$ f(x) = \begin{cases} e^{-x}, & \text{if } x>0, \\ 0, & \text{otherwise,} \end{cases} $
and that of Brand II follows a gamma distribution with the probability density function
$ g(x) = \begin{cases} \frac{x}{4} e^{-x/2}, & \text{if } x>0, \\ 0, & \text{otherwise.} \end{cases} $
The company uses the batteries of Brands I and II in proportion of $20\%$ and $80\%$ respectively, in its mobiles. The probability that a randomly selected mobile has the battery life more that $2$ years is
Understanding Battery Life Probabilities
This problem asks for the probability that a randomly selected mobile battery lasts longer than 2 years. The company uses two brands of batteries, Brand I and Brand II, with different lifetime distributions and in different proportions. We need to calculate the overall probability by considering the probability of each brand and the proportion in which they are used.
Brand I: Exponential Distribution
The lifetime of Brand I batteries follows an exponential distribution with the probability density function (PDF):
This is an exponential distribution with a rate parameter $\lambda = 1$. The probability that a Brand I battery lasts more than 2 years is calculated by integrating the PDF from 2 to infinity:
The integral of $e^{-x}$ is $-e^{-x}$. Evaluating this from 2 to infinity:
So, the probability that a Brand I battery lasts more than 2 years is $e^{-2}$.
Brand II: Gamma Distribution
The lifetime of Brand II batteries follows a gamma distribution with the PDF:
This PDF corresponds to a Gamma distribution with shape parameter $k=2$ and rate parameter $\beta = 1/2$ (or scale $\theta = 2$). The probability that a Brand II battery lasts more than 2 years is found by integrating the PDF from 2 to infinity:
We can solve this integral using integration by parts ($\int u dv = uv - \int v du$). Let $u = \frac{x}{4}$ and $dv = e^{-x/2} dx$. Then $du = \frac{1}{4} dx$ and $v = \int e^{-x/2} dx = -2e^{-x/2}$.
Applying the integration by parts formula:
Evaluate the first term:
Using L'Hopital's rule for the limit $\lim_{b \to \infty} (-\frac{b}{2} e^{-b/2}) = \lim_{b \to \infty} (-\frac{1}{2} / (-\frac{1}{2} e^{-b/2})) = \lim_{b \to \infty} e^{-b/2} = 0$.
So the first term is $0 - (-1 \cdot e^{-1}) = e^{-1}$.
Evaluate the integral part:
Combining the parts, the probability for Brand II is $e^{-1} + e^{-1} = 2e^{-1}$.
Therefore, $ P(\text{Life} > 2 | \text{Brand II}) = 2e^{-1} $.
Calculating Overall Probability
The company uses Brand I batteries in proportion $P(\text{Brand I}) = 20\% = \frac{1}{5}$ and Brand II batteries in proportion $P(\text{Brand II}) = 80\% = \frac{4}{5}$.
We use the Law of Total Probability to find the overall probability that a randomly selected battery lasts more than 2 years:
Substitute the calculated probabilities and proportions:
This result represents the total probability that a mobile battery, chosen randomly considering the usage proportions of both brands, will have a life of more than 2 years.
Important Questions from Random Variables
- Consider a discrete random variable $X$ with the probability mass function
$ P(X = 0) = \frac{\theta}{3}, \ P(X = 1) = 1 - \frac{\theta}{2}, \ P(X = 2) = \frac{\theta}{6}, $
where $\theta \in (0,1)$ is an unknown parameter. In a random sample of size $90$ from this distribution, the observed counts for $X = 0, 1$ and $2$ are $20, 60$ and $10$, respectively. Then, the maximum likelihood estimate of $\theta$ is - Let $X$ be a random sample of size $1$ from the probability density function
$ f(x|\theta) = \begin{cases} \frac{3}{\theta^3} (\theta - x)^2, & \text{if } 0<x<\theta, \\ 0, & \text{otherwise.} \end{cases} $
If $ \left(\frac{X}{1-\lambda_1}, \frac{X}{1-\lambda_2}\right) $ is a confidence interval for $\theta$ with confidence coefficient $1 - \alpha$, where $\lambda_i \in (0,1), \ i = 1,2, \ \lambda_1<\lambda_2$, and $\alpha \in (0,1)$, then which of the following statements is true? - Let $X_1, X_2, . . ., X_n$ be a random sample from a continuous distribution with the common probability density function
$ f(x|\theta) = \begin{cases} \frac{2\theta^2}{x^{\theta+1}}, & \text{if } x>2, \\ 0, & \text{otherwise,} \end{cases} $
where $\theta (> 0)$ is an unknown parameter. Suppose $P(Y>\chi^2_{m,\beta}) = \beta$, where $Y \sim \chi^2_m$. For testing $H_0: \theta = 1$ against $H_1 : \theta>1$, a uniformly most powerful test of size $\alpha, \ 0<\alpha<1$, will reject $H_0$ if - Suppose we want to estimate the population mean $\bar{Y}$ of a variable for a finite population of size $85$, with $34$ Statisticians and $51$ Biologists. We consider the following sampling scheme:
A stratified random sample with $2$ strata of Statisticians (Stratum-1) and Biologists (Stratum-2), where $12$ Statisticians and $15$ Biologists are drawn from Stratum-1 and Stratum-2, respectively, using SRSWOR scheme.
Denote $\bar{y}_S, \bar{y}_B$, and $\bar{y}$ as the mean of the variable among the Statistician sample, Biologist sample, and the combined sample, respectively. Which of the following is an unbiased estimator of $\bar{Y}$? Let $X$ be a random variable with the cumulative distribution function
$F_X(x) = \begin{cases} 0, & \text{if } x<0, \\ \frac{x+2}{5}, & \text{if } 0 \le x<2, \\ 1, & \text{if } x \ge 2. \end{cases}$
Then, which of the following statements are true?
