Maharashtra Board Class 12 Maths Part 2 Chapter 4 Time Series 4.1 Solutions

Exercise 4.1 Solutions Commerce Maths

Question 1. The following data gives the production of bleaching powder (in '000 tonnes) for the years 1962 to 1972.

Fit a trend line by graphical method to the above data.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख वर्ष 1962 से 1972 तक ब्लीचिंग पाउडर के उत्पादन के रुझान को दर्शाता है। क्षैतिज अक्ष वर्षों (1962-1972) को दर्शाती है, जबकि ऊर्ध्वाधर अक्ष उत्पादन ('000 टन में) को दर्शाती है। भूखंड पर बिंदु प्रत्येक वर्ष के उत्पादन को दिखाते हैं, और एक "ट्रेंड लाइन" खींची गई है जो समय के साथ उत्पादन में सामान्य प्रवृत्ति को दर्शाती है, जो एक आरोही पैटर्न दिखा रही है।
In simple words: A line graph showing the production of bleaching powder over the years, with a trend line indicating the general upward movement of production.

🎯 Exam Tip: For graphical methods, ensure clear labeling of axes, accurate plotting of data points, and a visually representative trend line for full marks.

Question 2. Use the method of least squares to fit a trend line to the data in problem 1 above. Also, obtain the trend value for the year 1975.
Answer: Solution:

n = 11, let the trend line the y = a + bu ......(1) Σy = na + bΣu ........(i)
Σuy = aΣu + bΣu² .........(ii) Substituting the values of Σy, Σu, Σuy, & Σu², we get 46 = 11a + 0
\(\therefore\) a = 4.18 And
114 = 0 + b(110)
\(\therefore\) b = 1.04 By (I) the equation of the trends line is
y = 4.18 + 1.04u
Where u = t - 1967 ........(iii)
For the year 1975 we have u = 8
Substituting in (iii) we get
Y= 4.18 + 1.04(8) = 12.5
Trend value for the year 1975 is 12.5 (in '000 tonnes).
In simple words: We used the least squares method to find a trend line equation (y = a + bu) from the given data. By substituting the calculated values of 'a' and 'b', and using the year 1975 to find 'u', we predicted the production for 1975 as 12.5 thousand tonnes.

🎯 Exam Tip: Carefully set up the table for least squares calculations. Ensure correct summation of all columns (Σy, Σu, Σu², Σuy) as errors here will propagate throughout the calculation of 'a' and 'b'.

Question 3. Obtain the trend line for the above data using 5 yearly moving averages.
Answer: Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ वर्ष 1962 से 1972 तक के उत्पादन डेटा को 5-वर्षीय चलती औसत विधि का उपयोग करके रुझान रेखा के साथ दर्शाता है। क्षैतिज अक्ष वर्षों को और ऊर्ध्वाधर अक्ष उत्पादन को दिखाता है। चलती औसत का उपयोग करके प्राप्त रुझान रेखा, मूल डेटा में अल्पकालिक उतार-चढ़ाव को सुचारू करती है, जिससे दीर्घकालिक पैटर्न स्पष्ट रूप से दिखाई देता है।
In simple words: This table calculates the 5-yearly moving average for the production data to smooth out fluctuations and reveal the underlying trend. The diagram visually represents this smoothed trend.

🎯 Exam Tip: When calculating moving averages, ensure the average is centered correctly. For an odd number of periods (like 5-yearly), the average is placed against the middle year of the period.

Question 4. The following table shows the index of industrial production for the period from 1976 to 1985, using the year 1976 as the base year.

Fit a trend line to the above data by graphical method.
Answer: Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ 1976 से 1985 तक औद्योगिक उत्पादन सूचकांक के डेटा को दर्शाता है, जिसमें एक ट्रेंड लाइन ग्राफिकल विधि का उपयोग करके फिट की गई है। क्षैतिज अक्ष वर्षों को और ऊर्ध्वाधर अक्ष उत्पादन सूचकांक को दिखाता है। ट्रेंड लाइन, जो डेटा बिंदुओं के माध्यम से खींची गई है, समय के साथ औद्योगिक उत्पादन सूचकांक में सामान्य वृद्धि का संकेत देती है।
In simple words: A line graph plots the industrial production index over the years, and a trend line is visually drawn to show the general direction or pattern of the data.

🎯 Exam Tip: When fitting a trend line graphically, aim to draw a line that passes through or as close as possible to the majority of the data points, balancing points above and below the line.

Question 5. Fit a trend line to the data in problem 4 above by the method of least squares. Also, obtain the trend value for the index of industrial production for the year 1987.
Answer: Solution:

\[u = \frac{t-1980.5}{1/2}\]

n = 10, Σu = 0, Σy = 42, Σu² = 330, Σuy = 148
Let the trend line be y = a + bu ......(i)
where \(u = \frac{t-1980.5}{1/2}\)
i.e. u = 2t - 3961
Σy = na + bΣu ......(ii)
Σuy = aΣu + bΣu² .........(iii)
Substituting the values of Σy, n, Σu, Σuy & Σu² We get
42 = 10a + 0
\(\therefore\) a = 4.2 and
148 = 0 + 5.330
\(\therefore\) b = 0.4485
\(\therefore\) by (i) the equation of the trends line is
Y = 4.2 + 0.4485u .........(iv)
where u = 2t - 3961
For the year 1987,
u = 13 by (iv) we have
Y = 4.2 + 0.4485(13) = 10.0305
\(\therefore\) The trend value for the year 1987 is 10.0305
In simple words: We applied the least squares method to the given industrial production data, calculating 'a' and 'b' for the trend line equation. We then used this equation to predict the industrial production index for 1987, which came out to be 10.0305.

🎯 Exam Tip: When the number of years is even, the origin for 'u' is taken between the two middle years, and 'u' is expressed in terms of half-yearly intervals (e.g., \(u = \frac{t - \text{midpoint year}}{1/2}\)). Be precise with these calculations.

Question 6. Obtain the trend values for the data in problem 4 using 4-yearly centered moving averages.
Answer: Solution:

In simple words: We calculated the 4-yearly moving total and then the 4-yearly moving average. Since it's an even period, we further calculated a 2-unit moving total of these averages to center them, resulting in the 4-yearly centered moving average, which provides the trend values.

🎯 Exam Tip: For even-period moving averages, remember the crucial step of centering the moving averages by taking a 2-period moving average of the initial moving averages. This ensures the trend values align with specific time points.

Question 7. The following table gives the production of steel (in millions of tonnes) for the years 1976 to 1986.

Fit a trend line to the above data by graphical method.
Answer: Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ 1976 से 1986 तक स्टील उत्पादन (मिलियन टन में) के डेटा को दर्शाता है। क्षैतिज अक्ष वर्षों को और ऊर्ध्वाधर अक्ष उत्पादन को दिखाता है। डेटा बिंदुओं को प्लॉट किया गया है, और एक "ट्रेंड लाइन" खींची गई है जो समय के साथ स्टील उत्पादन में सामान्य प्रवृत्ति को दर्शाती है, जिससे उतार-चढ़ाव के बावजूद समग्र पैटर्न का पता चलता है।
In simple words: A line graph showing steel production over several years, with a visually drawn trend line that indicates the overall pattern or direction of production despite yearly changes.

🎯 Exam Tip: When representing data graphically, ensure your scale is appropriate for both axes, making the trend discernible. Use distinct markers for actual data points before drawing the trend line.

Question 8. Fit a trend line to the data in Problem 7 by the method of least squares. Also, obtain the trend value for the year 1990.
Answer: Solution:

\[u = \frac{t-1981}{1}\]

n = 10, Σu = 0, ΣY = 62, Σu² = 110, Σuy = 87
Let the equation of the trend line be
Y = a + bu
where u = t - 1981 ......(i)
ΣY = na + bΣu ........(ii)
Σuy = aΣu + bΣu² .........(iii)
Substituting the values of Σy, n, Σu, Σuy, Σu² in (ii) & (iii)
62 = 11a + 0
\(\therefore\) a = 5.6364 And
87 = 0 + 5(110)
\(\therefore\) b = 0.7909
\(\therefore\) by (i) equation of the trend line is y = 5.6364 + 0.7909u
Where u = t-1981
For the year 1990,
u = 9
\(\therefore\) y = 5.6364 + 0.7909(9)
\(\therefore\) y = 12.7545 (in million tonnes)
In simple words: We used the least squares method to find the trend line equation for steel production. By calculating the coefficients 'a' and 'b' and then substituting the corresponding 'u' value for 1990, we estimated the steel production for that year to be 12.7545 million tonnes.

🎯 Exam Tip: Always double-check your arithmetic when calculating sums and products in the least squares table. A single error can lead to incorrect 'a' and 'b' values and, consequently, an inaccurate trend prediction.

Question 9. Obtain the trend values for the above data using 3-yearly moving averages.
Answer: Solution:

In simple words: We calculated the 3-yearly moving total for the production data and then divided each total by 3 to get the 3-yearly moving average. These averages represent the trend values, smoothing out short-term fluctuations.

🎯 Exam Tip: When computing odd-period moving averages, the average value is always placed in the middle of the period. Be careful with the first and last few entries, as they won't have a full period for calculation.

Question 10. The following table shows the production of gasoline in the U.S.A. for the years 1962 to 1976.

(i) Obtain trend values for the above data using 5-yearly moving averages.
(ii) Plot the original time series and trend values obtained above on the same graph.
Answer: Solution:
(i)

(ii)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ 1962 से 1976 तक USA में गैसोलीन उत्पादन की मूल समय श्रृंखला और 5-वर्षीय चलती औसत का उपयोग करके प्राप्त रुझान मूल्यों दोनों को दर्शाता है। क्षैतिज अक्ष वर्षों को और ऊर्ध्वाधर अक्ष उत्पादन (मिलियन बैरल में) को दर्शाता है। मूल डेटा बिंदु अस्थिरता दिखाते हैं, जबकि चलती औसत रुझान रेखा, मूल श्रृंखला को सुचारू करके, समय के साथ उत्पादन में एक स्पष्ट ऊपर की ओर रुझान दिखाती है।
In simple words: (i) The table calculates the 5-yearly moving average to smooth out the gasoline production data and find the trend values. (ii) The graph visually plots both the original, fluctuating production data and the smoother trend values from the 5-yearly moving average, showing how the trend line represents the general direction over time.

🎯 Exam Tip: When plotting both original and trend values, use different symbols or colors for clarity. The trend line should typically be smoother than the original data series, indicating the underlying pattern. Ensure the moving averages are correctly centered on the corresponding year.