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Probability distributions for Discrete RandomСтр 1 из 8Следующая ⇒
Chapter 3 Discrete random variables and probability distributions
Random variables Suppose that experiment of rolling two fair dice to be carried out. Let X be the sum of outcomes, then X can only assume the values 2, 3, 4, ……., 12 with the following probabilities:
The numerical value of random variable depends on the outcomes of the experiment. In this example, for instance, if it is (3, 2), then X is 5, and if it is (6, 6) then X is 12. In this example X is called a random variable. Definition: A random variable is a variable whose value is determined by the outcome of a random experiment. Notationally, we use capital letters, such as X, to denote the random variable and corresponding lowercase x to denote a possible value. Set of possible values of a random variable might be finite, infinite and countable, or uncountable. Definition: A random variable X is called a discrete random variable if it can take on no more than a countable number of values. Some examples of discrete random variable: 1. The number of employees working at a company. 2. The number of heads obtained in three tosses of a coin. 3. The number of customers visiting a bank during any given day. A random variable whose values are not countable is called a continuous random variable. Definition: A random variable X is called a continuous if it can take any value in an interval. Here are some examples of continuous random variables: 1. Prices of houses 2. The amount of oil imported. 3. Time taken by workers to learn a job. Probability distributions for Discrete Random Variables Let X be a discrete random, and x be one of its possible values. The probability that the random variable X takes the value x is denoted by Definition: The probability distribution function, P(x), of a discrete random variable X indicates that this variable takes the value x, as a function of x. That is Example 1: In the experiment of tossing a fair coin three times, let X be number of heads obtained. Determine and sketch the probability function of X.
Solution: Table 3.1 First, X is a variable and the number of heads in three tosses of a coin can have any of the values 0, 1, 2, or 3. We can make a list of the outcomes and the associated values of X. (Table 3.1). Note that, for each basic outcome there is only one value of X. However, several basic outcomes may yield the same value. We identify the events (i.e., the collections of the distinct values of X). (Table 3.2) Table 3.2.
The model of a fair coin entails that 8 basic outcomes are equally likely, so each is assigned the probability 1/8. The event [X=0] has a single outcome TTT, so its probability is 1/8. Similarly, the probabilities of [X=1], [X=2], and [X=3] are found to be 3/8, 3/8, and 1/8, respectively. Collecting these results, we obtain the probability distribution of X shown in table 3.3. Table 3.3. The probability distribution of X, the number of heads in 3 tosses of a coin.
Remark : When summed over all possible values of X, these probabilities must add up to 1.
The graphical representation of the probability of X, the number of heads in three tosses of a coin is shown in Fig. 3.1.
In the development of the probability distribution for a discrete random variable, the following two conditions must always be satisfied: Properties of probability function of discrete random variables: Let X be a discrete random variable with probability P(x). Then 1. 2. The individual probabilities sum to 1; that is Another representation of discrete probability distribution is also useful.
Cumulative probability function The cumulative probability function, where the function is evaluated of all values Properties of cumulative probability functions for discrete random variables: Let X be a discrete random variable with cumulative probability function 1. 2. If Example2: In the experiment of rolling a balanced die twice, let X be the minimum of the two numbers obtained. Determine and sketch the probability function and cumulative probability function of X. Solution: The possible values of X are 1, 2, 3, 4, 5, and 6, The sample space of this experiment consists of 36 basic outcomes. Hence the probability of any of them is 1/36. In our experiment P(X=1)=P{(1,1)(1,2)(2,1)(1,3)(3,1)(1,4)(4,1)(1,5)(5,1)(1,6)(6,1)}=11/36 P(X=2)=P{(2,2)(2,3)(3,2)(2,4)(4,2)(2,5)(5,2)(2,6)(6,2)}=9/36 P(X=3)=P{(3,3)(3,4)(4,3)(3,5)(5,3)(3,6)(6,3)}=7/36 P(X=4)=P{(4,4)(4,5)(5,4)(4,6)(6,4)}=5/36 P(X=5) =P {(5, 5) (5, 6) (6, 5)} =3/36 P(X=6) =P {(6, 6)} =1/36
Now let us form cumulative probability function. If
If
If
Continuing in this way we can write cumulative probability function as
The cumulative distribution function of X, It can be seen that the cumulative probability function increases in steps until the sum is 1. Example 3: A consumer agency surveyed all 2500 families living in a small town to collect data on the number of TV sets owned by them. The following table lists the frequency distribution of the data collected by this agency
a) Construct a probability distribution table. Draw a graph of the probability distribution. b) Calculate and draw the cumulative probability function. c) Find the probabilities: P(X=1), P(X>1),
Solution: a) In a chapter 2 we learned that the relative frequencies obtained from an experiment or a sample can be used as approximate probabilities. Using the relative frequencies, we can write the probability distribution on the discrete random variable X in the following table.
Figure 3.4 shows the graphical presentation of the probability distribution.
b) Let us form cumulative probability distribution function. If
If
Continuing in this way, we obtain
This function is plotted in Fig. 3.5.
c)
Exercises
1. Each of the following tables lists certain values of X and their probabilities. Determine if each of them satisfies the two conditions required for a valid probability distribution.
a) b) c)
2. For each case, list the values of x and P(x) and examine if the specification represents a probability distribution. If does not, state what properties are violated a) b) c) d) 3. The following table gives the probability distribution of a discrete random variable X.
a) Draw the probability function. b) Calculate and draw the cumulative probability function. c) Find d) Find the probability that x assumes a value less than 3 e) Find the probability that x assumes a value in the interval 2 to 4.
4. Despite all safety measures, accidents do happen at the factory. Let X denote the number of accidents that occur during a month at this factory. The following table lists the probability distribution of X.
a) Draw the probability function. b) Calculate and draw the cumulative probability function. c) Determine the probability that the number of accidents that will occur during a given month at this company is exactly 4. d) What is the probability that number of accidents will be at least 2? e) What is the probability that number of accidents will be less than 3? f) What is the probability that number of accidents will be between 2 to 4? g) Two month are chosen at random. What is the probability that on both of these months there will be fewer than two accidents? 5. Let X be the number of shopping trips made by family during a month. The following table lists the frequency distribution of X of 1000 families.
a) Draw the probability function. b) Calculate and draw the cumulative probability function. c) Find the following probabilities
6. The following table lists the probability distribution of the number of phone calls received per 10-minute period at an office.
Let X denote the number of phone calls received during a certain 10-minute period at this office. a) Find the probabilities: b) Two 10-minute period are chosen at random. What is the probability that at least one of them there will be at least one received call? 7. In successive rolls of a fair die, let X be the number of rolls until the first 6. Determine the probability function. 8. In a tennis championship, player A competes against player B in consecutive sets and the game continues until one player wins three sets. Assume that, for each set P (A wins) =0.4, P (B wins) =0.6, and the outcomes of different sets are independent. Let X stand for the number of sets played. a) List the possible values of X and identify the basic outcomes associated with each value. b) Obtain the probability distribution of X.
Answers 1. a) this is not a valid probability distribution; b) this is not a valid probability distribution; c) this is a valid probability distribution; 2. a) this is a valid probability distribution; b) this is not a valid probability distribution; c) this is a valid probability distribution; d) this is not a valid probability distribution; 3. c) 0.13; 0.16; 0.62; 0.38; d) 0.38; e) 0.72; 4. c) 0.10; d) 0.45; e) 0.75; f) 0.45; g) 0.3025; 5. c) 0.18; 0.51; 0.70; 0.49; 6. a) 0.26; 0.38; 0.28; 0.78; b) 0.9856; 7. P (5) =0.3456;
Expected value Once we have constructed the probability distribution for a random variable, we often want to compute the mean or expected value of the random variable. The mean of discrete random variable X, denoted either or (or expected) value of a discrete random variable is the value that we expect to observe per repetition, on average, if we perform an experiment a large number of times. For example, we may expect a house salesperson to sell on average, 3.50 houses per month. It does not mean that every month this salesperson will sell exactly 3.50 houses. (Actually he (or she) can not sell exactly 3.50 houses). This simply means that if we observe for many months, this salesperson will sell a different number of houses different months. However, the average of all sold houses in these months will be 3.50. Definition: The mean (or expected value) of discrete random variable X is defined as Here the sum extends over all distinct values x of X. In order to compute the expected value of a discrete random variable we must multiply each value of the random variable by the corresponding value of its probability function. We then add the resulting terms. Example: Sales show that five is the maximum number of cars sold on a given day at car selling company. Table 3.4 shows probability distribution of cars sold per day. Find the expected number of cars sold. Table 3.4
Solution: To find the expected number (or mean) of cars sold, we multiply each value of x by its probability and add these results.
In fact, it is impossible for company to sell exactly 1.50 cars in any given day. But we examine selling cars at this company for many days into the future, and see that, the expected value of 1.50 cars provides a good estimate of the mean or average daily sales volume. The expected value can be important to the managers from both planning and decision making points of view. For example, suppose that this company will be open 40 days during next 2-month. How many cars should the owner expect to be sold during this time? While we can not specify the exact value of 1.50 cars, it provides an expected sale of
Answers 1. a) 1.67; 0.906; b) 7.19; 1.102; 2. 1; 1; 1; 3. 0.44; 0.852; 4. 3.12; 1.09; 5. a) b) 26.4; 2.06; 8. a) 23.5; b) 92.75 and 9.63; c) 75.5; 834.75; 28.89. Random variables Let X and Y be a pair of discrete random variables with probability function The mean of random variable X is
The mean of random variable Y is
The mean, or expectation of any function
As an example let us calculate means of X, Y, and g(X, Y) for the Example above. The mean of X is:
It means, on average we expect that each student eats 1.1 snacks per day during final examination week. The mean of Y is:
It means, on average, we expect that each student has 1.55 tests per day during final examination week.
Covariance Suppose that X and Y are pair of random variables and they are dependent. We use covariance to measure the nature and strength of the relationship between them. Definition: Let X be a random variable with mean
An equivalent expression for
If Let us calculate table 3.8. Using an equivalent expression for
It means that there is a weak negative association between number of tests taken a day during a final examination week and number of eaten snacks.
Exercises 1. Shown below is the joint probability distribution for two random variables X and Y.
a) Find b) Specify the marginal probability distributions for X and Y. c) Compute the mean and variance for X and Y. d) Are X and Y independent random variables? Justify your answer. 2. There is a relationship between the number of lines in a newspaper advertisement for an apartment and the volume of interest from the potential renters. Let volume of interest be denoted by the random variable X, with the value 0 for little interest, 1 for moderate interest, and 2 for heavy interest. Let Y be the number of lines in a newspaper. Their joint probabilities are shown in the table
a) Find and interpret b) Find the joint cumulative probability function at X=2, Y=4, and interpret your result. c) Find and interpret the conditional probability function for Y, given X=0. d) Find and interpret the conditional probability function for X, given Y=4. e) If the randomly selected advertisement contains 5 lines, what is the probability that it has heavy interest from the potential renters? f) Find expected number of volume of interest. g) Find and interpret covariance between X and Y. h) Are the number of lines in the advertisement and volume of interest independent of one another? 3. Students at a university were classified according to the years at the university (X) and number of visits to a museum in the last year. (Y=0 for no visits, 1 for one visit, 2 for two visits, 3 for more than two visits). The accompanying table shows joint probabilities.
a) Find and interpret b) Find and interpret the mean number of X. c) Find and interpret the mean number of Y. d) If the randomly selected student is a e) If the randomly selected student has 1 visit to a museum, what is the probability that he (or she) is a f) Are number of visits to a museum and years at the university independent of each other? 4. It was found that 20% of all people both watched the show regularly and could correctly identify the advertised product. Also, 27% of all people regularly watched the show and 53% of all people could correctly identify the advertised product. Define a pair of random variables as follows: X=1 if regularly watch the show; X=0 otherwise Y=1 if product correctly identified; Y=0 otherwise. a) Find the joint probability function of X and Y. b) Find the conditional probability function of Y, given X=0. c) If randomly selected person could identify the product correctly, what is the probability that he (or she) regularly watch the show? d) Find and interpret the covariance between X and Y. Answers 1. a) 0.08; 0.18; 0.30; b)
2. a) 0.16; b) 0.76; c) d) g) 0.109; h) No; 3. a) 0.04; b) 2.39; c) 1.63; d) 3/13; e) 3/11; f) No; 4. a) b)
The binomial distribution An experiment that satisfies the following four conditions is called a binomial experiment: 1. There are n identical trials. In other words, the given experiment is repeated n times. All these repetitions are performed under identical conditions. 2. Each trial has two and only two outcomes. These outcomes are usually called a success and a failure. 3. The probability of success is denoted by p and that of failure by q, and 4. The trials are independent. In other words, the outcome of one trial does not affect the outcome of another trial. It is important to note that one of the two outcomes of a trial is called a success and the other a failure. Note that a success does not mean that the corresponding outcome is considered favorable or desirable. Similarly, a failure does not necessarily refer to an unfavorable or undesirable outcome. Success and failure simply the names used to denote the two possible outcomes of a trial. The random variable x that represents the number of successes in n trial for a binomial experiment is called a binomial random variable. Binomial formula: For a binomial experiment, the probability of exactly x successes in n trials is given by the binomial formula:
where To find the probability of x successes in n trials for a binomial experiment, the only values needed are those of n and q. These are called the parameters of the binomial distribution or simply the binomial parameters. Example: A certain drug is effective in 30 per cent of the cases in which it has been prescribed. If a doctor is now administering this drug to four patients, what is the probability that it will be effective for at least three of the patients? Solution: We can consider the administration of the drug to each patient as a trial. Thus, this experiment has four trials. There are only two outcomes for each trial: the drug is effective or the drug is not effective. The event “effective for at least three” can be broken down into two mutually exclusive events (outcomes), “ effective for three or effective for four”. If we use the term “success” instead of “effective” we can say that
P (at least 3 successes) =P (3 successes or 4 successes)= =P (3 successes)+ P (4 successes)= P(x=3)+P(x=4).
Now we can find Hence,
Hence, we have
Practically interpreted, this number means that if the drug is administrated to 10 000 sets of four patients, in about 837 of the 10 000 sets will be effective for at least three patients out of four.
Example: It is known from past data that despite all efforts, 2% of the packages mailed through post office do not arrive at their destinations within the specified time. A corporation mailed 10 packages through post office. a) Find the probability that exactly one of these 10 packages will not arrive at this destination within the specified time. b) Find the probability that at most one of these 10 packages will not arrive at this destination within the specified time. Solution: Let us call it a success if a package does not arrive at its destination within the specified time and a failure if it does arrive within the specified time. Then n=10; p=0.02; q=0.98 a) For this part, x = number of successes=1 n-x = number of failures=10-1=9 Substituting all values in the binomial formula, we obtain:
Thus, there is a 0.1667 probability that exactly one of the 10 packages mailed will not arrive at its destination within the specified time. b) The probability that at most one of the 10 packages will not arrive at its destination within the specified time is given by the sum of the probabilities of x=0 and x=1. Thus,
Thus, the probability that at most one of the 10 packages will not arrive at its destination within the specified time is 0.9838. Exercises
1. Let N=14, S=6, and n=5. Using the hypergeometric probability distribution formula, find a)
2. Let N=16, S=10, and n=5. Using the hypergeometric probability distribution formula, find a)
3. There are 25 goods, 5 of which are defective. We randomly select 4 goods. What is the probability that three of those four goods are nondefective and one is defective? 4. A committee of two members is to be formed from the list of 8 candidates. Of the 8 candidates, 5 are management and 3 are economics department students. Find the probability that a) both candidates are managers b) neither of the candidates are managers c) at most one of the candidates is a manager. 5. A company buys keyboards from another company. The keyboards are received in shipments of 100 boxes, each box containing 20 keyboards. The quality control department first randomly selects one box from each shipment, and then randomly selects five keyboards from that box. The shipment is accepted if not more than one of the five keyboards is defective. The quality control inspector selects a box from a recently received shipment of keyboards. Unknown to the inspector, this box contains six defective keyboards. a) What is the probability that this shipment will be accepted? b) What is the probability that this shipment will not be accepted?
Answers 1. a) 0.0599; b) 0.0030; c) 0.2378; 2. a) 0.0577; b) 0.0014; c) 0.0357; 3. 0.4506; 4. a) 0.3571; b) 0.1071; c) 0.6429; 5. a) 0.5165; b) 0.4835. Exercises 1. Using the Poisson formula, find the following probabilities a) b) 2. Let x be a Poisson random variable. Using the Poisson probabilities table, write the probability distribution of x for each of the following. Find the mean and standard deviation for each of these probability distributions. a) 3. An average of 7.5 crimes are reported per day to police in a city. Use the Poisson formula to find the probability that a) exactly 3 crimes will be reported to a police on a certain day b) at least 2 crimes will be reported to a police on a certain day. 4. A mail-order company receives an average 1.3 complaints per day. Find the probability that it will receive a) exactly 3 complaints b) 2 to 3 complaints c) more than 3 complaints d) less than 3 complaints on a certain day. 5. An average of 4.5 customers come to the bank per half hour. a) Find the probability that exactly 2 customers will come to this bank during a given hour; b) Find the probability that during a given hour, the number of customers who will come to the bank is at most 2. 6. An average of 0.6 accidents occur per month at a large company. a) Find the probability that no accident will occur at this company during a given month. b) Find the mean, variance, and standard deviation of the number of accidents that will occur at this company during a given month.
Answers 1. a) 0.1991; b) 0.0771; 2. a) 3. a) 0.03888; b) 0.9953; 4. a) 0.0998; b) 0.3301; c) 0.0431; d) 0.8569; 5. a) 0.0050; b) 0.0062; 6. a) 0.5488; b) Chapter 3 Discrete random variables and probability distributions
Random variables Suppose that experiment of rolling two fair dice to be carried out. Let X be the sum of outcomes, then X can only assume the values 2, 3, 4, ……., 12 with the following probabilities:
The numerical value of random variable depends on the outcomes of the experiment. In this example, for instance, if it is (3, 2), then X is 5, and if it is (6, 6) then X is 12. In this example X is called a random variable. Definition: A random variable is a variable whose value is determined by the outcome of a random experiment. Notationally, we use capital letters, such as X, to denote the random variable and corresponding lowercase x to denote a possible value. Set of possible values of a random variable might be finite, infinite and countable, or uncountable. Definition: A random variable X is called a discrete random variable if it can take on no more than a countable number of values. Some examples of discrete random variable: 1. The number of employees working at a company. 2. The number of heads obtained in three tosses of a coin. 3. The number of customers visiting a bank during any given day. A random variable whose values are not countable is called a continuous random variable. Definition: A random variable X is called a continuous if it can take any value in an interval. Here are some examples of continuous random variables: 1. Prices of houses 2. The amount of oil imported. 3. Time taken by workers to learn a job. Probability distributions for Discrete Random Variables Let X be a discrete random, and x be one of its possible values. The probability that the random variable X takes the value x is denoted by Definition: The probability distribution function, P(x), of a discrete random variable X indicates that this variable takes the value x, as a function of x. That is Example 1: In the experiment of tossing a fair coin three times, let X be number of heads obtained. Determine and sketch the probability function of X.
Solution: Table 3.1 First, X is a variable and the number of heads in three tosses of a coin can have any of the values 0, 1, 2, or 3. We can make a list of the outcomes and the associated values of X. (Table 3.1). Note that, for each basic outcome there is only one value of X. However, several basic outcomes may yield the same value. We identify the events (i.e., the collections of the distinct values of X). (Table 3.2) Table 3.2.
The model of a fair coin entails that 8 basic outcomes are equally likely, so each is assigned the probability 1/8. The event [X=0] has a single outcome TTT, so its probability is 1/8. Similarly, the probabilities of [X=1], [X=2], and [X=3] are found to be 3/8, 3/8, and 1/8, respectively. Collecting these results, we obtain the probability distribution of X shown in table 3.3. Table 3.3. The probability distribution of X, the number of heads in 3 tosses of a coin.
Remark : When summed over all possible values of X, these probabilities must add up to 1.
The graphical representation of the probability of X, the number of heads in three tosses of a coin is shown in Fig. 3.1.
In the development of the probability distribution for a discrete random variable, the following two conditions must always be satisfied: Properties of probability function of discrete random variables: Let X be a discrete random variable with probability P(x). Then 1. 2. The individual probabilities sum to 1; that is Another representation of discrete probability distribution is also useful.
Cumulative probability function The cumulative probability function, where the function is evaluated of all values Properties of cumulative probability functions for discrete random variables: Let X be a discrete random variable with cumulative probability function 1. 2. If Example2: In the experiment of rolling a balanced die twice, let X be the minimum of the two numbers obtained. Determine and sketch the probability function and cumulative probability function of X. Solution: The possible values of X are 1, 2, 3, 4, 5, and 6, The sample space of this experiment consists of 36 basic outcomes. Hence the probability of any of them is 1/36. In our experiment P(X=1)=P{(1,1)(1,2)(2,1)(1,3)(3,1)(1,4)(4,1)(1,5)(5,1)(1,6)(6,1)}=11/36 P(X=2)=P{(2,2)(2,3)(3,2)(2,4)(4,2)(2,5)(5,2)(2,6)(6,2)}=9/36 P(X=3)=P{(3,3)(3,4)(4,3)(3,5)(5,3)(3,6)(6,3)}=7/36 P(X=4)=P{(4,4)(4,5)(5,4)(4,6)(6,4)}=5/36 P(X=5) =P {(5, 5) (5, 6) (6, 5)} =3/36 P(X=6) =P {(6, 6)} =1/36
Now let us form cumulative probability function. If
If
If
Continuing in this way we can write cumulative probability function as
The cumulative distribution function of X, It can be seen that the cumulative probability function increases in steps until the sum is 1. Example 3: A consumer agency surveyed all 2500 families living in a small town to collect data on the number of TV sets owned by them. The following table lists the frequency distribution of the data collected by this agency
a) Construct a probability distribution table. Draw a graph of the probability distribution. b) Calculate and draw the cumulative probability function. c) Find the probabilities: P(X=1), P(X>1),
Solution: a) In a chapter 2 we learned that the relative frequencies obtained from an experiment or a sample can be used as approximate probabilities. Using the relative frequencies, we can write the probability distribution on the discrete random variable X in the following table.
Figure 3.4 shows the graphical presentation of the probability distribution.
b) Let us form cumulative probability distribution function. If
If
Continuing in this way, we obtain
This function is plotted in Fig. 3.5.
c)
Exercises
1. Each of the following tables lists certain values of X and their probabilities. Determine if each of them satisfies the two conditions required for a valid probability distribution.
a) b) c)
2. For each case, list the values of x and P(x) and examine if the specification represents a probability distribution. If does not, state what properties are violated a) b) c) d) 3. The following table gives the probability distribution of a discrete random variable X.
a) Draw the probability function. b) Calculate and draw the cumulative probability function. c) Find d) Find the probability that x assumes a value less than 3 e) Find the probability that x assumes a value in the interval 2 to 4.
4. Despite all safety measures, accidents do happen at the factory. Let X denote the number of accidents that occur during a month at this factory. The following table lists the probability distribution of X.
a) Draw the probability function. b) Calculate and draw the cumulative probability function. c) Determine the probability that the number of accidents that will occur during a given month at this company is exactly 4. d) What is the probability that number of accidents will be at least 2? e) What is the probability that number of accidents will be less than 3? f) What is the probability that number of accidents will be between 2 to 4? g) Two month are chosen at random. What is the probability that on both of these months there will be fewer than two accidents? 5. Let X be the number of shopping trips made by family during a month. The following table lists the frequency distribution of X of 1000 families.
a) Draw the probability function. b) Calculate and draw the cumulative probability function. c) Find the following probabilities
6. The following table lists the probability distribution of the number of phone calls received per 10-minute period at an office.
Let X denote the number of phone calls received during a certain 10-minute period at this office. a) Find the probabilities: b) Two 10-minute period are chosen at random. What is the probability that at least one of them there will be at least one received call? 7. In successive rolls of a fair die, let X be the number of rolls until the first 6. Determine the probability function. 8. In a tennis championship, player A competes against player B in consecutive sets and the game continues until one player wins three sets. Assume that, for each set P (A wins) =0.4, P (B wins) =0.6, and the outcomes of different sets are independent. Let X stand for the number of sets played. a) List the possible values of X and identify the basic outcomes associated with each value. b) Obtain the probability distribution of X.
Answers 1. a) this is not a valid probability distribution; b) this is not a valid probability distribution; c) this is a valid probability distribution; 2. a) this is a valid probability distribution; b) this is not a valid probability distribution; c) this is a valid probability distribution; d) this is not a valid probability distribution; 3. c) 0.13; 0.16; 0.62; 0.38; d) 0.38; e) 0.72; 4. c) 0.10; d) 0.45; e) 0.75; f) 0.45; g) 0.3025; 5. c) 0.18; 0.51; 0.70; 0.49; 6. a) 0.26; 0.38; 0.28; 0.78; b) 0.9856; 7. P (5) =0.3456;
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number of failures in n trials.
and 
separately. Since the drug is effective in 30% of the cases, we say that the probability that the drug is effective in any single case is p=0.3.
, then the equation of the particular binomial distribution is
.
.
.
; b) 
; c) 
; c) 
for 
for 
; b) 
; 
b) 
; 
.