BCOC-134 Solved Assignment 2025

Answer:

• For 0 to 20 items, the cost is Rs. 25 per item with no rebate.
• For 21 to 50 items, the cost is Rs. 25 per item minus a rebate of Re. 1 per item.
• For more than 50 items, the cost is Rs. 25 per item minus a rebate of Rs. 2 per item.

1. Cost function $C(x)$$C(x)$C(x)C(x)$C(x)$:
   • For $0<x\le 20$$0<x\le 20$0 < x <= 200 < x \leq 20$0<x\le 20$: $C(x)=25x$$C(x)=25x$C(x)=25 xC(x) = 25x$C(x)=25x$ (no rebate).
   • For $21\le x\le 50$$21\le x\le 50$21 <= x <= 5021 \leq x \leq 50$21\le x\le 50$: $C(x)=25x-1x=24x$$C(x)=25x-1x=24x$C(x)=25 x-1x=24 xC(x) = 25x – 1x = 24x$C(x)=25x-1x=24x$ (Re. 1 rebate per item).
   • For $x>50$$x>50$x > 50x > 50$x>50$: $C(x)=25x-2x=23x$$C(x)=25x-2x=23x$C(x)=25 x-2x=23 xC(x) = 25x – 2x = 23x$C(x)=25x-2x=23x$ (Rs. 2 rebate per item).
   So, the piecewise cost function is:
   $$C(x)=\{\begin{array}{ll}25x& \text{if}0<x\le 20,\\ 24x& \text{if}21\le x\le 50,\\ 23x& \text{if}x>50.\end{array}$$$$C(x)=\left\{\begin{array}{l}25x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }0<x\le 20,\\ 24x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }21\le x\le 50,\\ 23x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }x>50.\end{array}\right.$$C(x)={[25 x,”if “0 < x <= 20″,”],[24 x,”if “21 <= x <= 50″,”],[23 x,”if “x > 50.]:}C(x) = \begin{cases} 25x & \text{if } 0 < x \leq 20, \\ 24x & \text{if } 21 \leq x \leq 50, \\ 23x & \text{if } x > 50. \end{cases}$$C(x)=\{\begin{array}{ll}25x& \text{if }0<x\le 20,\\ 24x& \text{if }21\le x\le 50,\\ 23x& \text{if }x>50.\end{array}$$
2. Points of discontinuity:
   • A function is discontinuous where the value jumps at the boundaries between intervals. We need to check the left-hand limit and right-hand limit at $x=20$$x=20$x=20x = 20$x=20$ and $x=50$$x=50$x=50x = 50$x=50$.
   • At $x=20$$x=20$x=20x = 20$x=20$:
     • Left limit (as $x\to {20}^{-}$$x\to {20}^{-}$x rarr20^(-)x \to 20^-$x\to {20}^{-}$): $C(20)=25\cdot 20=500$$C(20)=25\cdot 20=500$C(20)=25*20=500C(20) = 25 \cdot 20 = 500$C(20)=25\cdot 20=500$.
     • Right limit (as $x\to {20}^{+}$$x\to {20}^{+}$x rarr20^(+)x \to 20^+$x\to {20}^{+}$): $C(x)=24x$$C(x)=24x$C(x)=24 xC(x) = 24x$C(x)=24x$, so $C(21)=24\cdot 21=504$$C(21)=24\cdot 21=504$C(21)=24*21=504C(21) = 24 \cdot 21 = 504$C(21)=24\cdot 21=504$.
     • Since $500\ne 504$$500\ne 504$500!=504500 \neq 504$500\ne 504$, there is a discontinuity at $x=20$$x=20$x=20x = 20$x=20$.
   • At $x=50$$x=50$x=50x = 50$x=50$:
     • Left limit (as $x\to {50}^{-}$$x\to {50}^{-}$x rarr50^(-)x \to 50^-$x\to {50}^{-}$): $C(50)=24\cdot 50=1200$$C(50)=24\cdot 50=1200$C(50)=24*50=1200C(50) = 24 \cdot 50 = 1200$C(50)=24\cdot 50=1200$.
     • Right limit (as $x\to {50}^{+}$$x\to {50}^{+}$x rarr50^(+)x \to 50^+$x\to {50}^{+}$): $C(x)=23x$$C(x)=23x$C(x)=23 xC(x) = 23x$C(x)=23x$, so $C(51)=23\cdot 51=1173$$C(51)=23\cdot 51=1173$C(51)=23*51=1173C(51) = 23 \cdot 51 = 1173$C(51)=23\cdot 51=1173$.
     • Since $1200\ne 1173$$1200\ne 1173$1200!=11731200 \neq 1173$1200\ne 1173$, there is a discontinuity at $x=50$$x=50$x=50x = 50$x=50$.
   The points at which the cost function is not continuous are $x=20$$x=20$x=20x = 20$x=20$ and $x=50$$x=50$x=50x = 50$x=50$.

A stereo manufacturer determines that in order to sell x units of a new stereo, the price per unit, in rupees, must be $p(x)=1000-x$$p(x)=1000-x$p(x)=1000-xp(x) = 1000 – x$p(x)=1000-x$. The manufacturer also determines that the total cost of producing x units is given by $C(x)=3000+20x$$C(x)=3000+20x$C(x)=3000+20 xC(x) = 3000 + 20x$C(x)=3000+20x$.

Answer:

A sum of money is deposited by Krishna which compounds interest annually. The amount at the end of 2 years is Rs. 5000 and at the end of 3 years is 5200. Find the money deposited and the rate of interest.

Answer:

• After 2 years, the amount is Rs. 5000: $P(1+r{)}^2=5000$$P(1+r{)}^2=5000$P(1+r)^(2)=5000P(1 + r)^2 = 5000$P(1+r{)}^2=5000$
• After 3 years, the amount is Rs. 5200: $P(1+r{)}^3=5200$$P(1+r{)}^3=5200$P(1+r)^(3)=5200P(1 + r)^3 = 5200$P(1+r{)}^3=5200$

The profits (in Rs. Lakhs) earned by 100 companies during the 1987-88 are shown below. Compute (a) Mean, (B) Variance, and (c) Standard deviation by using items and their squares.

Answer:

Step 1: Calculate Midpoints and Totals

(a) Mean

(b) Variance

(c) Standard Deviation

Final Answer

What do you mean by statistics? Explain its importance to economics and business. Also discuss the various functions of statistics.

Answer:

Importance to Economics and Business

• Economics: Statistics is crucial for analyzing economic data such as GDP, inflation rates, and unemployment. It helps in forecasting economic trends, testing economic theories, and formulating policies based on empirical evidence.
• Business: In business, statistics aids in market research, demand forecasting, quality control, and financial analysis. It enables businesses to make informed decisions, optimize operations, and assess risks and profitability.

Functions of Statistics

1. Data Collection: Gathering reliable data from various sources.
2. Data Organization: Arranging data in tables, charts, or graphs for clarity.
3. Data Analysis: Using measures like averages and correlations to interpret data.
4. Data Interpretation: Drawing conclusions and identifying trends or relationships.
5. Forecasting: Predicting future trends based on historical data.
6. Decision Making: Providing a basis for strategic planning and policy formulation.

What is weighted average? Under what conditions weighted average is preferable to a simple average?

Answer:

• When data points have different levels of significance or relevance (e.g., in grading systems where exams and assignments have different weights).
• When the frequency or size of data points varies (e.g., calculating an average price based on the quantity sold).
• When recent data is more relevant than older data (e.g., in financial indices like stock market averages).
• When the goal is to reflect the true impact of each value in a dataset more accurately.

The total cost of a firm is $C=\frac{1}{3}{x}^3-6x^2+40x+15$$C=\frac{1}{3}{x}^3-6x^2+40x+15$C=(1)/(3)x^(3)-6x^(2)+40 x+15C = \frac{1}{3}x^3 – 6x^2 + 40x + 15$C=\frac{1}{3}{x}^3-6x^2+40x+15$. Find the equilibrium output if price is fixed at Rs. 20 per unit.

Answer:

Step 1: Total Revenue

Step 2: Profit Function

Step 3: Maximize Profit

Step 4: Solve the Quadratic Equation

• $x=\frac{12+8}{2}=\frac{20}{2}=10$$x=\frac{12+8}{2}=\frac{20}{2}=10$x=(12+8)/(2)=(20)/(2)=10x = \frac{12 + 8}{2} = \frac{20}{2} = 10$x=\frac{12+8}{2}=\frac{20}{2}=10$
• $x=\frac{12-8}{2}=\frac{4}{2}=2$$x=\frac{12-8}{2}=\frac{4}{2}=2$x=(12-8)/(2)=(4)/(2)=2x = \frac{12 – 8}{2} = \frac{4}{2} = 2$x=\frac{12-8}{2}=\frac{4}{2}=2$

Step 5: Determine the Maximum

• At $x=2$$x=2$x=2x = 2$x=2$: $-2\cdot 2+12=8$$-2\cdot 2+12=8$-2*2+12=8-2 \cdot 2 + 12 = 8$-2\cdot 2+12=8$ (positive, minimum)
• At $x=10$$x=10$x=10x = 10$x=10$: $-2\cdot 10+12=-8$$-2\cdot 10+12=-8$-2*10+12=-8-2 \cdot 10 + 12 = -8$-2\cdot 10+12=-8$ (negative, maximum)

Final Answer

Find the effective discount rate when nominal rate of discount is 10% compounded continuously.

Answer:

Calculation:

Final Answer:

A businessman sells 2000 items per month at a price of Rs. 10 each. It is estimated that monthly sales will increase by 250 items for each Re. 0.25 reduction in price. Find the demand function corresponding to this estimate.

Answer:

Given:

• Initial sales: 2000 items per month at a price of Rs. 10 each.
• For each Re. 0.25 reduction in price, monthly sales increase by 250 items.

Step 1: Determine the Rate of Change

• A Re. 0.25 reduction in price (i.e., $\mathrm{\Delta }p=-0.25$$\mathrm{\Delta }p=-0.25$Delta p=-0.25\Delta p = -0.25$\mathrm{\Delta }p=-0.25$) leads to an increase of 250 items (i.e., $\mathrm{\Delta }q=250$$\mathrm{\Delta }q=250$Delta q=250\Delta q = 250$\mathrm{\Delta }q=250$).
• The slope of the demand function (rate of change of $q$$q$qq$q$ with respect to $p$$p$pp$p$) is:

Step 2: Form the Demand Function

Step 3: Use the Initial Condition

Step 4: Write the Demand Function

Final Answer

Why is matrix multiplication not commutative?

Answer:

Write short notes on the following:

Answer:

a) Partition Values

b) Correlation

Differentiate between the following:

Answer:

• Absolute Measures of Dispersion:
  • These measure the spread of data in the original units of the dataset (e.g., meters, dollars).
  • They are not unitless and depend on the scale of the data.
  • Examples: Range (difference between max and min values), Variance (average of squared deviations from the mean), Standard Deviation (square root of variance), Mean Absolute Deviation (average of absolute deviations from the mean).
  • Use: Useful for understanding the actual spread in the context of the data’s units.
  • Limitation: Not suitable for comparing dispersion across datasets with different units or scales.
  • Example: For a dataset of heights in cm, the standard deviation might be 5 cm, indicating spread in centimeters.
• Relative Measures of Dispersion:
  • These are unitless measures, expressed as ratios or percentages, making them independent of the data’s scale.
  • They allow comparison of dispersion across datasets with different units or means.
  • Examples: Coefficient of Variation (standard deviation divided by the mean, often expressed as a percentage), Coefficient of Range (range divided by the sum of max and min values), Coefficient of Quartile Deviation (difference between third and first quartiles divided by their sum).
  • Use: Ideal for comparing variability between datasets with different units or scales.
  • Example: A coefficient of variation of 10% indicates the standard deviation is 10% of the mean, regardless of units.

• Variance:
  • An absolute measure of dispersion that quantifies how much data points deviate from the mean.
  • Formula: For a sample, $s^2=\frac{1}{n-1}\sum _{i=1}^n(x_i-\overline{x}{)}^2$$s^2=\frac{1}{n-1}\sum _{i=1}^n (x_i-\overline{x}{)}^2$s^(2)=(1)/(n-1)sum_(i=1)^(n)(x_(i)- bar(x))^(2)s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i – \bar{x})^2$s^2=\frac{1}{n-1}\sum _{i=1}^n(x_i-\overline{x}{)}^2$, where $\overline{x}$$\overline{x}$bar(x)\bar{x}$\overline{x}$ is the mean and $n$$n$nn$n$ is the sample size.
  • Units: Expressed in the square of the data’s units (e.g., if data is in meters, variance is in square meters).
  • Use: Indicates the spread of data but is sensitive to the scale of measurement.
  • Limitation: Hard to interpret for datasets with different units or scales, and the squared units can be unintuitive.
  • Example: For test scores (60, 70, 80), variance might be 100 points², indicating spread in squared units.
• Coefficient of Variation (CV):
  • A relative measure of dispersion, defined as the ratio of the standard deviation to the mean, often expressed as a percentage.
  • Formula: $CV=\left(\frac{\text{Standard Deviation}}{\text{Mean}}\right)\times 100\mathrm{\%}$$CV=\left(\frac{\text{Standard Deviation}}{\text{Mean}}\right)\times 100\mathrm{\%}$CV=((“Standard Deviation”)/(“Mean”))xx100%CV = \left( \frac{\text{Standard Deviation}}{\text{Mean}} \right) \times 100\%$CV=\left(\frac{\text{Standard Deviation}}{\text{Mean}}\right)\times 100\mathrm{\%}$.
  • Units: Unitless, as it’s a ratio, making it scale-invariant.
  • Use: Allows comparison of variability across datasets with different units or means (e.g., comparing variability of heights vs. weights).
  • Limitation: Less meaningful if the mean is close to zero, as it can inflate the CV.
  • Example: For test scores with a mean of 70 and standard deviation of 10, $CV=\left(\frac{10}{70}\right)\times 100\mathrm{\%}\approx 14.29\mathrm{\%}$$CV=\left(\frac{10}{70}\right)\times 100\mathrm{\%}\approx 14.29\mathrm{\%}$CV=((10)/(70))xx100%~~14.29%CV = \left( \frac{10}{70} \right) \times 100\% \approx 14.29\%$CV=\left(\frac{10}{70}\right)\times 100\mathrm{\%}\approx 14.29\mathrm{\%}$, showing relative variability.