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SYLLABUS: Statistical data – Ungrouped data – Grouped data – Discrete data – Continuous data – Arithmetic mean – Variance – Standard deviation – Fitting a straight line using the method of least squares – Simple problems – Engineering applications (not for examinations).
Arithmetic Mean:
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Data
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Data is a collection of information assembled by observations, research measurements, or analysis. This data is divided into Ungrouped and Grouped.
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Ungrouped data
Ungrouped data is a type of data that has not been organized or grouped into specific categories. It is also known as raw data or unorganized data. It is simply a collection of observations or measurements without any structure or organization.
To find Arithmetic Mean
The arithmetic mean of n observations x1, x2, . . . . xn is defined as
\[\bar{x}\ =\ \frac{x_1\ +\ x_2\ +\ ……….\ +\ x_n}{n}\]
\[=\frac{\Sigma_{i=1}^{n} x_i}{n}\]
\[\color {purple} {Example\ 1\ .}\ \color{red}{Calculate\ the\ arithmetic mean\ of}\ 14,\ 26,\ 28,\ 20,\ 32,\ 30\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 6\ \hspace{20cm}\]
\[\Sigma x_i\ =\ 14\ +\ 26\ +\ +\ 28\ +\ 20\ +\ 32\ +\ 30\]
\[\Sigma x_i\ =\ 150\]
\[\bar{x}\ =\ \frac{\Sigma x_i}{n}\]
\[=\ \frac{150}{6}\]
\[\bar{x}\ =\ 25\]
\[\color {purple} {Example\ 2\ .}\ \color{red}{Calculate\ the\ arithmetic mean\ of}\ 14,\ 26,\ 28,\ 20,\ 35,\ and\ 33\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 6\ \hspace{20cm}\]
\[\Sigma x_i\ =\ 14\ +\ 26\ +\ +\ 28\ +\ 20\ +\ 35\ +\ 33\]
\[\Sigma x_i\ =\ 156\]
\[\bar{x}\ =\ \frac{\Sigma x_i}{n}\]
\[=\ \frac{156}{6}\]
\[\bar{x}\ =\ 26\]
\[\color {purple} {Example\ 3\ .}\ \color{red}{Calculate\ the\ arithmetic\ mean\ of}\ 10,\ 12,\ 14,\ 16\ and\ 18\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2024,\ October\ 2024,\ (Supp)\ June\ 25\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 5\ \hspace{20cm}\]
\[\Sigma x_i\ =\ 10\ +\ 12\ +\ 14\ +\ 16\ +\ 18\]
\[\Sigma x_i\ =\ 70\]
\[\bar{x}\ =\ \frac{\Sigma x_i}{n}\]
\[=\ \frac{70}{5}\]
\[\bar{x}\ =\ 14\]
\[\color {purple} {Example\ 4\ .}\ The\ arithmetic\ mean\ of\ 10\ numbers\ is\ 20\ ,\ \color{red}{find\ the\ sum\ of\ the}\ \hspace{10cm}\]\[\color{red}{values}\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023\ October\ 2024\ Supp(June)\ 25\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 10\ and\ \bar{x}\ =\ 20\ \hspace{17cm}\]
\[\bar{x} \ =\ \frac{\Sigma X}{n}\]
\[20\ =\ \frac{\Sigma X}{10}\]
\[\therefore\ \Sigma X\ =\ 200\]
\[\color {purple} {Example\ 5\ .}\ The\ arithmetic\ mean\ of\ 10\ numbers\ is\ 30\ ,\ \color{red}{find\ the\ total\ of\ the}\ \hspace{10cm}\]\[\color{red}{numbers.}\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2025\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 10\ and\ \bar{x}\ =\ 30\ \hspace{17cm}\]
\[\bar{x} \ =\ \frac{\Sigma X}{n}\]
\[30\ =\ \frac{\Sigma X}{10}\]
\[\therefore\ \Sigma X\ =\ 300\]
\[\color {purple} {Example\ 6\ .}\ The\ arithmetic\ mean\ of\ 6\ values\ is\ 45\ .\ If\ 3\ is\ added\ to\ each\ of\ the\ \hspace{5cm}\\ numbers,\ \color{red}{then\ find\ the\ arithmetic\ mean\ of\ the\ new\ set\ of\ values.}\ \hspace{3cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 6\ and\ \bar{x}\ =\ 45\ \hspace{17cm}\]
\[\bar{x} \ =\ \frac{\Sigma X}{n}\]
\[45\ =\ \frac{\Sigma X}{6}\]
\[\therefore\ \Sigma X\ =\ 270\]
\[corrected\ \Sigma X\ =\ 270\ +\ 18\]
\[corrected\ \Sigma X\ =\ 288\]
\[correct\ mean\ =\ \frac{288}{6}\ =\ 48\]
Grouped data
In ungrouped form, a discrete data can be represented as x1, x2, x3, ….. xn where each xi is an element of the data and n is the number of elements in the data. The number of times an element xi occurs in a data is called the frequency of the element and it is denoted by fi
| … | … | … |
The arithmetic mean of a grouped frequency distribution is defined as
\[\bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\]
Where fi is frequency, xi = 1,2,.. n is the mid values of the each of the class intervals and N = Σfi , total frequency.
\[\color {purple} {Example\ 7\ .}\ Find\ the\ arithmetic\ mean\ of\ the\ following\ frequency\ distribution\ \hspace{5cm}\]
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| f | 5 | 9 | 12 | 17 | 14 | 10 | 6 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| x | f | fx |
| 1 | 5 | 5 |
| 2 | 9 | 18 |
| 3 | 12 | 36 |
| 4 | 17 | 68 |
| 5 | 14 | 70 |
| 6 | 10 | 60 |
| 7 | 6 | 42 |
| N = 73 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{299}{73}\ =\ 4.09\]
A common way to group data is by creating a frequency distribution table. This involves dividing the data into intervals or classes, and then counting how many observations fall into each interval. This can be useful for identifying patterns and trends in the data, and for creating graphs such as histograms to visualize the distribution of the data.
| Class Interval | No. of elements in the Class Interval fi | Mid-value xi | fixi |
| … | … | … | … |
\[\bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\]
\[\color {purple} {Example\ 8\ .}\ Find\ the\ arithmetic\ mean\ of\ the\ following\ data\ \hspace{10cm}\]
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 |
| Frequency | 2 | 5 | 1 | 3 |
\[\hspace{5cm}\ October\ 2024,\ Supp\ June\ 2025\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| Marks | No. of students fi | Mid-value xi | fixi |
| 0-10 | 2 | 5 | 10 |
| 10-20 | 5 | 15 | 75 |
| 20-30 | 1 | 25 | 25 |
| 30-40 | 3 | 35 | 105 |
| N = 11 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{215}{11}\ =\ 19.54\]
\[\color {purple} {Example\ 9\ .}\ Find\ the\ arithmetic\ mean\ of\ the\ following\ data\ \hspace{10cm}\]
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 2 | 6 | 9 | 7 | 4 | 2 |
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| Marks | No. of students fi | Mid-value xi | fixi |
| 0-10 | 2 | 5 | 10 |
| 10-20 | 6 | 15 | 90 |
| 20-30 | 9 | 25 | 225 |
| 30-40 | 7 | 35 | 245 |
| 40-50 | 4 | 45 | 180 |
| 50-60 | 2 | 55 | 110 |
| N = 30 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{860}{30}\ =\ 28.67\]
\[\color {purple} {Example\ 10\ .}\ Find\ the\ arithmetic\ mean\ of\ the\ following\ data\ \hspace{10cm}\]
| Class-interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 3 | 5 | 16 | 18 | 12 | 7 | 4 |
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| Marks | No. of students fi | Mid-value xi | fixi |
| 0-10 | 3 | 5 | 15 |
| 10-20 | 5 | 15 | 75 |
| 20-30 | 16 | 25 | 400 |
| 30-40 | 18 | 35 | 630 |
| 40-50 | 12 | 45 | 540 |
| 50-60 | 7 | 55 | 385 |
| 60-70 | 4 | 65 | 260 |
| N = 65 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{2305}{65}\ =\ 35.46\]
\[\color {purple} {Example\ 11\ .}\ Find\ the\ arithmetic\ mean\ of\ the\ following\ data\ \hspace{10cm}\]
| Class-interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 11 | 14 | 15 | 20 | 15 | 13 | 12 |
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| Marks | No. of students fi | Mid-value xi | fixi |
| 0-10 | 11 | 5 | 55 |
| 10-20 | 14 | 15 | 210 |
| 20-30 | 15 | 25 | 375 |
| 30-40 | 20 | 35 | 700 |
| 40-50 | 15 | 45 | 675 |
| 50-60 | 13 | 55 | 715 |
| 60-70 | 12 | 65 | 780 |
| N = 100 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{3510}{100}\ =\ 35.10\]
Variance:
The mean of the squared differences of all elements from their arithmetic mean is called the variance. It is denoted by σ² and is defined as
\[\sigma^2\ =\ \frac{\Sigma x_i^2}{n}\ -\ (\frac{\Sigma x_i}{n})^2\ (For\ ungrouped\ Data)\]
\[\sigma^2\ =\ \frac{\Sigma x_i^2\ f_i}{n}\ -\ (\frac{\Sigma x_i\ f_i}{n})^2\ (For\ grouped\ Data)\]
Standard Deviation
A positive square root of the variance is called standard deviation. It is denoted by σ.
\[\sigma\ =\ \sqrt{\frac{\Sigma x_i^2}{n}\ -\ (\frac{\Sigma x_i}{n})^2}\ (For\ ungrouped\ Data)\]
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ x_i^2}{N}\ -\ (\frac{\Sigma f_i\ x_i}{N}})^2\ \ (For\ grouped\ Data)\]
\[\color {purple} {Example\ 12\ .}\ \text{If the standard deviation of a data is 6.8,}\ \color{red}{find\ its\ variance}\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ Given\ \sigma\ =\ 6.8\ \hspace{20cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ variance\ =\ \sigma^2\]
\[\hspace{3cm}\ =\ (6.8)^2\]
\[\hspace{2cm}\ \boxed{variance\ =\ 46.24}\]
\[\color {purple} {Example\ 13\ .}\ \text{If the standard deviation of a data is 8,}\ \color{red}{find\ its\ variance}\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 25\]
\[\color {blue} {Soln:}\ Given\ \sigma\ =\ 8\ \hspace{20cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ variance\ =\ \sigma^2\]
\[\hspace{3cm}\ =\ (8)^2\]
\[\hspace{2cm}\ \boxed{variance\ =\ 64}\]
\[\color {purple} {Example\ 14\ .}\ \text{If the variance of a data is 100,}\ \color{red}{then\ find\ its\ standard\ deviation}\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ Given\ variance\ =\ 100\ \hspace{20cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ variance\ =\ \sigma^2\]
\[\sigma\ =\ \sqrt{variance}\]
\[\sigma\ =\ \sqrt{100}\]
\[\hspace{3cm}\ =\ 10\]
\[\hspace{2cm}\ \boxed{standard\ deviation\ =\ 10}\]
\[\color {purple} {Example\ 15\ .}\ Find\ the\ variance\ of\ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 8\ \hspace{20cm}\]
| xi | xi 2 |
| -3 | 9 |
| -2 | 4 |
| -1 | 1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
| 3 | 9 |
\[\sigma^2\ =\ \frac{\Sigma x_i^2}{n}\ -\ (\frac{\Sigma x_i}{n})^2\ (For\ ungrouped\ Data)\]
\[=\ \frac{1}{7}\ (28)\ =\ 4\]
\[\sigma^2\ =\ 4\]
\[\color {purple} {Example\ 16\ .}\ Find\ the\ standard\ deviation\ of\ 2,\ 7,\ 3,\ 12\ and\ 9\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2024\ Supp(June)\ 25\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 5\ \hspace{20cm}\]
| xi | xi 2 |
| 2 | 4 |
| 7 | 49 |
| 3 | 9 |
| 12 | 144 |
| 9 | 81 |
\[\sigma\ =\ \sqrt{\frac{\Sigma x_i^2}{n}\ -\ (\frac{\Sigma x_i}{n})^2}\ (For\ ungrouped\ Data)\]
\[=\ \sqrt{\frac{287}{5}\ -\ (\frac{33}{5})^2}\]
\[=\ \sqrt{57.4\ -\ (6.6)^2}\]
\[=\ \sqrt{57.4\ -\ 43.56}\]
\[=\ \sqrt{13.84}\]
\[\sigma\ =\ 3.72\ (approximately)\]
\[\color {purple} {Example\ 17\ .}\ \text{Calculate the standard deviation for the following data}\ \hspace{10cm}\]
| Items | 5 | 15 | 25 | 35 |
| Frequency | 2 | 1 | 1 | 3 |
\[\hspace{5cm}\ October\ 2023,\ April\ 2024\ April\ 25\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 4\ \hspace{20cm}\]
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ x_i^2}{N}\ -\ (\frac{\Sigma f_i\ x_i}{N}})^2\]
| xi | fi | xi 2 | fi xi 2 | |
| 5 | 2 | 10 | 25 | 50 |
| 15 | 1 | 15 | 225 | 225 |
| 25 | 1 | 25 | 625 | 625 |
| 35 | 3 | 105 | 1225 | 3675 |
| N=7 |
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ x_i^2}{N}\ -\ (\frac{\Sigma f_i\ x_i}{N}})^2\]
\[=\ \sqrt{\frac{4575}{7}\ -\ (\frac{155}{7}})^2\]
\[=\ \sqrt{653.57\ -\ 490.18}\]
\[=\ \sqrt{163.3904}\]
\[\sigma\ =\ 12.78\]
\[\color {purple} {Example\ 18\ .}\ Find\ the\ standard\ deviation\ of\ the\ following\ data:\ \hspace{10cm}\]
| Class interval | 4 – 8 | 8 – 12 | 12 – 16 | 16 – 20 |
| frequency | 3 | 6 | 4 | 7 |
\[\hspace{5cm}\ October\ 2024\ June(Supp)\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ x_i^2}{N}\ -\ (\frac{\Sigma f_i\ x_i}{N})^2}, \ where\ N\ =\ \Sigma {f_i}\]
| Marks | fi | xi | fixi | xi 2 | fixi2 |
| 4 -8 | 3 | 6 | 18 | 36 | 108 |
| 8 -12 | 6 | 10 | 60 | 100 | 600 |
| 12 – 16 | 4 | 14 | 56 | 196 | 784 |
| 16 – 20 | 7 | 18 | 126 | 324 | 2268 |
| N = 20 |
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ x_i^2}{N}\ -\ (\frac{\Sigma f_i\ x_i}{N})^2}\]
\[=\ \sqrt{\frac{3760}{20}\ -\ (\frac{-260}{20})^2}\]
\[=\ \sqrt{188\ -\ (13)^2}\]
\[=\ \sqrt{188\ -\ 169}\]
\[=\ \sqrt{19}\]
\[\sigma\ =\ 4.36\ (approximately)\]
CURVE FITTING
Fitting a straight line using the method of least squares
Statement of principle of least squares
Let y = f(x) be the functional relationship sought between the variables (x,y). Let (xi,yi), i = 1,2,3,…,n. be the observed set of values of the variables (x,y). di = yi – f(xi) which is the difference between the observed value of y and the value of y. The principle of least squares states that the parameters involved in y = f(x) should be chosen in such a way that Σ is minimum.
Let us consider the fitting of the straight line y = ax + b — (1) to a set of ‘n’ points
(xi,yi), i = 1,2,3,…,n. Here a and b are arbitrary constants and can be found by using Normal equations.
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (1)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (2)\]
\[\color {purple} {Example\ 19\ .}\ Write\ down\ the\ normal\ equations\ to\ fit\ a\ straight\ line\ y\ =\ a\ x\ +\ b.\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023,\ April\ 2024,\ October\ 2024,\ April\ 25,\ Supp(June)\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
The Normal equations are
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (1)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (2)\]
\[\color {purple} {Example\ 20\ .}\ Fit\ a\ straight\ line\ to\ the\ following\ data\ by\ the\ method\ of\ least\ squares.\ \hspace{8cm}\]
| x | 0 | 1 | 2 | 3 | 4 |
| y | 1 | 1 | 3 | 4 | 6 |
\[\hspace{5cm}\ April\ 2024,\ April\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ line\ be\ y\ =\ a\ x\ +\ b\ ——–\ (1)\]
\[The\ Normal\ equations\ are\]
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (2)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (3)\]
| xi | yi | xi 2 | xi yi |
| 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
| 2 | 3 | 4 | 6 |
| 3 | 4 | 9 | 12 |
| 4 | 6 | 16 | 24 |
\[(2)\ becomes\ a\ (10)\ +\ 5b\ =\ 15\]
\[2\ a\ +\ b\ =\ 3\ ——–\ (4)\]
\[(3)\ becomes\ a\ (30)\ +\ b\ (10)\ =\ 43\]
\[3\ a\ +\ b\ =\ 4.3\ ——–\ (5)\]
\[Solving\ (4)\ and\ (5)\]
\[2\ a\ +\ b\ =\ 3\]
\[3\ a\ +\ b\ =\ 4.3\]
———————————-
\[a\ =\ 1.3\]
\[put\ a\ =\ 1.3\ in\ (4)\]
\[2(1.3)\ +\ b\ =\ 3\]
\[2.6\ +\ b\ =\ 3\]
\[b\ =\ 3\ -\ 2.6\]
\[b\ =\ 0.4\]
\[Eqn\ (1)\ becomes\ y\ =\ 1.3\ x\ +\ 0.4\]
\[\color {purple} {Example\ 21\ .}\ Using\ the\ method\ of\ least\ square\ ,\ fit\ a\ straight\ line\ to\ the\ following\ data:\ \hspace{8cm}\]
| x | 0 | 1 | 2 | 3 | 4 |
| y | 10 | 14 | 19 | 26 | 30 |
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ line\ be\ y\ =\ a\ x\ +\ b\ ——–\ (1)\]
\[The\ Normal\ equations\ are\]
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (2)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (3)\]
| xi | yi | xi 2 | xi yi |
| 0 | 10 | 0 | 0 |
| 1 | 14 | 1 | 14 |
| 2 | 19 | 4 | 38 |
| 3 | 26 | 9 | 78 |
| 4 | 30 | 16 | 120 |
\[(2)\ becomes\ a\ (10)\ +\ 5b\ =\ 99\]
\[2\ a\ +\ b\ =\ 19.8\ ——–\ (4)\]
\[(3)\ becomes\ a\ (30)\ +\ b\ (10)\ =\ 250\]
\[3\ a\ +\ b\ =\ 25\ ——–\ (5)\]
\[Solving\ (4)\ and\ (5)\]
\[2\ a\ +\ b\ =\ 19.8\]
\[3\ a\ +\ b\ =\ 25\]
———————————-
\[a\ =\ 5.2\]
\[put\ a\ =\ 5.2\ in\ (4)\]
\[2(5.2)\ +\ b\ =\ 19.8\]
\[10.4\ +\ b\ =\ 19.8\]
\[b\ =\ 19.8\ -\ 10.4\]
\[Eqn\ (1)\ becomes\ y\ =\ 5.2\ x\ +\ 9.4\]
\[\color {purple} {Example\ 22\ .}\ Fit\ a\ straight\ line\ to\ the\ following\ data:\ \hspace{15cm}\]
| x | 0 | 1 | 2 | 3 | 4 |
| y | 10 | 14 | 19 | 26 | 31 |
\[\hspace{5cm}\ October\ 2024,\ Supp (June)\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ line\ be\ y\ =\ a\ x\ +\ b\ ——–\ (1)\]
\[The\ Normal\ equations\ are\]
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (2)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (3)\]
| xi | yi | xi 2 | xi yi |
| 0 | 10 | 0 | 0 |
| 1 | 14 | 1 | 14 |
| 2 | 19 | 4 | 38 |
| 3 | 26 | 9 | 78 |
| 4 | 31 | 16 | 124 |
\[(2)\ becomes\ a\ (10)\ +\ 5b\ =\ 100\]
\[2\ a\ +\ b\ =\ 20\ ——–\ (4)\]
\[(3)\ becomes\ a\ (30)\ +\ b\ (10)\ =\ 254\]
\[3\ a\ +\ b\ =\ 25.4\ ——–\ (5)\]
\[Solving\ (4)\ and\ (5)\]
\[2\ a\ +\ b\ =\ 20\]
\[3\ a\ +\ b\ =\ 25.4\]
———————————-
\[a\ =\ 5.4\]
\[put\ a\ =\ 5.4\ in\ (4)\]
\[2(5.4)\ +\ b\ =\ 20\]
\[10.8\ +\ b\ =\ 20\]
\[b\ =\ 20\ -\ 10.8\]
\[b\ =\ 9.2\]
\[Eqn\ (1)\ becomes\ y\ =\ 5.4\ x\ +\ 9.2\]
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