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.

  1. The distribution function is given by \[F(x)=\begin{cases}0& x\leq 5\\ \frac{x-5}{7}& 5<x\leq 12\\ 1&x\geq 12\end{cases}\]
  2. The probability that one of these spots last at least 7 seconds is
  3. The probability that one of these spots last more than 6 seconds, given that it will not last more than 10 seconds is
  4. 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 \(b>2\).

  1. \(b\) 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.

  1. \(\lambda\) in this case is equal to
  2. 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
  3. 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 \(\lambda=0.5\).

  1. \(P(X>2)=e^{-1}\)

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.

  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.

  1. The expected time that a car will be waiting until being inspected is 25 minutes.

Question 7

Let \(Z\sim\mathcal{N}(0,1)\)

  1. \(P(0<Z\leq 2.05)=0.9790\)
  2. \(P(-1.22\leq Z < 1.05)=0.7419\)
  3. \(P(Z\geq -2.05)=0.0202\)
  4. The value for \(k\) such that \(P(|Z|>k)=0.05\) is equal to 1.96.
  5. 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

  1. \(P(6<X\leq 12)=0.8849\)
  2. \(P(0\leq X \leq 8) = 0.5404\)
  3. \(P(X\leq -4)=0.0228\)
  4. \(P(|X-6|>10)=0.0456\)
  5. \(k\) 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 \(P(a\leq X\leq 15)=0.7332\)

  1. \(a\) 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.

  1. The percentage of days in which the amount of the deposits is in between 105 and 135 euro is 96.99%
  2. The probability that the amount of deposits is above its mean, in the days that it never surpasses 125 is 0.32.
  3. 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. \(X\sim \mathcal{N}(250,100)\).

  1. The probability of a package weight between 230.4 and 269.6 grams is (use 2 decimals, like 0.12)
  2. 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)
  3. Find and interpret the value for \(k\) such that \(P(X<k)=0.9\) Answer: grs (use only 2 decimals, like 123.45)
  4. 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 \(Z\sim\mathcal{N}(0,1)\)

  1. \(P(X<Z+1)=0.5\)
  2. The value for \(a\) such that \(P(X>Z+a)=0.1\) is equal to

Question 13

Check the truthfulness of each sentence

  1. If \(Y\sim U(0,1)\) and \(0<x<z<1\), then: \[P(x<Y<z)=z-x\]
  2. If \(X\sim U(0,1)\) with \(0<a<b<1\), then \[P(X<a|X<b)=\frac{a}{b}\]
  3. If \(X\) is normally distributed, then the media coincides with the mean.
  4. If \(X\sim \mathcal{N}(0,1)\), and \(Y\sim\mathcal{N}(1,2)\), with \(X\) and \(Y\) independent r.v.s, then \(2X+Y\sim\mathcal{N}(1,6)\)
  5. Letting \(\Phi\) be the distribution function of a r.v. distributed normal standard, then \(\Phi(-x)=\Phi(x)\) \(\forall x\in\mathbb{R}\).
  6. Letting \(\phi\) represent the density of a r.v. distributed normal standard, then \(\phi(-x)=1-\phi(x)\) \(\forall x\in\mathbb{R}\).