GSEB Class 12 Statistics Solutions Chapter 4 Time Series Exercise 4.2

GSEB Solutions Class 12 Statistics Part 1 Chapter 4 Time Series Ex 4.2

 

Question 1. The information about death rate of a state in different years is given in the following table. Fit a linear equation to find trend and hence estimate the death rate for the year 2017.

Year	2009	2010	2011	2012	2013	2014	2015
Death rate	7.6	6.9	7.1	7.3	7.2	6.9	6.9

Answer: For this problem, we have 7 years of data, so \( n = 7 \). We assign time values \( t = 1, 2, 3, 4, 5, 6, 7 \). The death rate is represented by \( y \). Our goal is to fit a linear trend equation, which is given by \( \hat{y} = a + bt \). To find the values of \( a \) and \( b \), we first create a table with \( t \), \( y \), \( ty \), and \( t^2 \):

Year	t	Death rate y	t \(\cdot\) y	t\(^2\)
2009	1	7.6	7.6	1
2010	2	6.9	13.8	4
2011	3	7.1	21.3	9
2012	4	7.3	29.2	16
2013	5	7.2	36.0	25
2014	6	6.9	41.4	36
2015	7	6.9	48.3	49
n = 7	\( \Sigma t = 28 \)	\( \Sigma y = 49.9 \)	\( \Sigma ty = 197.6 \)	\( \Sigma t^2 = 140 \)

First, we calculate the mean of \( t \) and \( y \): \( \bar{t} = \frac{\Sigma t}{n} = \frac{28}{7} = 4 \) \( \bar{y} = \frac{\Sigma y}{n} = \frac{49.9}{7} = 7.13 \) (approximately) Next, we calculate the slope \( b \) using the formula: \( b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2} \) Substituting the values: \( b = \frac{7(197.6) - (28)(49.9)}{7(140) - (28)^2} \) \( b = \frac{1383.2 - 1397.2}{980 - 784} \) \( b = \frac{-14}{196} \) \( b = -0.07 \) Now, we calculate the y-intercept \( a \) using the formula: \( a = \bar{y} - b\bar{t} \) Substituting the values of \( \bar{y} \), \( b \), and \( \bar{t} \): \( a = 7.13 - (-0.07)(4) \) \( a = 7.13 + 0.28 \) \( a = 7.41 \) So, the linear trend equation is \( \hat{y} = 7.41 - 0.07t \). To estimate the death rate for the year 2017: The year 2017 corresponds to \( t = 9 \) (since 2009 is \( t=1 \), 2015 is \( t=7 \), 2016 is \( t=8 \), and 2017 is \( t=9 \)). Substitute \( t = 9 \) into the trend equation: \( \hat{y} = 7.41 - 0.07(9) \) \( \hat{y} = 7.41 - 0.63 \) \( \hat{y} = 6.78 \) Therefore, the estimated death rate for the year 2017 is 6.78.
In simple words: We used the given yearly death rates to create a line that shows the general trend. This line helps us guess what the death rate might be in the future. For the year 2017, the estimated death rate is 6.78.

🎯 Exam Tip: Remember to clearly show all steps for calculating \( \bar{t} \), \( \bar{y} \), \( b \), and \( a \). Double-check calculations, especially when dealing with negative values for \( b \), as sign errors are common.

 

Question 2. The data about Cost Inflation Index (CII) declared by the central government are as follows. The year 1981-82 is the base for this index. Find the estimate of this index for the year 2015-16 by fitting the linear equation to these data.

Year	2007-08	2008-09	2009-10	2010-11	2011-12	2012-13	2013-14	2014-15
CII	551	582	632	711	785	852	939	1024

Answer: In this problem, we have 8 years of data, so \( n = 8 \). The time values are assigned as \( t = 1, 2, ..., 8 \). The Cost Inflation Index (CII) is represented by \( y \). We need to fit a linear trend equation of the form \( \hat{y} = a + bt \). To determine \( a \) and \( b \), we construct a table including \( t \), \( y \), \( ty \), and \( t^2 \):

Year	Time t	CII y	t \(\cdot\) y	t\(^2\)
2007-08	1	551	551	1
2008-09	2	582	1164	4
2009-10	3	632	1896	9
2010-11	4	711	2844	16
2011-12	5	785	3925	25
2012-13	6	852	5112	36
2013-14	7	939	6573	49
2014-15	8	1024	8192	64
n = 8	\( \Sigma t = 36 \)	\( \Sigma y = 6076 \)	\( \Sigma ty = 30257 \)	\( \Sigma t^2 = 204 \)

First, calculate the mean of \( t \) and \( y \): \( \bar{t} = \frac{\Sigma t}{n} = \frac{36}{8} = 4.5 \) \( \bar{y} = \frac{\Sigma y}{n} = \frac{6076}{8} = 759.5 \) Next, calculate the slope \( b \) using the formula: \( b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2} \) Substituting the values: \( b = \frac{8(30257) - (36)(6076)}{8(204) - (36)^2} \) \( b = \frac{242056 - 218736}{1632 - 1296} \) \( b = \frac{23320}{336} \) \( b = 69.4 \) (approximately) Now, calculate the y-intercept \( a \) using the formula: \( a = \bar{y} - b\bar{t} \) Substituting the values of \( \bar{y} \), \( b \), and \( \bar{t} \): \( a = 759.5 - 69.4(4.5) \) \( a = 759.5 - 312.3 \) \( a = 447.2 \) So, the linear trend equation is \( \hat{y} = 447.2 + 69.4t \). To estimate the CII for the year 2015-16: The year 2014-15 corresponds to \( t=8 \). Therefore, the year 2015-16 corresponds to \( t=9 \). Substitute \( t = 9 \) into the trend equation: \( \hat{y} = 447.2 + 69.4(9) \) \( \hat{y} = 447.2 + 624.6 \) \( \hat{y} = 1071.8 \) Hence, the estimated CII for the year 2015-16 is 1071.8.
In simple words: We used the given yearly Cost Inflation Index numbers to find a pattern. This pattern helps us predict the index for a future year. For the year 2015-16, the predicted Cost Inflation Index is 1071.8.

🎯 Exam Tip: When performing calculations, round \( b \) and \( a \) to one decimal place as shown, but maintain precision in intermediate steps if possible to avoid rounding errors. Clearly state the final linear trend equation and the estimated value.

 

Question 3. The number of two wheelers registered (in thousand) in a city in different years is as follows. Use the method of fitting linear equation to these data to obtain the estimates for the number of vehicles registered in the year 2016 and 2017. Also find the trend values for each year.

Year	2010	2011	2012	2013	2014	2015
No. of vehicles (thousand)	69	75	82	91	101	115

Answer: In this problem, we have 6 years of data, so \( n = 6 \). We assign time values \( t = 1, 2, ..., 6 \). The number of vehicles (in thousand) is represented by \( y \). We need to fit a linear trend equation of the form \( \hat{y} = a + bt \). To find \( a \) and \( b \), we prepare a table with \( t \), \( y \), \( ty \), and \( t^2 \):

Year	Time t	No. of vehicles (in '000) y	t \(\cdot\) y	t\(^2\)	Trend values \( \hat{y} = 57.12 + 9.06t \) (t = 1, 2, 3, ..., 6)
2010	1	69	69	1	66.18
2011	2	75	150	4	75.24
2012	3	82	246	9	84.30
2013	4	91	364	16	93.36
2014	5	101	505	25	102.42
2015	6	115	690	36	111.48
n = 6	\( \Sigma t = 21 \)	\( \Sigma y = 533 \)	\( \Sigma ty = 2024 \)	\( \Sigma t^2 = 91 \)	\( \Sigma \hat{y} = 532.98 \approx 533 \)

First, calculate the mean of \( t \) and \( y \): \( \bar{t} = \frac{\Sigma t}{n} = \frac{21}{6} = 3.5 \) \( \bar{y} = \frac{\Sigma y}{n} = \frac{533}{6} = 88.83 \) (approximately) Next, calculate the slope \( b \) using the formula: \( b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2} \) Substituting the values: \( b = \frac{6(2024) - (21)(533)}{6(91) - (21)^2} \) \( b = \frac{12144 - 11193}{546 - 441} \) \( b = \frac{951}{105} \) \( b = 9.06 \) (approximately) Now, calculate the y-intercept \( a \) using the formula: \( a = \bar{y} - b\bar{t} \) Substituting the values of \( \bar{y} \), \( b \), and \( \bar{t} \): \( a = 88.83 - 9.06(3.5) \) \( a = 88.83 - 31.71 \) \( a = 57.12 \) So, the linear trend equation is \( \hat{y} = 57.12 + 9.06t \). To estimate the number of vehicles for the year 2016: The year 2015 corresponds to \( t=6 \). Therefore, the year 2016 corresponds to \( t=7 \). Substitute \( t = 7 \) into the trend equation: \( \hat{y} = 57.12 + 9.06(7) \) \( \hat{y} = 57.12 + 63.42 \) \( \hat{y} = 120.54 \) (thousand) To estimate the number of vehicles for the year 2017: The year 2017 corresponds to \( t=8 \). Substitute \( t = 8 \) into the trend equation: \( \hat{y} = 57.12 + 9.06(8) \) \( \hat{y} = 57.12 + 72.48 \) \( \hat{y} = 129.6 \) (thousand) The trend values for each year are: For \( t=1 \) (2010): \( \hat{y} = 57.12 + 9.06(1) = 66.18 \) (thousand) For \( t=2 \) (2011): \( \hat{y} = 57.12 + 9.06(2) = 75.24 \) (thousand) For \( t=3 \) (2012): \( \hat{y} = 57.12 + 9.06(3) = 84.30 \) (thousand) For \( t=4 \) (2013): \( \hat{y} = 57.12 + 9.06(4) = 93.36 \) (thousand) For \( t=5 \) (2014): \( \hat{y} = 57.12 + 9.06(5) = 102.42 \) (thousand) For \( t=6 \) (2015): \( \hat{y} = 57.12 + 9.06(6) = 111.48 \) (thousand)
In simple words: We calculated a trend line for the number of two-wheelers registered each year. This line helps us predict how many vehicles will be registered in future years like 2016 and 2017. It also gives us an idea of the expected number of vehicles for the past years based on the trend.

🎯 Exam Tip: When asked to find trend values for each year, ensure you calculate and present them clearly for all given years. Pay attention to the units (e.g., "thousand vehicles") in your final answer for estimates.

 

Question 4. The average age of women (in years) at the time of marriage obtained from the data of different census surveys in India are given in the following table. Fit an equation for a linear trend from the data and show it on a graph. Find the estimate for the value of the given variable for the year 2021 using the linear equation.

Year of census survey	1971	1981	1991	2001	2011
Average age of women at marriage (years)	17.7	18.7	19.3	20.2	22.2

Answer: For this problem, we have 5 census survey years, so \( n = 5 \). We assign time values \( t = 1, 2, ..., 5 \). The average age of women at marriage is represented by \( y \). We need to fit a linear trend equation of the form \( \hat{y} = a + bt \). To calculate \( a \) and \( b \), we construct a table with \( t \), \( y \), \( ty \), and \( t^2 \):

Year of census survey	Time t	Age of women at marriage (year) y	t \(\cdot\) y	t\(^2\)
1971	1	17.7	17.7	1
1981	2	18.7	37.4	4
1991	3	19.3	57.9	9
2001	4	20.2	80.8	16
2011	5	22.2	111.0	25
n = 5	\( \Sigma t = 15 \)	\( \Sigma y = 98.1 \)	\( \Sigma ty = 304.8 \)	\( \Sigma t^2 = 55 \)

First, calculate the mean of \( t \) and \( y \): \( \bar{t} = \frac{\Sigma t}{n} = \frac{15}{5} = 3 \) \( \bar{y} = \frac{\Sigma y}{n} = \frac{98.1}{5} = 19.62 \) Next, calculate the slope \( b \) using the formula: \( b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2} \) Substituting the values: \( b = \frac{5(304.8) - (15)(98.1)}{5(55) - (15)^2} \) \( b = \frac{1524 - 1471.5}{275 - 225} \) \( b = \frac{52.5}{50} \) \( b = 1.05 \) Now, calculate the y-intercept \( a \) using the formula: \( a = \bar{y} - b\bar{t} \) Substituting the values of \( \bar{y} \), \( b \), and \( \bar{t} \): \( a = 19.62 - 1.05(3) \) \( a = 19.62 - 3.15 \) \( a = 16.47 \) So, the linear trend equation is \( \hat{y} = 16.47 + 1.05t \). To estimate the average age of women at marriage for the year 2021: The year 2011 corresponds to \( t=5 \). Therefore, the year 2021 corresponds to \( t=6 \) (since each step is 10 years and represents 1 unit of \( t \)). Substitute \( t = 6 \) into the trend equation: \( \hat{y} = 16.47 + 1.05(6) \) \( \hat{y} = 16.47 + 6.30 \) \( \hat{y} = 22.77 \) (years)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख विवाह के समय महिलाओं की औसत आयु में रुझान (ट्रेंड) को दर्शाता है। क्षैतिज अक्ष (X-अक्ष) पर समय (वर्ष) को 1 सेमी = 10 वर्ष के पैमाने पर दर्शाया गया है, और ऊर्ध्वाधर अक्ष (Y-अक्ष) पर आयु (वर्ष) को 1 सेमी = 1 वर्ष के पैमाने पर दिखाया गया है। इसमें मूल डेटा बिंदुओं (Original Series) को एक रेखा से जोड़ा गया है और इसके साथ ही एक सीधी ट्रेंड रेखा (Trend) भी खींची गई है जो डेटा के समग्र पैटर्न को दिखाती है। ट्रेंड रेखा मूल डेटा बिंदुओं के करीब से गुजरती है, जो बताती है कि विवाह की औसत आयु समय के साथ धीरे-धीरे बढ़ रही है। जब t = 6 (जो वर्ष 2021 को दर्शाता है), तो ट्रेंड के अनुसार विवाह के समय महिलाओं की अनुमानित औसत आयु लगभग 22.9 वर्ष है।
In simple words: We used the average marriage ages over different census years to create a line that shows how this age is changing over time. This line helps us guess what the average marriage age might be in the future. For the year 2021, we estimate the average age of marriage to be about 22.77 years. The graph visually shows this upward trend.

🎯 Exam Tip: For graphical representation questions, ensure your graph axes are labeled correctly (X-axis for time, Y-axis for the variable), and include a clear scale. Mark both the original data points and the fitted trend line distinctly. Accuracy in plotting is key for full marks.