3 Probability Questions: Exponential distribution and Normal

Question

1. A battery has an average lifespan of $2,000$ hours. It is exponentially distributed. $X \sim \text{exp}(\lambda = \frac{1}{2000})$. $10$ toy cars are using this kind of battery (each has its own) what is the probability at least 3 of the toy cars will still drive after $1,700$ hours?

2. You have a machine that creates nails which have in average a diameter of 4 cm. and a standart deviation of $0.1$ cm. Suppose the distribution of the diameter is Normal ($\sim N$) what is the percentage of the nails that will have a diameter over $4.012$

3. a light bulb is distributed exponentially with a lifespan of 1400 hours. The light bulb is inside a street lamp. what is the probability you would need to replace the light bulb 3 times in the next 2000 hours?
   I did not quite understand, $\lambda = \frac{1}{1400}$ and and so the question is what is the probability that it would turn off 3 times? thank you.

My solution:

1. I first calculated what is the probability that a certain car will keep driving after $1,700$ hours. $X \sim exp(\frac{1}{2000})$ . $\lambda = \frac{1700}{2000}$

And so the probability is $\frac{1}{2000} e^{\frac{1700}{2000}} \approx 2.137e(-4)$. and then what I do is little Bernoulli experiments:

$P(X \geq 3) = 1- P(X \leq 2) = 1-P(X=0)-P(X=1)-P(X=2)$

$P(X=i) = \binom{10}{i} (2.137*10^{-4})^i \cdot (1-(2.137*10^{-4}))^{10-i}$

Did I get it right?

2. $\mu = 4$ and $\sigma = 0.1$ and we need to calculate: $P(X > 4.012) = 1- P(X<4.012) = 1-P(Z < \frac{4.012 - 4}{0.1}) = 1- P(Z < 0.12) = 0.4522 $
   but in the answers there are only:

• 0.454
• 0.546
• 0.884
• 0.116

So which is right? is it the closest one? is it even the right answer (what I got)

Thank you!

Answer 1

1. not very good. The probability

$\mathbb{P}[X>1700]=e^{-\frac{1700}{2000}}\approx 42.74\%$

The rest of the procedure is ok

2. your calculation is ok...I do not know how they approximated the result but your is fine. Chose the closest to yours

Answer 2

At 1. the probability that a toy car will still drive after (at least) 1,700 hours is $P(X\geq 1700)=1-P(X\leq x)=1-\int_{0}^{1700} \frac{1}{2000} \cdot e^{-\frac{x}{2000}} \, dx=e^{-1700/2000}\approx 0.42741$.

In general it is $P(X\geq x)=e^{-\frac{x}{\lambda}}.$ Then indeed you can use the binomial distribution like you have done.

If I use this table I get $1-P(X\leq 0.12)=1 - 0.54776=0.45224 \approx0.4522$. This is the same result you have got (rounded to 4 decimal places). So the first option is the closest result.

Answer 3

For 2), you can derive it from the definition of Normal rvs. If $X \sim N(\mu, \sigma^2)$, then (set $z=\frac{x-\mu}{\sigma}$, and if $x=s, z=\frac{s-\mu}{\sigma}=t$) $$ P(X>s) = \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{s}^{\infty} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} dx = \frac{1}{\sqrt{2 \pi}}\int_{t}^{\infty}e^{-\frac{z^2}{2}}dz = \Phi(t) = \Phi(\frac{s-\mu}{\sigma}) $$ where $s=4.012$.