Solve these questions step by step and show me formulas of mean and variance

Given a sample of 160 people:

• Type O blood: 40 people
• Type A blood: 60 people
• Type B blood: 36 people
• Type AB blood: 24 people Total people: $40 + 60 + 36 + 24 = 160$.

The frequency distribution is:

$$\begin{array}{|c|c|c|} \hline \text{Blood Type} & \text{Frequency (Number of People)} & \text{Relative Frequency (Probability)} \\ \hline \text{O} & 40 & \frac{40}{160} = 0.25 \\ \text{A} & 60 & \frac{60}{160} = 0.375 \\ \text{B} & 36 & \frac{36}{160} = 0.225 \\ \text{AB} & 24 & \frac{24}{160} = 0.15 \\ \hline \text{Total} & 160 & 1.00 \\ \hline \end{array}$$

• a) Probability a person has type O blood:

  $$P(\text{Type O}) = \frac{\text{Number of Type O}}{\text{Total Number of People}} = \frac{40}{160} = \frac{1}{4}$$ $$P(\text{Type O}) = \boxed{0.25}$$

• b) Probability a person has type A or type B blood: Since having type A and type B blood are mutually exclusive events:

  $$P(\text{Type A or Type B}) = P(\text{Type A}) + P(\text{Type B})$$ $$P(\text{Type A or Type B}) = \frac{60}{160} + \frac{36}{160} = \frac{96}{160}$$ $$P(\text{Type A or Type B}) = \frac{3}{5} = \boxed{0.60}$$

• c) Probability a person has neither type A nor type O blood: This means the person has type B or type AB blood.

  $$P(\text{Neither A nor O}) = P(\text{Type B or Type AB}) = P(\text{Type B}) + P(\text{Type AB})$$ $$P(\text{Neither A nor O}) = \frac{36}{160} + \frac{24}{160} = \frac{60}{160}$$ $$P(\text{Neither A nor O}) = \frac{3}{8} = \boxed{0.375}$$

• d) Probability a person does not have type AB blood: This is the complement of having type AB blood.

  $$P(\text{Not Type AB}) = 1 - P(\text{Type AB})$$ $$P(\text{Not Type AB}) = 1 - \frac{24}{160} = 1 - \frac{3}{20} = \frac{17}{20}$$ $$P(\text{Not Type AB}) = \boxed{0.85}$$

Q2. Mean and Variance of a Continuous Random Variable

The probability density function (PDF) is given by:

$$f(x) = \begin{cases} \frac{5}{2} \cos x & \text{if } -\frac{\pi}{2} \le x \le \frac{\pi}{2} \\ 0 & \text{otherwise} \end{cases}$$

Formulas:
For a continuous random variable $X$ with PDF $f(x)$:

• The mean (expected value) is given by:

$$E[X] = \mu = \int_{-\infty}^{\infty} x f(x) dx$$

• The variance is given by:

$$Var(X) = \sigma^2 = E[X^2] - (E[X])^2$$

where $E[X^2]$ is the expected value of $X^2$:

$$E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) dx$$

$$E[X] = \int_{-\pi/2}^{\pi/2} x \left(\frac{5}{2} \cos x\right) dx$$

The integrand $g(x) = x \cos x$ is an odd function because $g(-x) = (-x)\cos(-x) = -x\cos x = -g(x)$.
Since the interval of integration $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is symmetric about 0, the integral of an odd function over this interval is 0.

$$E[X] = \frac{5}{2} \int_{-\pi/2}^{\pi/2} x \cos x \, dx = \frac{5}{2} \times 0$$ $$E[X] = \boxed{0}$$

$$E[X^2] = \int_{-\pi/2}^{\pi/2} x^2 \left(\frac{5}{2} \cos x\right) dx$$

The integrand $h(x) = x^2 \cos x$ is an even function because $h(-x) = (-x)^2\cos(-x) = x^2\cos x = h(x)$.
For an even function integrated over a symmetric interval, we can write:

$$E[X^2] = \frac{5}{2} \times 2 \int_{0}^{\pi/2} x^2 \cos x \, dx = 5 \int_{0}^{\pi/2} x^2 \cos x \, dx$$

We use integration by parts for $\int x^2 \cos x \, dx$:
$\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx$
For $\int 2x \sin x \, dx$:
$\int 2x \sin x \, dx = -2x \cos x - \int (-2 \cos x) \, dx = -2x \cos x + 2 \sin x$
Substituting back:
$\int x^2 \cos x \, dx = x^2 \sin x - (-2x \cos x + 2 \sin x) = x^2 \sin x + 2x \cos x - 2 \sin x$
Now, evaluate the definite integral from 0 to $\frac{\pi}{2}$:

$$\left[ x^2 \sin x + 2x \cos x - 2 \sin x \right]_{0}^{\pi/2}$$ $$= \left( \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + 2\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - 2 \sin\left(\frac{\pi}{2}\right) \right) - (0)$$ $$= \left( \frac{\pi^2}{4}(1) + \pi(0) - 2(1) \right) - 0$$ $$= \frac{\pi^2}{4} - 2$$

Therefore,

$$E[X^2] = 5 \left( \frac{\pi^2}{4} - 2 \right)$$ $$E[X^2] = \frac{5\pi^2}{4} - 10$$

$$Var(X) = E[X^2] - (E[X])^2$$ $$Var(X) = \left( \frac{5\pi^2}{4} - 10 \right) - (0)^2$$ $$Var(X) = \boxed{\frac{5\pi^2}{4} - 10}$$