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Probability distributions - Multiple choice test 1

Take a multiple choice test on probability distributions.

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In this test you will be asked questions about the following topics:

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When you give a wrong answer, it is marked with nav_button and the right answer is marked with nav_button.

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Question 1. Suppose you buy a lottery ticket. You can either win 1 dollar (with probability $frac{1}{10}$) or win nothing (with probability $frac{9}{10}$). The amount you win is a random variable and it has a:

nav_button Gamma distribution

nav_button normal distribution

nav_button exponential distribution

nav_button Bernoulli distribution

nav_button Explanation | nav_button Next question

Explanation. Denote by X the amount you win. X can take only two values (1 and 0), with probabilities $frac{1}{10}$ and $frac{9}{10}$.This corresponds to the definition of a Bernoulli random variable. For more details, see the lecture entitled Bernoulli distribution.

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Question 2. Consider a Bernoulli random variable X, with support [eq1] and probability mass function:[eq2]The variance of X is:

nav_button [eq3]

nav_button [eq4]

nav_button [eq5]

nav_button [eq6]

nav_button Explanation | nav_button Next question

Explanation. The variance of a Bernoulli random variable X is calculated as follows:[eq7]For more details, see the lecture entitled Bernoulli distribution.

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Question 3. Consider a Bernoulli random variable X, with support [eq8] and probability mass function:[eq9]The moment generating function of X is:

nav_button [eq10]

nav_button [eq11]

nav_button [eq12]

nav_button [eq13]

nav_button Explanation | nav_button Next question

Explanation. The moment generating function of a Bernoulli random variable X is calculated as follows:[eq14]For more details, see the lecture entitled Bernoulli distribution.

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Question 4. Suppose you independently flip a coin $3$ times and the outcome of each toss can be either head (with probability $frac{1}{2}$) or tails (also with probability $frac{1}{2}$). Denote by X the number of times the outcome is tails (out of the $3$ tosses). The random variable X has a:

nav_button Bernoulli distribution

nav_button Poisson distribution

nav_button binomial distribution

nav_button exponential distribution

nav_button Explanation | nav_button Next question

Explanation. X is the sum of $3$ independent Bernoulli random variables (that take value 1 in case the outcome is tails and 0 in case the outcome is head). As we explained in the lecture entitled binomial distribution, the sum of n independent Bernoulli random variables is a binomial random variable.

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Question 5. Suppose you independently play a game $4$ times. Each time you play, the probability of winning is $frac{1}{2}$. Denote by X the number of games you win (out of the $4$ games you play). Denote by [eq15] the probability mass function of X. Then, when [eq16]:

nav_button [eq17]

nav_button [eq18]

nav_button [eq19]

nav_button [eq20]

nav_button Explanation | nav_button Next question

Explanation. X is the sum of $4$ independent Bernoulli random variables that take value 1 in case you win the game (with probability $frac{1}{2}$) and 0 in case you lose (with probability $frac{1}{2}$). As we explained in the lecture entitled binomial distribution, the sum of n independent Bernoulli random variables is a binomial random variable. The probability mass function of a binomial random variable is:[eq21]Therefore, in our case, when [eq22]:[eq23]See the lecture entitled binomial distribution for further details.

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Question 6. Suppose you independently throw a dart n times. Each time you throw a dart, the probability of hitting the target is p. Denote by X the number of times you hit the target (out of the n total times you throw a dart). The expected number of hits is:

nav_button [eq24]

nav_button [eq25]

nav_button [eq26]

nav_button [eq27]

nav_button Explanation | nav_button Next question

Explanation. X is the sum of n independent Bernoulli random variables that take value 1 in case you hit the target (with probability $p $) and 0 in case you miss it (with probability $1-p$). As we explained in the lecture entitled binomial distribution, the sum of n independent Bernoulli random variables is a binomial random variable. The expected value of a binomial random variable is:[eq28]See the lecture entitled binomial distribution for further details.

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Question 7. Suppose you independently throw a dart $3$ times. Each time you throw a dart, the probability of missing the target is $frac{1}{4}$. Denote by X the number of times you hit the target (out of the $3$ total times you throw a dart). The variance of X is:

nav_button [eq29]

nav_button [eq30]

nav_button [eq31]

nav_button [eq32]

nav_button Explanation | nav_button Next question

Explanation. X is the sum of n independent Bernoulli random variables that take value 1 in case you hit the target (with probability $frac{3}{4}$) and 0 in case you miss it (with probability $frac{1}{4}$). As we explained in the lecture entitled binomial distribution, the sum of n independent Bernoulli random variables is a binomial random variable. The variance of a binomial random variable is:[eq33]See the lecture entitled binomial distribution for further details.

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Question 8. The probability that a working steel cutting machine breaks down during a time interval (given that it has not broken down before that time interval) is approximately proportional to the length of that time interval. Denote by X the total time elapsed before the machine breaks down. X can be assumed to have a:

nav_button Gamma distribution

nav_button Poisson distribution

nav_button Exponential distribution

nav_button Uniform distribution

nav_button Explanation | nav_button Next question

Explanation. Denote by X the time we need to wait before a certain event occurs. Suppose X is unknown (it is a random variable). Take any time interval [eq34]. If the probability[eq35]is (approximately) proportional to $Delta t$, then X has an exponential distribution. In other words, X has an exponential distribution if the conditional probability that the event occurs during a time interval is proportional to the length of that time interval. See the lecture entitled Exponential distribution for more details on the exponential distribution.

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Question 9. Let X be an exponential random variable with parameter $lambda $. The distribution function of X is:

nav_button [eq36]

nav_button [eq37]

nav_button [eq38]

nav_button [eq39]

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Explanation. The probability density function of an exponential random variable is:[eq40]The distribution function is the integral of the probability density function:[eq41]See the lecture entitled Exponential distribution for more details on the distribution function of the exponential distribution.

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Question 10. The probability that a new customer enters our shop during a given minute is approximately $1%$, irrespective of how many customers have entered the shop during the previous minutes. Assume that the total time we need to wait before a new customer enters our shop (denote it by X) has an exponential distribution. Then, the probability that no customer enters the shop during the next hour is:

nav_button [eq42]

nav_button [eq43]

nav_button [eq44]

nav_button [eq45]

nav_button Explanation | nav_button Next question

Explanation. Time is measured in minutes. Therefore, the probability that no customer enters the shop during the next hour is:[eq46]where [eq47] is the distribution function of X. Since X is an exponential random variable with rate parameter $1%$, its distribution function is:[eq48]Therefore:[eq49]See the lecture entitled Exponential distribution for more details.

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Question 11. Denote by X the time (in years) that will elapse before a certain firm goes bankrupt. X has an exponential distribution and its expected value is $10$ years. The probability density function of X is:

nav_button [eq50]

nav_button [eq51]

nav_button [eq52]

nav_button [eq53]

nav_button Explanation | nav_button Next question

Explanation. The probability density function of an exponential random variable X with rate parameter $lambda $ is:[eq54]and its expected value is:[eq55]In this case the expected value is 10 years. Therefore:[eq56]which implies:[eq57]Thus, the probability density function of X is:[eq58]See the lecture entitled Exponential distribution for more details on the exponential distribution and its expected value.

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Question 12. The time between the arrival of one customer and the arrival of the next customer has an exponential distribution and is independent of previous arrivals. The number of customers that will arrive during the next hour has:

nav_button a Poisson distribution

nav_button a Gamma distribution

nav_button an exponential distribution

nav_button a binomial distribution

nav_button Explanation | nav_button Next question

Explanation. If a certain event can occur many times within a unit of time and the time elapsed between two successive occurrences of the event has an exponential distribution (independent of previous occurrences), then the total number of occurrences within a unit of time has a Poisson distribution. See the lecture entitled Poisson distribution for more details on the Poisson distribution.

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Question 13. Let X be a Poisson random variable with rate parameter $lambda =2$. Then:

nav_button [eq59]

nav_button [eq60]

nav_button [eq61]

nav_button [eq62]

nav_button Explanation | nav_button Next question

Explanation. For $xin U{2124} _{+}$, the probability mass function of a Poisson random variable is:[eq63]Therefore, for $lambda =2$:[eq64]See the lecture entitled Poisson distribution for more details on the Poisson distribution.

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Question 14. The time between the arrival of one customer and the arrival of the next customer has an exponential distribution with parameter $lambda =2$ and is independent of previous arrivals. Time is expressed in hours. Denote by X the number of customers that will arrive during the next hour. The expected value of X is:

nav_button [eq65]

nav_button [eq66]

nav_button [eq67]

nav_button [eq68]

nav_button Explanation | nav_button Next question

Explanation. If a certain event can occur many times within a unit of time and the time elapsed between two successive occurrences of the event has an exponential distribution with parameter $lambda $ (independent of previous occurrences), then the total number of occurrences within a unit of time has a Poisson distribution with parameter $lambda $. Therefore X has a Poisson distribution with parameter $lambda $. The expected value of a Poisson random variable with parameter $lambda $ is:[eq69]See the lecture entitled Poisson distribution for more details on the Poisson distribution.

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Question 15. The sum of two independent exponential random variables with common parameter $lambda $ has:

nav_button an exponential distribution

nav_button a binomial distribution

nav_button a Gamma distribution

nav_button a Poisson distribution

nav_button Explanation

Explanation. The sum of two independent exponential random variables with common parameter $lambda $ has a Gamma distribution. See the lecture entitled Exponential distribution for a proof of this fact.