1 answer

For each situation described, determine an appropriate distribution to model the situation, and (if the information...

Question:

For each situation described, determine an appropriate distribution to model the situation, and (if the information is includ

For each situation described, determine an appropriate distribution to model the situation, and (if the information is included) the parameter values. Additionally, calculate any probabilities that are mentioned. i) The lifetime of a car battery. Advertisers of the brand say their battery tends to last 3.5 years. If a battery has lasted 3 years, what is the probability that the battery lasts until the claimed average? ii) The weight of mice fed with a new nutrient mix. The data looks bell shaped and centered around 40g. iii) The number of calls per hour received by a service center. Ticketing software shows an average of 15 calls per hour. What is the probability of seeing at least 1 customer in a two hour time frame?

Answers

i) The lifetime can be modeled using an exponential distribution. The claim of 3.5 years of life can be used as the average life time of battery.

Let X years be the lifetime of any given battery. We can say that X has an Exponential distribution with mean 3.5 years and parameter \lambda=\frac{1}{\text{mean}}=\frac{1}{3.5}

ans: Let X years be the lifetime of any given battery. X has an Exponential distribution with parameter \lambda=\frac{1}{3.5}

X\sim \text{Exponential}\left(\frac{1}{3.5} \right )

The cdf of X is

\begin{align*} P(X\le x)&=1-e^{-\lambda x}\\ &=1-e^{-\frac{1}{3.5} x},\quad x>0 \end{align*}

The conditional probability that the battery lasts until the claimed average (which is 3.5 years or more ) given that it has lasted 3 years is

\begin{align*} P(X>3.5\mid X>3)&=\frac{P(X>3.5\cap X>3)}{P(X>3)}\quad\text{using the formula for conditional probability}\\ &=\frac{P(X>3.5)}{P(X>3)}\text{ as }(X>3.5\text{ and }X>3)\text{ is same as }X>3.5\\ &=\frac{1-P(X<3.5)}{(1-P(X<3))}\\ &=\frac{1-(1-e^{\frac{1}{3.5}\times 3.5})}{1-(1-e^{\frac{1}{3.5}\times 3})}\\ &=\frac{e^{\frac{1}{3.5}\times 3.5}}{e^{\frac{1}{3.5}\times 3}}\\ &=0.8669 \end{align*}

ans: If a battery lasted 3 years, the probability that the battery lasts until the claimed average is 0.8669

ii) Since the data looks bell shaped and centered around 40g , we can assume a normal distribution with mean 40 g

ans:

Let X be the weight of a randomly selected mouse fed with the new nutrient mix. We can say that X is normally distributed with mean \mu=40 and standard deviation \sigma

X\sim N(40,\sigma^2)

iii) The number of calls per hour can be modeled using a Poisson process with a rate 15 calls per hour

ans: Let N(t) be the number of calls in a time period t hours.

We can say that N(t) is a Poisson process with rate \lambda=15

N(t)\sim \text{Poisson}(15t)

The pmf of N(t) is

\begin{align*} P(N(t)=n)&=\frac{(\lambda t)^ne^{-\lambda t}}{n!}\\ &=\frac{(15t)^ne^{-15t}}{n!},\quad n=0,1,\ldots \end{align*}

The probability of seeing at least 1 customer (customer call?) in a 2 hour time frame (t=2) is

\begin{align*} P(N(t=2)\ge 1)&=1-P(N(t=2)=0)\\ &=1-\frac{(15\times 2)^0e^{-15\times 2}}{0!}\\ &=1-0.0000\\ &=1 \end{align*}

ans: The probability of seeing at least 1 customer in a 2 hour time frame is 1

.

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