class, and $$h$$ is the class width.

Here, $$L = 50000.5$$, $$F = 15$$, $$f = 15$$, and $$h = 10000$$.

$$\text{Median} = 50000.5 + \left( \frac{25.5 – 15}{15} \right) \times 10000 = 50000.5 + \left( \frac{10.5}{15} \right) \times 10000$$

$$= 50000.5 + 7000 = 57000.5$$

4. Compute absolute deviations from the median for each class midpoint:

– Class 30,000-40,000: Midpoint = 35000, Absolute deviation = |35000 – 57000.5| = 22000.5

– Class 40,001-50,000: Midpoint = 45000, Absolute deviation = |45000 – 57000.5| = 12000.5

– Class 50,001-60,000: Midpoint = 55000, Absolute deviation = |55000 – 57000.5| = 2000.5

– Class 60,001-70,000: Midpoint = 65000, Absolute deviation = |65000 – 57000.5| = 8000.5

5. Calculate mean deviation:

Using frequencies for weighted sum:

Mean Deviation =

$$\frac{(22000.5 \times 5) + (12000.5 \times 10) + (2000.5 \times 15) + (8000.5 \times 20)}{50}$$

$$= \frac{110002.5 + 120005 + 30007.5 + 160010}{50} = \frac{420025}{50} = 8400.5$$

Hence, the mean deviation about the median is \$8,400.5.

Key Concept: Real-world Application of Range, Range and Data Interpretation

c) City B has greater income inequality than City A.

[Solution Description] The range for City A is computed as \$150,000 – \$40,000 = \$110,000. For City B, the range is \$200,000 – \$50,000 = \$150,000. A larger range suggests more significant income disparity or inequality within the city. Since City B’s range is greater, it implies that there is greater income inequality in City B compared to City A.

Key Concept: Grouped Data Calculation

b) 8

[Solution Description] To find the mean deviation about the mean:

Step 1: Calculate the midpoints ($$x_i$$) of each class interval:

\begin{align*}

&10{-}20: \frac{10+20}{2} = 15

&20{-}30: \frac{20+30}{2} = 25

&30{-}40: \frac{30+40}{2} = 35

&40{-}50: \frac{40+50}{2} = 45

\end{align*}

Step 2: Calculate the sum of frequencies $$N = 4 + 6 + 8 + 2 = 20$$.

Step 3: Calculate the mean $$\bar{x}$$ using the formula:

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{(4 \times 15) + (6 \times 25) + (8 \times 35) + (2 \times 45)}{20} = \frac{60 + 150 + 280 + 90}{20} = \frac{580}{20} = 29$$

Step 4: Find the absolute deviations $$|x_i – \bar{x}|$$ and multiply by their frequencies:

\begin{align*}

|15-29| \cdot 4 &= 14 \cdot 4 = 56

|25-29| \cdot 6 &= 4 \cdot 6 = 24

|35-29| \cdot 8 &= 6 \cdot 8 = 48

|45-29| \cdot 2 &= 16 \cdot 2 = 32

\end{align*}

Step 5: Calculate the mean deviation about the mean:

$$M.D.(x) = \frac{\sum f_i |x_i – \bar{x}|}{N} = \frac{56 + 24 + 48 + 32}{20} = \frac{160}{20} = 8$$

Key Concept: Formula Identification

c) $$\sigma^2 = \frac{\sum f_i (x_i – \bar{x})^2}{N}$$

[Solution Description] The formula for variance in a frequency distribution is $$\sigma^2 = \frac{\sum f_i (x_i – \bar{x})^2}{N}$$, where $$f_i$$ represents the frequency of each data point, $$x_i$$ represents each data point, and $$N$$ is the total number of observations.

Key Concept: Advanced Variance Calculation, Impact of Assumed Mean and Scale

c) 8

[Solution Description]

First, recognize that reducing the class interval halves $$h$$ from 10 to 5.

The formula for variance $$\sigma^2$$ using step-deviation is:

$$\sigma^2 = \left(\frac{h}{N}\right)^2 \sum {f_i y_i^2}$$

Given $$h = 5$$, $$N = 50$$, and $$\sum f_i y_i^2 = 800$$:

$$\sigma^2 = \left(\frac{5}{50}\right)^2 \cdot 800$$

Calculate the inner expression:

$$\left(\frac{5}{50}\right)^2 \cdot 800 = \frac{1}{100} \cdot 800 = 8$$

Therefore, the variance is 8.

Key Concept: Correcting Mean

d) $$20 + \frac{10}{n}$$

[Solution Description] The initial sum of all observations can be found by multiplying the mean with the number of observations, assuming $$n$$ to be the number of observations:

$$\text{Sum} = 20n$$. Correcting the errors in data, we need to add $$10 + 10 = 20$$ to the total sum since each value was under-recorded by 10. The new mean is calculated as:

$$\text{New Sum} = 20n + 20$$

$$\text{New Mean} = \frac{\text{New Sum}}{n} = \frac{20n + 20}{n} = 20 + \frac{20}{n}$$

Without knowing $$n$$, the new mean will be $$20 + \frac{20}{n}$$.

Key Concept: Advanced Frequency Distribution, Comparative Analysis

a) Dataset A

[Solution Description]

For Dataset A:

Calculating the mean, $$\bar{x}$$:

$$\bar{x}_A = \frac{50(4) + 60(3) + 70(3)}{10} = \frac{550}{10} = 55.$$

Variance calculation:

\begin{align*}

\sum f(x_i – \bar{x})^2_A &= 4(50 – 55)^2 + 3(60 – 55)^2 + 3(70 – 55)^2

&= 4(25) + 3(25) + 3(225) = 100 + 75 + 675 = 850.

\end{align*}

Standard deviation $$\sigma_A$$:

$$\sigma_A = \sqrt{\frac{850}{10}} = \sqrt{85} = 9.22.$$

For Dataset B:

Calculating the mean, $$\bar{x}$$:

$$\bar{x}_B = \frac{45(2) + 55(5) + 65(3)}{10} = \frac{565}{10} = 56.5.$$

Variance calculation:

\begin{align*}

\sum f(x_i – \bar{x})^2_B &= 2(45 – 56.5)^2 + 5(55 – 56.5)^2 + 3(65 – 56.5)^2

&= 2(132.25) + 5(2.25) + 3(72.25) = 264.5 + 11.25 + 216.75 = 492.5.

\end{align*}

Standard deviation $$\sigma_B$$:

$$\sigma_B = \sqrt{\frac{492.5}{10}} = \sqrt{49.25} = 7.01.$$

Comparing $$\sigma_A$$ and $$\sigma_B$$, Dataset A has a higher standard deviation than Dataset B, indicating more variability.

Key Concept: Continuous Frequency Distribution, Real-world Application

b) 87.75

[Solution Description]

To find the variance for the given continuous frequency distribution, we start by identifying the midpoints ($$m_i$$) of each class interval. The midpoints are calculated as follows:

\begin{align*}

m_1 &= \frac{140 + 150}{2} = 145,

m_2 &= \frac{150 + 160}{2} = 155,

m_3 &= \frac{160 + 170}{2} = 165,

m_4 &= \frac{170 + 180}{2} = 175,

m_5 &= \frac{180 + 190}{2} = 185.

\end{align*}

Next, calculate the mean $$\bar{x}$$ using the formula:

$$\bar{x} = \frac{\sum f_i m_i}{N} = \frac{5 \times 145 + 45 \times 155 + 80 \times 165 + 55 \times 175 + 15 \times 185}{200}$$

Solving further:

\begin{align*}

\bar{x} &= \frac{725 + 6975 + 13200 + 9625 + 2775}{200},

&= \frac{33300}{200},

&= 166.5.

\end{align*}

Now, use the variance formula for continuous frequency distribution:

$$\sigma^2 = \frac{\sum f_i (x_i – \bar{x})^2}{N}$$

Calculating $$(x_i – \bar{x})^2$$ and then $$\sum f_i (x_i – \bar{x})^2$$:

\begin{align*}

&f_1(145 – 166.5)^2 = 5 \times 462.25 = 2311.25,

&f_2(155 – 166.5)^2 = 45 \times 132.25 = 5951.25,

&f_3(165 – 166.5)^2 = 80 \times 2.25 = 180,

&f_4(175 – 166.5)^2 = 55 \times 72.25 = 3973.75,

&f_5(185 – 166.5)^2 = 15 \times 342.25 = 5133.75.

\end{align*}

Summing up all these products gives us:

$$\sum f_i (x_i – \bar{x})^2 = 2311.25 + 5951.25 + 180 + 3973.75 + 5133.75 = 17550$$

Finally, calculating the variance:

$$\sigma^2 = \frac{17550}{200} = 87.75$$

So the variance for the distribution is 87.75.

Key Concept: Range and Data Transformation, Real-world Application of Range

c) 69g

[Solution Description]

Original weights: 540g, 550g, 570g, 590g, 600g

Step 1: Calculate the original range.

$$\text{Original Range} = \text{Max} – \text{Min} = 600g – 540g = 60g$$

Step 2: Increase each weight by 15\%.

New weights:

\begin{align*}

540g \times 1.15 &= 621g

550g \times 1.15 &= 632.5g

570g \times 1.15 &= 655.5g

590g \times 1.15 &= 678.5g

600g \times 1.15 &= 690g

\end{align*}

Step 3: Calculate the new range.

$$\text{New Range} = 690g – 621g = 69g$$

Therefore, the new range is 69g.

Key Concept: Variance Calculation, Real-world Application of Dispersion

c) 7291.67

[Solution Description]

To calculate the variance, we first find the mean ($$\bar{x}$$) of the sales data:

$$\bar{x} = \frac{200 + 250 + 300 + 350 + 400 + 450}{6} = 325$$

Next, calculated each squared deviation from the mean:

$$(200 – 325)^2 = 15625,\quad (250 – 325)^2 = 5625,\quad (300 – 325)^2 = 625$$

$$(350 – 325)^2 = 625,\quad (400 – 325)^2 = 5625,\quad (450 – 325)^2 = 15625$$

Sum these squared deviations:

$$\sum (x_i – \bar{x})^2 = 15625 + 5625 + 625 + 625 + 5625 + 15625 = 43750$$

Then, the variance ($$\sigma^2$$) is:

$$\sigma^2 = \frac{43750}{6} \approx 7291.67$$

The variance of approximately 7291.67 indicates a high variability in the monthly sales, showing inconsistency over time.

Key Concept: Limitation Recall

b) It only considers extreme values

[Solution Description] Range is considered a limited measure of dispersion because it only considers extreme values, i.e., the maximum and minimum values. It does not account for all data points, which can provide more accurate insights into data variability.