When observed over a long period of time, a time series data can predict trends that can forecast increase, decrease, or stagnation of a variable under consideration. The table below shows the sale of an item in a district during 1996–2001:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]Determine the equation of the straight-line trend.
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When calculating a straight-line trend using least squares:
- Assign coded time values (\( t \)) to simplify calculations.
- Use the formulas \( a = \frac{\sum y}{N} \) and \( b = \frac{\sum (t \cdot y)}{\sum t^2} \).
Solution and Explanation
Step 1: Define coded time variables (\( t \)). Let \( t = -2, -1, 0, 1, 2, 3 \) correspond to the years 1996 through 2001.
Step 2: Present the data in a table:
Step 3: Apply the least squares formula for a linear trend:
\[ y = a + bt, \]
where:
\[ a = \frac{\sum y}{N}, \quad b = \frac{\sum (t \cdot y)}{\sum t^2}. \]
Calculate the constants:
\[ a = \frac{\sum y}{N} = \frac{35.6}{6} = 5.93, \quad b = \frac{\sum (t \cdot y)}{\sum t^2} = \frac{22.4}{19} = 1.18. \]
Step 4: Formulate the trend equation:
\[ y = 5.93 + 1.18t. \]
Tabulate the trend values of the years and compute the expected sales trend for the year 2002.
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Solution and Explanation
Step 1: Utilize the trend equation \( y = 5.93 + 1.18t \).
Step 2: Calculate the trend values for \( t = -2, -1, 0, 1, 2, 3 \):
\[ \begin{array}{|c|c|c|} \hline \textbf{Year} & t & \textbf{Trend Value (y)} \\ \hline 1996 & -2 & 5.93 + 1.18(-2) = 3.57 \\ 1997 & -1 & 5.93 + 1.18(-1) = 4.75 \\ 1998 & 0 & 5.93 + 1.18(0) = 5.93 \\ 1999 & 1 & 5.93 + 1.18(1) = 7.11 \\ 2000 & 2 & 5.93 + 1.18(2) = 8.29 \\ 2001 & 3 & 5.93 + 1.18(3) = 9.47 \\ \hline \end{array} \]Step 3: Determine the trend value for the year 2002, where \( t = 4 \):
\[ y = 5.93 + 1.18(4) = 10.65. \]Final Answer:
- The trend values for the years 1996 through 2001 are \( 3.57, 4.75, 5.93, 7.11, 8.29, \) and \( 9.47 \), respectively.
- The projected trend for sales in 2002 is \( 10.65 \) lakh ₹.
Fit a straight-line trend by the method of least squares for the following data:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
\textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\
\hline
\textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\
\hline
\end{array}
\]
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]Show Hint
Solution and Explanation
Step 1: Assign \( t \): Let \( t = -3, -2, -1, 0, 1, 2, 3 \) for the years 2004 to 2010.
Step 2: Tabulate the data:
\[ \begin{array}{|c|c|c|c|c|} \hline \textbf{Year} & \textbf{Profit (y)} & t & t^2 & t \cdot y \\ \hline 2004 & 114 & -3 & 9 & -342 \\ 2005 & 130 & -2 & 4 & -260 \\ 2006 & 126 & -1 & 1 & -126 \\ 2007 & 144 & 0 & 0 & 0 \\ 2008 & 138 & 1 & 1 & 138 \\ 2009 & 156 & 2 & 4 & 312 \\ 2010 & 164 & 3 & 9 & 492 \\ \hline \textbf{Total} & 972 & 0 & 28 & 214 \\ \hline \end{array} \]Step 3: Use the least squares formula:
\[ y = a + bt, \]where:
\[ a = \frac{\sum y}{N}, \quad b = \frac{\sum (t \cdot y)}{\sum t^2}. \]Substituting the values:
\[ a = \frac{972}{7} = 138.86, \quad b = \frac{214}{28} = 7.64. \]Step 4: Write the equation:
\[ y = 138.86 + 7.64t. \]Top Questions on Probability Distribution
Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to 50% of learners were self-taught using internet resources and upskilled themselves.
A student may spend 1 hour to 6 hours in a day in upskilling self. The probability distribution of the number of hours spent by a student is given below:
\[ P(X = x) = \begin{cases} kx^2, & \text{for } x = 1, 2, 3, \\ 2kx, & \text{for } x = 4, 5, 6, \\ 0, & \text{otherwise.} \end{cases} \]
where \( x \) denotes the number of hours. Based on the above information, answer the following questions:
(i) Express the probability distribution given above in the form of a probability distribution table.
(ii) Find the value of \( k \).
(iii)(a) Find the mean number of hours spent by the student.
(iii)(b) Find \( P(1 < X < 6) \).- CBSE Class XII - 2024
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- CBSE Class XII - 2024
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Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to 50% learners were self-taught using internet resources and upskilled themselves. A student may spend 1 hour to 6 hours in a day upskilling self. The probability distribution of the number of hours spent by a student is given below:
\[ P(X = x) = \begin{cases} kx^2 & {for } x = 1, 2, 3, \\ 2kx & {for } x = 4, 5, 6, \\ 0 & {otherwise}. \end{cases} \]
Based on the above information, answer the following:
- CBSE Class XII - 2024
- Mathematics
- Probability Distribution
Questions number 19 and 20 are Assertion and Reason-based questions. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.- CBSE Class XII - 2024
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