Chapter III Part II - Statistics I - Exercises
This is a set of exercises created by the teaching Faculty for Statistics I, Chapter III, part II, continuous probability models, from the Lisbon Accounting and Business School.
Questions
Question 1
The length of short commercial spots (between 5 and 12 sec) on a streaming platform can be considered a random variable with uniform distribution.
- The distribution function is given by \[F(x)=\begin{cases}0& x\leq 5\\ \frac{x-5}{7}& 5<x\leq 12\\ 1&x\geq 12\end{cases}\]
- The probability that one of these spots last at least 7 seconds is
- The probability that one of these spots last more than 6 seconds, given that it will not last more than 10 seconds is
- The expected value and standard deviation for this r.v. is \(8.5\) and \(2\) respectively, both in seconds.
Question 2
Let the r.v. \(X\sim U(2,b)\) with \(b>2\).
- \(b\) must equal what in order for \(P(3\leq X\leq 5)=0.4\)?
Question 3
In a factory, the time to produce some item is a r.v. distributed exponentially with expected value of 5 minutes.
- \(\lambda\) in this case is equal to
- You now that the item has been in production for at least 2 minutes. The probability of taking at least 4 more minutes until completed is
- Choose randomly 5 items. The probability that two of them took less than 4 minutes to produce is 0.2757.
Question 4
Let \(X\) be a r.v. distributed exponentially with \(\lambda=0.5\).
- \(P(X>2)=e^{-1}\)
Question 5
On the consultation of Dr. Rotcod The time until the first appointment, and between consecutive appointments, are independent random variables distributed exponentially with parameter equal to 0.1.
- The probability of starting the first appointment after the first 10 minutes is 0.3679.
Question 6
The number of cars checked in the inspection center, in an hour, is a r.v. distributed Poisson with an expected value of 4.
- The expected time that a car will be waiting until being inspected is 25 minutes.
Question 7
Let \(Z\sim\mathcal{N}(0,1)\)
- \(P(0<Z\leq 2.05)=0.9790\)
- \(P(-1.22\leq Z < 1.05)=0.7419\)
- \(P(Z\geq -2.05)=0.0202\)
- The value for \(k\) such that \(P(|Z|>k)=0.05\) is equal to 1.96.
- The value for \(k\) such that \(P(|Z|<k)=0.90\) is equal to 1.645.
Question 8
If \(X\sim \mathcal{N}(\mu=6,\sigma^2=25)\) then
- \(P(6<X\leq 12)=0.8849\)
- \(P(0\leq X \leq 8) = 0.5404\)
- \(P(X\leq -4)=0.0228\)
- \(P(|X-6|>10)=0.0456\)
- \(k\) must be equal to 0.4 such that \(P(X>k)=0.90\).
Question 9
Let \(X\) a r.v. distributed \(\mathcal{N}(12,2)\). Considere \(P(a\leq X\leq 15)=0.7332\)
- \(a\) must be equal to 10.32.
Question 10
The number of deposits made in a fintech, every day, is a r.v. distributed \(\mathcal{N}(120,64)\) in euro.
- The percentage of days in which the amount of the deposits is in between 105 and 135 euro is 96.99%
- The probability that the amount of deposits is above its mean, in the days that it never surpasses 125 is 0.32.
- The mean and variance of the weekly total amount of deposits (i.e. in 5 days) are respectively 600 euro and 320 euro2.
Question 11
An item is produce in a way such that its weight (in grams) is a r.v. \(X\sim \mathcal{N}(250,100)\).
- The probability of a package weight between 230.4 and 269.6 grams is (use 2 decimals, like 0.12)
- The quality control department rejects packages with a weight farther away from the mean than 15.3 grams. The probability that a random package is rejected is (use three decimals, like 0.123)
- Find and interpret the value for \(k\) such that \(P(X<k)=0.9\) Answer: grs (use only 2 decimals, like 123.45)
- Randomly picking 4 packages, compute the probability that the total weight is above its mean. Justify. Answer:
Question 12
Being \(X\) and \(Z\) independent random variables, distributed \(X\sim\mathcal{N}(1,1)\) and \(Z\sim\mathcal{N}(0,1)\)
- \(P(X<Z+1)=0.5\)
- The value for \(a\) such that \(P(X>Z+a)=0.1\) is equal to
Question 13
Check the truthfulness of each sentence
- If \(Y\sim U(0,1)\) and \(0<x<z<1\), then: \[P(x<Y<z)=z-x\]
- If \(X\sim U(0,1)\) with \(0<a<b<1\), then \[P(X<a|X<b)=\frac{a}{b}\]
- If \(X\) is normally distributed, then the media coincides with the mean.
- If \(X\sim \mathcal{N}(0,1)\), and \(Y\sim\mathcal{N}(1,2)\), with \(X\) and \(Y\) independent r.v.s, then \(2X+Y\sim\mathcal{N}(1,6)\)
- Letting \(\Phi\) be the distribution function of a r.v. distributed normal standard, then \(\Phi(-x)=\Phi(x)\) \(\forall x\in\mathbb{R}\).
- Letting \(\phi\) represent the density of a r.v. distributed normal standard, then \(\phi(-x)=1-\phi(x)\) \(\forall x\in\mathbb{R}\).