Statistics

The two regression lines intersect at the mean values $(\bar x, \bar y)$.

Given: $3x + 4y = 65$ $3x + y = 32$

Subtracting: $3y = 33 \Rightarrow y = 11$

Putting $y = 11$ in $3x + y = 32$: $3x + 11 = 32 \Rightarrow 3x = 21 \Rightarrow x = 7$

So, $\bar x = 7,\qquad \bar y = 11$

Mean of x = 7 and mean of y = 11.

For paired data $(x,y)$ there are two regression coefficients, found by the method of least squares.

Regression coefficient of $y$ on $x$ (from the normal equations $\sum y=na+b\sum x$ and $\sum xy=a\sum x+b\sum x^2$):

$b_{yx}=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=r\,\frac{\sigma_y}{\sigma_x}$

Regression coefficient of $x$ on $y$:

$b_{xy}=\frac{n\sum xy-\sum x\sum y}{n\sum y^2-(\sum y)^2}=r\,\frac{\sigma_x}{\sigma_y}$

Also, $b_{yx}\cdot b_{xy}=r^2$, so $r=\pm\sqrt{b_{yx}\,b_{xy}}$ (with the sign of the regression coefficients), and both coefficients always have the same sign.

The correlation coefficient between $x$ and $y$ cannot be determined because no paired data for $x$ and $y$ are provided. A numerical value of $r$ requires the paired observations or sufficient summary statistics.

a) Karl Pearson’s coefficient of correlation

Here,

Karl Pearson’s coefficient of correlation is

Substituting the values:

So,

Interpretation: There is a weak negative correlation between price and demand. As price increases, demand tends to decrease slightly, but the relationship is not strong.

b) Equation of regression line $Y$ on $X$

The regression coefficient of $Y$ on $X$ is

Now,

The regression line is

Hence, the required regression line is

Exact form:

c) Number of trials of binomial distribution

Given:

Now,

Also,

Therefore,

The number of trials is 10.

Given:
$n=5$, $\bar X = \dfrac{262}{5}=52.4$, $\bar Y = \dfrac{226}{5}=45.2$

Karl Pearson’s coefficient of correlation: $r=\frac{\sum xy}{\sqrt{\sum x^2 \cdot \sum y^2}}$ where $x=X-\bar X$ and $y=Y-\bar Y$.

Now, $r=\frac{131.60}{\sqrt{59.20 \times 318.80}}$ $r=\frac{131.60}{\sqrt{18872.96}}$ $r=\frac{131.60}{137.38}$ $r=0.9579 \approx 0.958$

Interpretation:
The coefficient of correlation is $r=0.958$, which is very close to $+1$. Hence, there is a strong positive correlation between height and weight. That is, as height increases, weight also tends to increase.

Let age be $x$ (months) and weight be $y$ (kg). $n=5$.

Here $\sum x=17$, so $\bar x=3.4$ and $\bar y=8.4$.

Weight at 4 months (regression of $y$ on $x$)

$b_{yx}=\frac{5(165)-17(42)}{5(75)-17^2}=\frac{111}{86}=1.2907$

$y-8.4=1.2907(x-3.4)\ \Rightarrow\ y=4.0116+1.2907x$

At $x=4$: $y=4.0116+5.1628\approx 9.17$ kg.

Age when weight is 6 kg (regression of $x$ on $y$)

$b_{xy}=\frac{5(165)-17(42)}{5(382)-42^2}=\frac{111}{146}=0.7603$

$x-3.4=0.7603(y-8.4)\ \Rightarrow\ x=-2.9865+0.7603y$

At $y=6$: $x=-2.9865+4.5618\approx 1.58$ months.

The baby's weight at 4 months is about 9.17 kg, and a 6 kg baby is about 1.58 months old.

(c) 8

Geometric mean between $a$ and $b$ is $\sqrt{ab} = \sqrt{1 \times 64} = \sqrt{64} = 8$.

(b) 2.24

Using Karl Pearson's coefficient of skewness, $S_K(P)=\frac{\bar X-M_o}{\sigma}\Rightarrow \sigma=\frac{1.4}{0.625}=2.24$

Let height be $x$ and weight be $y$.

Here, $n=5,\quad \bar x=\frac{25}{5}=5,\quad \bar y=\frac{238}{5}=47.6$

Karl Pearson's correlation coefficient is $r=\frac{\Sigma dx\,dy}{\sqrt{\Sigma dx^2 \cdot \Sigma dy^2}}$ $r=\frac{50}{\sqrt{2.5 \times 1043.2}}$ $r=\frac{50}{\sqrt{2608}}$ $r=\frac{50}{51.0686}=0.9791 \approx 0.98$

Interpretation: Since $r=0.98$ is very close to $+1$, there is a very high degree of positive correlation between height and weight. That is, as the height of the students increases, their weight also increases.

The two regression lines intersect at the mean values $(\bar X,\bar Y)$.

Given: $3X+2Y-26=0 \Rightarrow 3X+2Y=26$ $6X+Y-31=0 \Rightarrow 6X+Y=31$

From the second equation: $Y=31-6X$

Substitute in the first: $3X+2(31-6X)=26$ $3X+62-12X=26$ $-9X=-36$ $X=4$

Then, $Y=31-6(4)=31-24=7$

Mean values: $\boxed{\bar X=4,\quad \bar Y=7}$

Given regression equations: $3X + 2Y - 26 = 0 \quad \text{... (i)}$ $6X + Y - 31 = 0 \quad \text{... (ii)}$

Step 1: Identify which equation is which.

For equation (i): $2Y = 26 - 3X \Rightarrow Y = 13 - \dfrac{3}{2}X$, so $b_{yx} = -\dfrac{3}{2}$

For equation (ii): $6X = 31 - Y \Rightarrow X = \dfrac{31}{6} - \dfrac{1}{6}Y$, so $b_{xy} = -\dfrac{1}{6}$

Step 2: Verify. $b_{xy} \times b_{yx} = \left(-\dfrac{1}{6}\right)\left(-\dfrac{3}{2}\right) = \dfrac{1}{4} \le 1 \quad \checkmark$

(The reverse assignment gives $b_{xy} \times b_{yx} = 4 > 1$, which is impossible.)

Answer: $b_{yx} = -\dfrac{3}{2} = -1.5, \qquad b_{xy} = -\dfrac{1}{6} \approx -0.167$

Also, $r = -\sqrt{0.25} = -0.5$ (negative, since both coefficients are negative).