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:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]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 \( b = \frac{\sum (t \cdot y)}{\sum t^2} \).
Step 1: Assign coded time variables (\( t \)) for convenience:
Let \( t = -2, -1, 0, 1, 2, 3 \) for the years 1996 to 2001, respectively.
Step 2: Tabulate the data:
Step 3: Use the least squares formula for straight-line trend:
\[ y = a + bt, \]
where:
\[ a = \frac{\sum y}{N}, \quad b = \frac{\sum (t \cdot y)}{\sum t^2}. \]
Substitute the values:
\[ a = \frac{\sum y}{N} = \frac{35.6}{6} = 5.93, \quad b = \frac{\sum (t \cdot y)}{\sum t^2} = \frac{22.4}{19} = 1.18. \]
Step 4: Write the equation:
\[ y = 5.93 + 1.18t. \]
Step 1: Use the trend equation \( y = 5.93 + 1.18t \).
Step 2: Compute 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: Compute the trend value for 2002 (\( t = 4 \)):
\[ y = 5.93 + 1.18(4) = 10.65. \]Final Answer:
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]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:
\[ y = 138.86 + 7.64t. \]X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|
P(X) | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
List-I | List-II |
---|---|
(A) k | (I) 7/10 |
(B) P(X < 3) | (II) 53/100 |
(C) P(X ≥ 2) | (III) 1/10 |
(D) P(2 < X ≤ 7) | (IV) 3/10 |
X | 0 | 1 | 2 | otherwise |
P(X) | k | 2k | 3k | 0 |
X | 0 | 1 | 2 | otherwise |
P(X) | k | 2k | 3k | 0 |
Then:
(A) \( k = \frac{1}{6} \)
(B) \( P(X < 2) = \frac{1}{2} \)
(C) \( E(X) = \frac{3}{4} \)
(D) \( P(1 < X \leq 2) = \frac{5}{6} \)
Choose the correct answer from the options given below:
X | 3 | 4 | 5 |
---|---|---|---|
P(X) | 0.5 | 0.2 | 0.3 |