It is often the case that a number is naturally associated to the outcome of a random experiment: the number of boys in a three-child family, the number of defective light bulbs in a case of 100 bulbs, the length of time until the next customer arrives at the drive-through window at a bank. Such a number varies from trial to trial of the corresponding experiment, and does so in a way that cannot be predicted with certainty; hence, it is called a random variable. In this chapter and the next we study such variables.
A random variableA numerical value generated by a random experiment. is a numerical quantity that is generated by a random experiment.
We will denote random variables by capital letters, such as X or Z, and the actual values that they can take by lowercase letters, such as x and z.
Table 4.1 "Four Random Variables" gives four examples of random variables. In the second example, the three dots indicates that every counting number is a possible value for X. Although it is highly unlikely, for example, that it would take 50 tosses of the coin to observe heads for the first time, nevertheless it is conceivable, hence the number 50 is a possible value. The set of possible values is infinite, but is still at least countable, in the sense that all possible values can be listed one after another. In the last two examples, by way of contrast, the possible values cannot be individually listed, but take up a whole interval of numbers. In the fourth example, since the light bulb could conceivably continue to shine indefinitely, there is no natural greatest value for its lifetime, so we simply place the symbol $\infty $ for infinity as the right endpoint of the interval of possible values.
Table 4.1 Four Random Variables
Experiment | Number X | Possible Values of X |
---|---|---|
Roll two fair dice | Sum of the number of dots on the top faces | 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 |
Flip a fair coin repeatedly | Number of tosses until the coin lands heads | 1, 2, 3,4, … |
Measure the voltage at an electrical outlet | Voltage measured | 118 ≤ x ≤ 122 |
Operate a light bulb until it burns out | Time until the bulb burns out | 0 ≤ x < ∞ |
A random variable is called discreteA random variable with a finite or countable number of possible values. if it has either a finite or a countable number of possible values. A random variable is called continuousA random variable whose possible values contain an interval of decimal numbers. if its possible values contain a whole interval of numbers.
The examples in the table are typical in that discrete random variables typically arise from a counting process, whereas continuous random variables typically arise from a measurement.
Classify each random variable as either discrete or continuous.
Classify each random variable as either discrete or continuous.
Classify each random variable as either discrete or continuous.
Classify each random variable as either discrete or continuous.
Identify the set of possible values for each random variable. (Make a reasonable estimate based on experience, where necessary.)
Identify the set of possible values for each random variable. (Make a reasonable estimate based on experience, where necessary.)
Associated to each possible value x of a discrete random variable X is the probability $P\left(x\right)$ that X will take the value x in one trial of the experiment.
The probability distributionA list of each possible value and its probability. of a discrete random variable X is a list of each possible value of X together with the probability that X takes that value in one trial of the experiment.
The probabilities in the probability distribution of a random variable X must satisfy the following two conditions:
A fair coin is tossed twice. Let X be the number of heads that are observed.
Solution:
The possible values that X can take are 0, 1, and 2. Each of these numbers corresponds to an event in the sample space $S=\left\{hh,ht,th,tt\right\}$ of equally likely outcomes for this experiment: X = 0 to $\left\{tt\right\}$, X = 1 to $\left\{ht,th\right\}$, and X = 2 to $\left\{hh\right\}.$ The probability of each of these events, hence of the corresponding value of X, can be found simply by counting, to give
$$\begin{array}{cccc}x& 0& 1& 2\\ P\left(x\right)& 0.25& 0.50& 0.25\end{array}$$This table is the probability distribution of X.
“At least one head” is the event X ≥ 1, which is the union of the mutually exclusive events X = 1 and X = 2. Thus
$$P\left(X\ge 1\right)=P\left(1\right)+P\left(2\right)=0.50+0.25=0.75$$A histogram that graphically illustrates the probability distribution is given in Figure 4.1 "Probability Distribution for Tossing a Fair Coin Twice".
Figure 4.1 Probability Distribution for Tossing a Fair Coin Twice
A pair of fair dice is rolled. Let X denote the sum of the number of dots on the top faces.
Solution:
The sample space of equally likely outcomes is
$$\begin{array}{cccccc}11& 12& 13& 14& 15& 16\\ 21& 22& 23& 24& 25& 26\\ 31& 32& 33& 34& 35& 36\\ 41& 42& 43& 44& 45& 46\\ 51& 52& 53& 54& 55& 56\\ 61& 62& 63& 64& 65& 66\end{array}$$The possible values for X are the numbers 2 through 12. X = 2 is the event {11}, so $P\left(2\right)=1\u221536.$ X = 3 is the event {12,21}, so $P\left(3\right)=2\u221536.$ Continuing this way we obtain the table
$$\begin{array}{cccccccccccc}x& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ P\left(x\right)& \frac{1}{36}& \frac{2}{36}& \frac{3}{36}& \frac{4}{36}& \frac{5}{36}& \frac{6}{36}& \frac{5}{36}& \frac{4}{36}& \frac{3}{36}& \frac{2}{36}& \frac{1}{36}\end{array}$$This table is the probability distribution of X.
The event X ≥ 9 is the union of the mutually exclusive events X = 9, X = 10, X = 11, and X = 12. Thus
$$P\left(X\ge 9\right)=P\left(9\right)+P\left(10\right)+P\left(11\right)+P\left(12\right)=\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{10}{36}=0.2\stackrel{-}{7}$$Before we immediately jump to the conclusion that the probability that X takes an even value must be 0.5, note that X takes six different even values but only five different odd values. We compute
$$\begin{array}{lll}\hfill P\left(X\text{\hspace{0.17em}is\hspace{0.17em}even}\right)& =& P\left(2\right)+P\left(4\right)+P\left(6\right)+P\left(8\right)+P\left(10\right)+P\left(12\right)\\ & =& \frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36}=\frac{18}{36}=0.5\end{array}$$A histogram that graphically illustrates the probability distribution is given in Figure 4.2 "Probability Distribution for Tossing Two Fair Dice".
Figure 4.2 Probability Distribution for Tossing Two Fair Dice
The meanThe number $\mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)$, measuring its average upon repeated trials. (also called the expected valueIts mean.) of a discrete random variable X is the number
$$\mathit{\mu}=E\left(X\right)=\mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)$$The mean of a random variable may be interpreted as the average of the values assumed by the random variable in repeated trials of the experiment.
Find the mean of the discrete random variable X whose probability distribution is
$$\begin{array}{ccccc}x& \text{\u2212}2& 1& 2& 3.5\\ P\left(x\right)& 0.21& 0.34& 0.24& 0.21\end{array}$$Solution:
The formula in the definition gives
$$\begin{array}{lll}\hfill \mathit{\mu}& =& \mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)\\ & =& \left(\text{\u2212}2\right)\xb70.21+\left(1\right)\xb70.34+\left(2\right)\xb70.24+\left(3.5\right)\xb70.21=1.135\end{array}$$A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200, and third prize is $100. Let X denote the net gain from the purchase of one ticket.
Solution:
If a ticket is selected as the first prize winner, the net gain to the purchaser is the $300 prize less the $1 that was paid for the ticket, hence X = 300 − 1 = 299. There is one such ticket, so P(299) = 0.001. Applying the same “income minus outgo” principle to the second and third prize winners and to the 997 losing tickets yields the probability distribution:
$$\begin{array}{ccccc}x& 299& 199& 99& \text{\u2212}1\\ P\left(x\right)& 0.001& 0.001& 0.001& 0.997\end{array}$$Let W denote the event that a ticket is selected to win one of the prizes. Using the table
$$P\left(W\right)=P\left(299\right)+P\left(199\right)+P\left(99\right)=0.001+0.001+0.001=0.003$$Using the formula in the definition of expected value,
$$E\left(X\right)=299\xb70.001+199\xb70.001+99\xb70.001+\left(\text{\u2212}1\right)\xb70.997=\text{\u2212}0.4$$The negative value means that one loses money on the average. In particular, if someone were to buy tickets repeatedly, then although he would win now and then, on average he would lose 40 cents per ticket purchased.
The concept of expected value is also basic to the insurance industry, as the following simplified example illustrates.
A life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year.
Solution:
Let X denote the net gain to the company from the sale of one such policy. There are two possibilities: the insured person lives the whole year or the insured person dies before the year is up. Applying the “income minus outgo” principle, in the former case the value of X is 195 − 0; in the latter case it is $195-\mathrm{200,000}=\text{\u2212}\mathrm{199,805}.$ Since the probability in the first case is 0.9997 and in the second case is $1-0.9997=0.0003$, the probability distribution for X is:
$$\begin{array}{ccc}x& 195& \text{\u2212}\mathrm{199,805}\\ P\left(x\right)& 0.9997& 0.0003\end{array}$$Therefore
$$E\left(X\right)=\mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)=195\xb70.9997+\left(\text{\u2212}\mathrm{199,805}\right)\xb70.0003=135$$Occasionally (in fact, 3 times in 10,000) the company loses a large amount of money on a policy, but typically it gains $195, which by our computation of $E\left(X\right)$ works out to a net gain of $135 per policy sold, on average.
The variance, ${\mathit{\sigma}}^{2}$, of a discrete random variable X is the number
$${\mathit{\sigma}}^{2}=\mathrm{\Sigma}{\left(x-\mathit{\mu}\right)}^{2}\text{\hspace{0.17em}}P\left(x\right)$$which by algebra is equivalent to the formula
$${\mathit{\sigma}}^{2}=\left[\right.\mathrm{\Sigma}{x}^{2}\text{\hspace{0.17em}}P\left(x\right)\left]\right.-{\mathit{\mu}}^{2}$$The standard deviationThe number $\sqrt{{\displaystyle \mathrm{\Sigma}{(x-\mathit{\mu})}^{2}P\left(x\right)}}$ (also computed using $\sqrt{\left[{\displaystyle \mathrm{\Sigma}{x}^{2}P\left(x\right)}\text{\hspace{0.17em}}\right]-{\mathit{\mu}}^{2}}$), measuring its variability under repeated trials., σ, of a discrete random variable X is the square root of its variance, hence is given by the formulas
$$\mathit{\sigma}=\sqrt{\mathrm{\Sigma}{\left(x-\mathit{\mu}\right)}^{2}\text{\hspace{0.17em}}P\left(x\right)}=\sqrt{\left[\right.\mathrm{\Sigma}{x}^{2}\text{\hspace{0.17em}}P\left(x\right)\left]\right.-{\mathit{\mu}}^{2}}$$The variance and standard deviation of a discrete random variable X may be interpreted as measures of the variability of the values assumed by the random variable in repeated trials of the experiment. The units on the standard deviation match those of X.
A discrete random variable X has the following probability distribution:
$$\begin{array}{ccccc}x& \text{\u2212}1& 0& 1& 4\\ P\left(x\right)& 0.2& 0.5& a& 0.1\end{array}$$A histogram that graphically illustrates the probability distribution is given in Figure 4.3 "Probability Distribution of a Discrete Random Variable".
Figure 4.3 Probability Distribution of a Discrete Random Variable
Compute each of the following quantities.
Solution:
Using the formula in the definition of μ,
$$\mathit{\mu}=\mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)=\left(\text{\u2212}1\right)\xb70.2+0\xb70.5+1\xb70.2+4\xb70.1=0.4$$Using the formula in the definition of ${\mathit{\sigma}}^{2}$ and the value of μ that was just computed,
$$\begin{array}{lll}\hfill {\mathit{\sigma}}^{2}& =& \mathrm{\Sigma}{\left(x-\mathit{\mu}\right)}^{2}\text{\hspace{0.17em}}P\left(x\right)\\ & =& {\left(\text{\u2212}1-0.4\right)}^{2}\xb70.2+{\left(0-0.4\right)}^{2}\xb70.5+{\left(1-0.4\right)}^{2}\xb70.2+{\left(4-0.4\right)}^{2}\xb70.1\\ & =& 1.84\end{array}$$Determine whether or not the table is a valid probability distribution of a discrete random variable. Explain fully.
$\begin{array}{ccccc}x& \text{\u2212}2& 0& 2& 4\\ P\left(x\right)& 0.3& 0.5& 0.2& 0.1\end{array}$
$\begin{array}{cccc}x& 0.5& 0.25& 0.25\\ P\left(x\right)& \text{\u2212}0.4& 0.6& 0.8\end{array}$
$\begin{array}{cccccc}x& 1.1& 2.5& 4.1& 4.6& 5.3\\ P\left(x\right)& 0.16& 0.14& 0.11& 0.27& 0.22\end{array}$
Determine whether or not the table is a valid probability distribution of a discrete random variable. Explain fully.
$\begin{array}{cccccc}x& 0& 1& 2& 3& 4\\ P\left(x\right)& \text{\u2212}0.25& 0.50& 0.35& 0.10& 0.30\end{array}$
$\begin{array}{cccc}x& 1& 2& 3\\ P\left(x\right)& 0.325& 0.406& 0.164\end{array}$
$\begin{array}{cccccc}x& 25& 26& 27& 28& 29\\ P\left(x\right)& 0.13& 0.27& 0.28& 0.18& 0.14\end{array}$
A discrete random variable X has the following probability distribution:
$$\begin{array}{cccccc}x& 77& 78& 79& 80& 81\\ P\left(x\right)& 0.15& 0.15& 0.20& 0.40& 0.10\end{array}$$Compute each of the following quantities.
A discrete random variable X has the following probability distribution:
$$\begin{array}{cccccc}x& 13& 18& 20& 24& 27\\ P\left(x\right)& 0.22& 0.25& 0.20& 0.17& 0.16\end{array}$$Compute each of the following quantities.
If each die in a pair is “loaded” so that one comes up half as often as it should, six comes up half again as often as it should, and the probabilities of the other faces are unaltered, then the probability distribution for the sum X of the number of dots on the top faces when the two are rolled is
$$\begin{array}{ccccccc}x& 2& 3& 4& 5& 6& 7\\ P\left(x\right)& \frac{1}{144}& \frac{4}{144}& \frac{8}{144}& \frac{12}{144}& \frac{16}{144}& \frac{22}{144}\end{array}$$ $$\begin{array}{cccccc}x& 8& 9& 10& 11& 12\\ P\left(x\right)& \frac{24}{144}& \frac{20}{144}& \frac{16}{144}& \frac{12}{144}& \frac{9}{144}\end{array}$$Compute each of the following.
Borachio works in an automotive tire factory. The number X of sound but blemished tires that he produces on a random day has the probability distribution
$$\begin{array}{ccccc}x& 2& 3& 4& 5\\ P\left(x\right)& 0.48& 0.36& 0.12& 0.04\end{array}$$In a hamster breeder's experience the number X of live pups in a litter of a female not over twelve months in age who has not borne a litter in the past six weeks has the probability distribution
$$\begin{array}{cccccccc}x& 3& 4& 5& 6& 7& 8& 9\\ P\left(x\right)& 0.04& 0.10& 0.26& 0.31& 0.22& 0.05& 0.02\end{array}$$The number X of days in the summer months that a construction crew cannot work because of the weather has the probability distribution
$$\begin{array}{cccccc}x& 6& 7& 8& 9& 10\\ P\left(x\right)& 0.03& 0.08& 0.15& 0.20& 0.19\end{array}$$ $$\begin{array}{ccccc}x& 11& 12& 13& 14\\ P\left(x\right)& 0.16& 0.10& 0.07& 0.02\end{array}$$Let X denote the number of boys in a randomly selected three-child family. Assuming that boys and girls are equally likely, construct the probability distribution of X.
Let X denote the number of times a fair coin lands heads in three tosses. Construct the probability distribution of X.
Five thousand lottery tickets are sold for $1 each. One ticket will win $1,000, two tickets will win $500 each, and ten tickets will win $100 each. Let X denote the net gain from the purchase of a randomly selected ticket.
Seven thousand lottery tickets are sold for $5 each. One ticket will win $2,000, two tickets will win $750 each, and five tickets will win $100 each. Let X denote the net gain from the purchase of a randomly selected ticket.
An insurance company will sell a $90,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $478. Find the expected value to the company of a single policy if a person in this risk group has a 99.62% chance of surviving one year.
An insurance company will sell a $10,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $368. Find the expected value to the company of a single policy if a person in this risk group has a 97.25% chance of surviving one year.
An insurance company estimates that the probability that an individual in a particular risk group will survive one year is 0.9825. Such a person wishes to buy a $150,000 one-year term life insurance policy. Let C denote how much the insurance company charges such a person for such a policy.
An insurance company estimates that the probability that an individual in a particular risk group will survive one year is 0.99. Such a person wishes to buy a $75,000 one-year term life insurance policy. Let C denote how much the insurance company charges such a person for such a policy.
A roulette wheel has 38 slots. Thirty-six slots are numbered from 1 to 36; half of them are red and half are black. The remaining two slots are numbered 0 and 00 and are green. In a $1 bet on red, the bettor pays $1 to play. If the ball lands in a red slot, he receives back the dollar he bet plus an additional dollar. If the ball does not land on red he loses his dollar. Let X denote the net gain to the bettor on one play of the game.
A roulette wheel has 38 slots. Thirty-six slots are numbered from 1 to 36; the remaining two slots are numbered 0 and 00. Suppose the “number” 00 is considered not to be even, but the number 0 is still even. In a $1 bet on even, the bettor pays $1 to play. If the ball lands in an even numbered slot, he receives back the dollar he bet plus an additional dollar. If the ball does not land on an even numbered slot, he loses his dollar. Let X denote the net gain to the bettor on one play of the game.
The time, to the nearest whole minute, that a city bus takes to go from one end of its route to the other has the probability distribution shown. As sometimes happens with probabilities computed as empirical relative frequencies, probabilities in the table add up only to a value other than 1.00 because of round-off error.
$$\begin{array}{ccccccc}x& 42& 43& 44& 45& 46& 47\\ P\left(x\right)& 0.10& 0.23& 0.34& 0.25& 0.05& 0.02\end{array}$$Tybalt receives in the mail an offer to enter a national sweepstakes. The prizes and chances of winning are listed in the offer as: $5 million, one chance in 65 million; $150,000, one chance in 6.5 million; $5,000, one chance in 650,000; and $1,000, one chance in 65,000. If it costs Tybalt 44 cents to mail his entry, what is the expected value of the sweepstakes to him?
The number X of nails in a randomly selected 1-pound box has the probability distribution shown. Find the average number of nails per pound.
$$\begin{array}{cccc}x& 100& 101& 102\\ P\left(x\right)& 0.01& 0.96& 0.03\end{array}$$Three fair dice are rolled at once. Let X denote the number of dice that land with the same number of dots on top as at least one other die. The probability distribution for X is
$$\begin{array}{cccc}x& 0& u& 3\\ P\left(x\right)& p& \frac{15}{36}& \frac{1}{36}\end{array}$$Two fair dice are rolled at once. Let X denote the difference in the number of dots that appear on the top faces of the two dice. Thus for example if a one and a five are rolled, X = 4, and if two sixes are rolled, X = 0.
A fair coin is tossed repeatedly until either it lands heads or a total of five tosses have been made, whichever comes first. Let X denote the number of tosses made.
A manufacturer receives a certain component from a supplier in shipments of 100 units. Two units in each shipment are selected at random and tested. If either one of the units is defective the shipment is rejected. Suppose a shipment has 5 defective units.
Shylock enters a local branch bank at 4:30 p.m. every payday, at which time there are always two tellers on duty. The number X of customers in the bank who are either at a teller window or are waiting in a single line for the next available teller has the following probability distribution.
$$\begin{array}{ccccc}x& 0& 1& 2& 3\\ P\left(x\right)& 0.135& 0.192& 0.284& 0.230\end{array}$$ $$\begin{array}{cccc}x& 4& 5& 6\\ P\left(x\right)& 0.103& 0.051& 0.005\end{array}$$The owner of a proposed outdoor theater must decide whether to include a cover that will allow shows to be performed in all weather conditions. Based on projected audience sizes and weather conditions, the probability distribution for the revenue X per night if the cover is not installed is
$$\begin{array}{lll}\text{Weather}\hfill & \hfill x\hfill & \hfill P\left(x\right)\hfill \\ \text{Clear}\hfill & \hfill \text{\$3000}\hfill & \hfill \text{0}\text{.61}\hfill \\ \text{Threatening}\hfill & \hfill \text{\$2800}\hfill & \hfill \text{0}\text{.17}\hfill \\ \text{Light\hspace{0.17em}rain}\hfill & \hfill \text{\$1975}\hfill & \hfill \text{0}\text{.11}\hfill \\ \text{Show-cancelling\hspace{0.17em}rain}\hfill & \hfill \text{\$0}\hfill & \hfill \text{0}\text{.11}\hfill \end{array}$$The additional cost of the cover is $410,000. The owner will have it built if this cost can be recovered from the increased revenue the cover affords in the first ten 90-night seasons.
$\begin{array}{ccccc}x& 0& 1& 2& 3\\ P\left(x\right)& 1/8& 3/8& 3/8& 1/8\end{array}$
$\begin{array}{ccccc}x& \text{\u2212}1& 999& 499& 99\\ P\left(x\right)& \frac{4987}{5000}& \frac{1}{5000}& \frac{2}{5000}& \frac{10}{5000}\end{array}$
136
$\begin{array}{ccc}x& C& C\text{\u2212}\mathrm{150,000}\\ P\left(x\right)& 0.9825& 0.0175\end{array}$
$\begin{array}{ccc}x& \text{\u2212}1& 1\\ P\left(x\right)& \frac{20}{38}& \frac{18}{38}\end{array}$
101.02
$\begin{array}{ccccccc}x& 0& 1& 2& 3& 4& 5\\ P\left(x\right)& \frac{6}{36}& \frac{10}{36}& \frac{8}{36}& \frac{6}{36}& \frac{4}{36}& \frac{2}{36}\end{array}$
$\begin{array}{cccc}x& 0& 1& 2\\ P\left(x\right)& 0.902& 0.096& 0.002\end{array}$
The experiment of tossing a fair coin three times and the experiment of observing the genders according to birth order of the children in a randomly selected three-child family are completely different, but the random variables that count the number of heads in the coin toss and the number of boys in the family (assuming the two genders are equally likely) are the same random variable, the one with probability distribution
$$\begin{array}{ccccc}x& 0& 1& 2& 3\\ P\left(x\right)& 0.125& 0.375& 0.375& 0.125\end{array}$$A histogram that graphically illustrates this probability distribution is given in Figure 4.4 "Probability Distribution for Three Coins and Three Children". What is common to the two experiments is that we perform three identical and independent trials of the same action, each trial has only two outcomes (heads or tails, boy or girl), and the probability of success is the same number, 0.5, on every trial. The random variable that is generated is called the binomial random variableA random variable that counts successes in a fixed number of independent, identical trials of a success/failure experiment. with parameters n = 3 and p = 0.5. This is just one case of a general situation.
Figure 4.4 Probability Distribution for Three Coins and Three Children
Suppose a random experiment has the following characteristics.
Then the discrete random variable X that counts the number of successes in the n trials is the binomial random variable with parameters n and p. We also say that X has a binomial distribution with parameters n and p.
The following four examples illustrate the definition. Note how in every case “success” is the outcome that is counted, not the outcome that we prefer or think is better in some sense.
Often the most difficult aspect of working a problem that involves the binomial random variable is recognizing that the random variable in question has a binomial distribution. Once that is known, probabilities can be computed using the following formula.
If X is a binomial random variable with parameters n and p, then
$$P\left(x\right)=\frac{n!}{x!\left(n-x\right)!}{p}^{x}{q}^{n-x}$$where $q=1-p$ and where for any counting number m, $m!$ (read “m factorial”) is defined by
$$0!=1,\text{\hspace{1em}}1!=1,\text{\hspace{1em}}2!=1\xb72,\text{\hspace{1em}}3!=1\xb72\xb73$$and in general
$$m!=1\xb72\text{\hspace{0.17em}\xb7 \xb7 \xb7\hspace{0.17em}}\left(m\text{\u2212}1\right)\xb7m$$Seventeen percent of victims of financial fraud know the perpetrator of the fraud personally.
Solution:
The random variable X is binomial with parameters n = 5 and p = 0.17; $q=1-p=0.83.$ The possible values of X are 0, 1, 2, 3, 4, and 5.
$$\begin{array}{lll}\hfill P\left(0\right)& =& \frac{5!}{0!5!}\text{\hspace{0.17em}}{\left(0.17\right)}^{0}\text{\hspace{0.17em}}{\left(0.83\right)}^{5}\\ & =& \frac{1\xb72\xb73\xb74\xb75}{\left(1\right)\xb7\left(1\xb72\xb73\xb74\xb75\right)}\text{\hspace{0.17em}}1\xb7\left(0.3939040643\right)\\ & =& 0.3939040643\approx 0.3939\end{array}$$ $$\begin{array}{lll}\hfill P\left(1\right)& =& \frac{5!}{1!4!}\text{\hspace{0.17em}}{\left(0.17\right)}^{1}\text{\hspace{0.17em}}{\left(0.83\right)}^{4}\\ & =& \frac{1\xb72\xb73\xb74\xb75}{\left(1\right)\xb7\left(1\xb72\xb73\xb74\right)}\text{\hspace{0.17em}}\left(0.17\right)\xb7\left(0.47458321\right)\\ & =& 5\xb7\left(0.17\right)\xb7\left(0.47458321\right)=0.4033957285\approx 0.4034\end{array}$$ $$\begin{array}{lll}\hfill P\left(2\right)& =& \frac{5!}{2!3!}\text{\hspace{0.17em}}{\left(0.17\right)}^{2}\text{\hspace{0.17em}}{\left(0.83\right)}^{3}\\ & =& \frac{1\xb72\xb73\xb74\xb75}{\left(1\xb72\right)\xb7\left(1\xb72\xb73\right)}\text{\hspace{0.17em}}\left(0.0289\right)\xb7\left(0.571787\right)\\ & =& 10\xb7\left(0.0289\right)\xb7\left(0.571787\right)=0.165246443\approx 0.1652\end{array}$$The remaining three probabilities are computed similarly, to give the probability distribution
$$\begin{array}{ccccccc}x& 0& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.3939& 0.4034& 0.1652& 0.0338& 0.0035& 0.0001\end{array}$$The probabilities do not add up to exactly 1 because of rounding.
This probability distribution is represented by the histogram in Figure 4.5 "Probability Distribution of the Binomial Random Variable in ", which graphically illustrates just how improbable the events X = 4 and X = 5 are. The corresponding bar in the histogram above the number 4 is barely visible, if visible at all, and the bar above 5 is far too short to be visible.
Figure 4.5 Probability Distribution of the Binomial Random Variable in Note 4.29 "Example 7"
The average number of cases per day in which the victim knew the perpetrator is the mean of X, which is
$$\begin{array}{lll}\hfill \mathit{\mu}& =& \mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)\\ & =& 0\xb70.3939+1\xb70.4034+2\xb70.1652+3\xb70.0338+4\xb70.0035+5\xb70.0001\\ & =& 0.8497\end{array}$$Since a binomial random variable is a discrete random variable, the formulas for its mean, variance, and standard deviation given in the previous section apply to it, as we just saw in Note 4.29 "Example 7" in the case of the mean. However, for the binomial random variable there are much simpler formulas.
If X is a binomial random variable with parameters n and p, then
$$\mathit{\mu}=np\text{\hspace{1em}}{\mathit{\sigma}}^{2}=npq\text{\hspace{1em}}\mathit{\sigma}=\sqrt{npq}$$where $q=1-p$
Find the mean and standard deviation of the random variable X of Note 4.29 "Example 7".
Solution:
The random variable X is binomial with parameters n = 5 and p = 0.17, and $q=1-p=0.83.$ Thus its mean and standard deviation are
$$\mathit{\mu}=np=5\xb70.17=0.85\text{\hspace{1em}}\mathrm{\text{\hspace{0.17em}(exactly)}}$$and
$$\mathit{\sigma}=\sqrt{npq}=\sqrt{5\xb70.17\xb70.83}=\sqrt{.7055}\approx 0.8399$$In order to allow a broader range of more realistic problems Chapter 12 "Appendix" contains probability tables for binomial random variables for various choices of the parameters n and p. These tables are not the probability distributions that we have seen so far, but are cumulative probability distributions. In the place of the probability $P\left(x\right)$ the table contains the probability
$$P\left(X\le x\right)=P\left(0\right)+P\left(1\right)+\text{\hspace{0.17em}\xb7 \xb7 \xb7\hspace{0.17em}}+P\left(x\right)$$This is illustrated in Figure 4.6 "Cumulative Probabilities". The probability entered in the table corresponds to the area of the shaded region. The reason for providing a cumulative table is that in practical problems that involve a binomial random variable typically the probability that is sought is of the form $P\left(X\le x\right)$ or $P\left(X\ge x\right).$ The cumulative table is much easier to use for computing $P\left(X\le x\right)$ since all the individual probabilities have already been computed and added. The one table suffices for both $P\left(X\le x\right)$ or $P\left(X\ge x\right)$ and can be used to readily obtain probabilities of the form $P\left(x\right)$, too, because of the following formulas. The first is just the Probability Rule for Complements.
Figure 4.6 Cumulative Probabilities
If X is a discrete random variable, then
$$P\left(X\ge x\right)=1-P\left(X\le x\text{\u2212}1\right)\text{\hspace{1em}}\mathrm{\text{\hspace{0.17em}and}}\text{\hspace{1em}}P\left(x\right)=P\left(X\le x\right)-P\left(X\le x\text{\u2212}1\right)$$A student takes a ten-question true/false exam.
Solution:
Let X denote the number of questions that the student guesses correctly. Then X is a binomial random variable with parameters n = 10 and p = 0.50.
The probability sought is $P\left(6\right).$ The formula gives
$$P\left(6\right)=\frac{10!}{\left(6!\right)\left(4!\right)}{\left(.5\right)}^{6}.{5}^{4}=0.205078125$$Using the table,
$$P\left(6\right)=P\left(X\le 6\right)-P\left(X\le 5\right)=0.8281-0.6230=0.2051$$The student must guess correctly on at least 60% of the questions, which is $0.60\xb710=6$ questions. The probability sought is not $P\left(6\right)$ (an easy mistake to make), but
$$P\left(X\ge 6\right)=P\left(6\right)+P\left(7\right)+P\left(8\right)+P\left(9\right)+P\left(10\right)$$Instead of computing each of these five numbers using the formula and adding them we can use the table to obtain
$$P\left(X\ge 6\right)=1-P\left(X\le 5\right)=1-0.6230=0.3770$$which is much less work and of sufficient accuracy for the situation at hand.
An appliance repairman services five washing machines on site each day. One-third of the service calls require installation of a particular part.
Solution:
Let X denote the number of service calls today on which the part is required. Then X is a binomial random variable with parameters n = 5 and $p=1\u22153=0.\stackrel{-}{3}.$
Note that the probability in question is not $P\left(1\right)$, but rather P(X ≤ 1). Using the cumulative distribution table in Chapter 12 "Appendix",
$$P\left(X\le 1\right)=0.4609$$Determine whether or not the random variable X is a binomial random variable. If so, give the values of n and p. If not, explain why not.
Determine whether or not the random variable X is a binomial random variable. If so, give the values of n and p. If not, explain why not.
X is a binomial random variable with parameters n = 12 and p = 0.82. Compute the probability indicated.
X is a binomial random variable with parameters n = 16 and p = 0.74. Compute the probability indicated.
X is a binomial random variable with parameters n = 5, p = 0.5. Use the tables in Chapter 12 "Appendix" to compute the probability indicated.
X is a binomial random variable with parameters n = 5, $p=0.\stackrel{-}{3}.$ Use the table in Chapter 12 "Appendix" to compute the probability indicated.
X is a binomial random variable with the parameters shown. Use the tables in Chapter 12 "Appendix" to compute the probability indicated.
X is a binomial random variable with the parameters shown. Use the tables in Chapter 12 "Appendix" to compute the probability indicated.
X is a binomial random variable with the parameters shown. Use the special formulas to compute its mean μ and standard deviation σ.
X is a binomial random variable with the parameters shown. Use the special formulas to compute its mean μ and standard deviation σ.
X is a binomial random variable with the parameters shown. Compute its mean μ and standard deviation σ in two ways, first using the tables in Chapter 12 "Appendix" in conjunction with the general formulas $\mathit{\mu}={\displaystyle \mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)}$ and $\mathit{\sigma}=\sqrt{\left[{\displaystyle \mathrm{\Sigma}{x}^{2}\text{\hspace{0.17em}}P(x)}\text{\hspace{0.17em}}\right]-{\mathit{\mu}}^{2}}$, then using the special formulas $\mathit{\mu}=np$ and $\mathit{\sigma}=\sqrt{npq}.$
X is a binomial random variable with the parameters shown. Compute its mean μ and standard deviation σ in two ways, first using the tables in Chapter 12 "Appendix" in conjunction with the general formulas $\mathit{\mu}={\displaystyle \mathrm{\Sigma}x\text{\hspace{0.17em}}P\left(x\right)}$ and $\mathit{\sigma}=\sqrt{\left[{\displaystyle \mathrm{\Sigma}{x}^{2}\text{\hspace{0.17em}}P(x)}\text{\hspace{0.17em}}\right]-{\mathit{\mu}}^{2}}$, then using the special formulas $\mathit{\mu}=np$ and $\mathit{\sigma}=\sqrt{npq}.$
X is a binomial random variable with parameters n = 10 and $p=1\u22153.$ Use the cumulative probability distribution for X that is given in Chapter 12 "Appendix" to construct the probability distribution of X.
X is a binomial random variable with parameters n = 15 and $p=1\u22152.$ Use the cumulative probability distribution for X that is given in Chapter 12 "Appendix" to construct the probability distribution of X.
In a certain board game a player's turn begins with three rolls of a pair of dice. If the player rolls doubles all three times there is a penalty. The probability of rolling doubles in a single roll of a pair of fair dice is 1/6. Find the probability of rolling doubles all three times.
A coin is bent so that the probability that it lands heads up is 2/3. The coin is tossed ten times.
An English-speaking tourist visits a country in which 30% of the population speaks English. He needs to ask someone directions.
The probability that an egg in a retail package is cracked or broken is 0.025.
An appliance store sells 20 refrigerators each week. Ten percent of all purchasers of a refrigerator buy an extended warranty. Let X denote the number of the next 20 purchasers who do so.
Adverse growing conditions have caused 5% of grapefruit grown in a certain region to be of inferior quality. Grapefruit are sold by the dozen.
The probability that a 7-ounce skein of a discount worsted weight knitting yarn contains a knot is 0.25. Goneril buys ten skeins to crochet an afghan.
One-third of all patients who undergo a non-invasive but unpleasant medical test require a sedative. A laboratory performs 20 such tests daily. Let X denote the number of patients on any given day who require a sedative.
About 2% of alumni give money upon receiving a solicitation from the college or university from which they graduated. Find the average number monetary gifts a college can expect from every 2,000 solicitations it sends.
Of all college students who are eligible to give blood, about 18% do so on a regular basis. Each month a local blood bank sends an appeal to give blood to 250 randomly selected students. Find the average number of appeals in such mailings that are made to students who already give blood.
About 12% of all individuals write with their left hands. A class of 130 students meets in a classroom with 130 individual desks, exactly 14 of which are constructed for people who write with their left hands. Find the probability that exactly 14 of the students enrolled in the class write with their left hands.
A travelling salesman makes a sale on 65% of his calls on regular customers. He makes four sales calls each day.
A corporation has advertised heavily to try to insure that over half the adult population recognizes the brand name of its products. In a random sample of 20 adults, 14 recognized its brand name. What is the probability that 14 or more people in such a sample would recognize its brand name if the actual proportion p of all adults who recognize the brand name were only 0.50?
When dropped on a hard surface a thumbtack lands with its sharp point touching the surface with probability 2/3; it lands with its sharp point directed up into the air with probability 1/3. The tack is dropped and its landing position observed 15 times.
A professional proofreader has a 98% chance of detecting an error in a piece of written work (other than misspellings, double words, and similar errors that are machine detected). A work contains four errors.
A multiple choice exam has 20 questions; there are four choices for each question.
In spite of the requirement that all dogs boarded in a kennel be inoculated, the chance that a healthy dog boarded in a clean, well-ventilated kennel will develop kennel cough from a carrier is 0.008.
Investigators need to determine which of 600 adults have a medical condition that affects 2% of the adult population. A blood sample is taken from each of the individuals.
$$\begin{array}{ccccc}x& 0& 1& 2& 3\\ P\left(x\right)& 0.0173& 0.0867& 0.1951& 0.2602\end{array}$$ $$\begin{array}{ccccc}x& 4& 5& 6& 7\\ P\left(x\right)& 0.2276& 0.1365& 0.0569& 0.0163\end{array}$$ $$\begin{array}{ccccc}x& 8& 9& 10& \\ P\left(x\right)& 0.0030& 0.0004& 0.0000& \end{array}$$
0.0046
40
0.1019
0.0577