a) If dx1+x^2 + dy1+y^2 = 0 , show that x+y1 xy = c , where c is constant. [2] b) Which of the following distribution is consistent from the information given below. [2] Distribution 'A'…

a) Solution to the Differential Equation

Given: $$\frac{dx}{1+x^2}+\frac{dy}{1+y^2}=0$$

Step 1: Rewrite the equation $$\frac{dx}{1+x^2}=-\frac{dy}{1+y^2}$$

Step 2: Integrate both sides $$\int \frac{dx}{1+x^2}=-\int \frac{dy}{1+y^2}$$

The integrals are standard: $${\tan }^{-1}(x)=-{\tan }^{-1}(y)+c$$ where $c$ is the constant of integration.

Step 3: Rearrange the equation $${\tan }^{-1}(x)+{\tan }^{-1}(y)=c$$

Step 4: Use the addition formula for ${\tan }^{-1}$ Recall that: $${\tan }^{-1}(a)+{\tan }^{-1}(b)={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)\text{if}ab<1$$

Applying this to our equation: $${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)=c$$

Step 5: Take the tangent of both sides $$\frac{x+y}{1-xy}=\tan (c)$$

Since $\tan (c)$ is a constant, we denote it as $c$ (a new constant): $$\frac{x+y}{1-xy}=c$$

Conclusion: We have shown that $\frac{x+y}{1-xy}=c$, where $c$ is a constant.

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b) Checking Consistency of Distributions

Given:

• Distribution A: Mean (${\mu }_A$) = 95, Standard Deviation (${\sigma }_A$) = 10
• Distribution B: Mean (${\mu }_B$) = 90, Standard Deviation (${\sigma }_B$) = 9

Assumption: The question implies that the two distributions are samples from the same population or are related in some way. To check consistency, we compare their coefficients of variation (CV), which is a measure of relative variability.

Step 1: Calculate the Coefficient of Variation (CV) for both distributions $$CV=\frac{\sigma }{\mu }\times 100\%$$

For Distribution A: $$C{V}_A=\frac{10}{95}\times 100\%\approx 10.53\%$$

For Distribution B: $$C{V}_B=\frac{9}{90}\times 100\%=10\%$$

Step 2: Compare the CVs The CVs of both distributions are very close (10.53% vs. 10%), suggesting that the relative variability is similar. This indicates that the two distributions are consistent in terms of their spread relative to their means.

Conclusion: Since the coefficients of variation are nearly equal, Distribution A and Distribution B are consistent.

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c) Probability of Getting Exactly Two Heads When Tossing Three Coins

Given: Three coins are tossed simultaneously. We need to find the probability of getting exactly two heads.

Step 1: Determine the total number of possible outcomes When tossing three coins, each coin has 2 possible outcomes: Heads (H) or Tails (T). Thus, the total number of possible outcomes is: $$2^3=8$$

The sample space $S$ is: $$S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$$

Step 2: Identify favorable outcomes (exactly two heads) The outcomes with exactly two heads are: $$\{HHT,HTH,THH\}$$

Number of favorable outcomes = 3.

Step 3: Calculate the probability $$P(\text{exactly two heads})=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{3}{8}$$

Conclusion: The probability of getting exactly two heads when tossing three coins is $\frac{3}{8}$.