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Question 27 of 41
Q.

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 2019 and 2020:

Year201220132014201520162017
No. of vehicles (thousand)69758291101115

OR

Find the trend using five yearly moving averages from the following data of profit (in lakh ₹) of a trader in different years:

Year201020112012201320142015201620172018
Profit (Lakh ₹)454254605172817569
Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2023Subjective· 5mImportance★★★★★
66% · 27/41 Questions
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Coded t=−5,−3,−1,1,3,5t=-5,-3,-1,1,3,5: ∑y=533,∑t2=70,∑ty=317\sum y=533,\sum t^2=70,\sum ty=317; a=88.83,b=4.529a=88.83,b=4.529; y^=88.83+4.529t\hat y=88.83+4.529t; 2019(t=9t=9)≈129.6\approx129.6, 2020(t=11t=11)≈138.6\approx138.6. OR five-year moving averages listed below.

Main part: n=6n=6 (even), so code time in steps of 2: t=−5,−3,−1,1,3,5t=-5,-3,-1,1,3,5 for 2012–2017.

Yearyyttt2t^2tyty
201269-525-345
201375-39-225
201482-11-82
2015911191
201610139303
2017115525575
Total533070317

a=∑yn=5336=88.83,b=∑ty∑t2=31770=4.529.a=\frac{\sum y}{n}=\frac{533}{6}=88.83,\qquad b=\frac{\sum ty}{\sum t^2}=\frac{317}{70}=4.529.

Trend line: y^=88.83+4.529 t\hat y=88.83+4.529\,t (each unit of tt = half a year; origin mid-2014.5).

For 2019, t=9t=9; for 2020, t=11t=11:

y^2019=88.83+4.529(9)=88.83+40.76=129.59 thousand,\hat y_{2019}=88.83+4.529(9)=88.83+40.76=129.59\ \text{thousand},

y^2020=88.83+4.529(11)=88.83+49.81=138.65 thousand.\hat y_{2020}=88.83+4.529(11)=88.83+49.81=138.65\ \text{thousand}.

OR part — five-yearly moving averages of profit (lakh ₹):

YearProfit5-year moving total5-year moving average (trend)
201045——
201142——

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