Probability and Binomial Distribution
By: Olivia and Anna
Part 1:
For this chapter, there were many different ideas that were covered. Random variables are the object in a survey that is being measured or the quantity that can change. An example of a random variable could be a pineapple in a survey where the weight and size of a shipment of pineapples are being calculated. A discrete variable is a variable that only a finite or countable number of values. A continuous variable is one that can take any of the number of values in a line interval. An example of a discrete variable would be a time of day where the value can only be whole numbers and not a decimal. A continuous variable would be something like a GPA where you can have numbers such as 3.3 or 2.8 that aren’t always whole numbers. A probability distribution is an assignment of probabilities to each value of a random variable. The features of a probability distribution are that it has a probability assigned to each distinct value of the random variable, and the sum of all the probabilities must equal 1. To find the mean of a probability distribution you use the formula=sumxP(x).The mean tells you the average probability of the set of data. The standard deviation is the spread of all the data. To calculate standard deviation you do: sum (x- )^2 x P(x), then take the square root of that. The key concepts of a binomial distribution are: it must be a fixed number of trials, which are all independent, there are only two outcomes- success or failure, the probability for success is the same for each event, and the goal is the find the probability of r successes out of n trials. In a probability distribution, (n) represents the number of trials in an event, (p) represents the probability of a certain event happening, and (q) is the probability that the same event won’t happen. The three methods for computing binomial probabilities are: with a formula: P(r)=n!r!(n-r)!prq(n-r),with the table which has calculations for 1-20 trails, the number of events for each trial, and the probabilities of .1-.95 of the event happening, and on a calculator: where you use (binompdf) of n,p,q. To calculate the mean for a probability distribution, you multiply the number of trials for the distribution (n) with the probability of the event happening (p). To find the standard deviation, you multiply the number of trials (n), the probability of the event happening (p), and the probability of the event not happening (q), then take the square root of that value.
Example:
Bob has to take a 10 question multiple-choice quiz and each question has five answers, but he hasn’t been in class lately. What is the probability that he will get all answers right? Wrong?
The probability that he will get all the answers correct is .000. The probability that he’ll get all of them wrong is .107.
For this problem, the (n) would be 10, which is the number of questions, (p) is the .20, the probability that he’ll get the answer for one question correct, and (q) is .80, the probability that he will get the answer for one question wrong.
To compute the binomial probability, we used the binomial distribution table which was a quick way to find the probabilities for the answer to each question.
This is an example of a binomial distribution because there are a fixed number of trials: the 10 questions on the quiz, all the trials are independent: the answer you give for one question has no effect on the other questions, there are only two outcomes-- success or failure: he either gets the question right or wrong, and the probability of success is the same for each event: there is a 50% chance for each answer being correct.
Part 2: You struggle to get to work on time in the morning, and only make it about 75% of the time. What is the probability you make it to work on time at least 8 of the next 10 days? What’s the probability you’re not late at all? The probability you make it to work 8 of the next 10 days is 0.526. The probability you’re not late is 0.000.
For this chapter, there were many different ideas that were covered. Random variables are the object in a survey that is being measured or the quantity that can change. An example of a random variable could be a pineapple in a survey where the weight and size of a shipment of pineapples are being calculated. A discrete variable is a variable that only a finite or countable number of values. A continuous variable is one that can take any of the number of values in a line interval. An example of a discrete variable would be a time of day where the value can only be whole numbers and not a decimal. A continuous variable would be something like a GPA where you can have numbers such as 3.3 or 2.8 that aren’t always whole numbers. A probability distribution is an assignment of probabilities to each value of a random variable. The features of a probability distribution are that it has a probability assigned to each distinct value of the random variable, and the sum of all the probabilities must equal 1. To find the mean of a probability distribution you use the formula=sumxP(x).The mean tells you the average probability of the set of data. The standard deviation is the spread of all the data. To calculate standard deviation you do: sum (x- )^2 x P(x), then take the square root of that. The key concepts of a binomial distribution are: it must be a fixed number of trials, which are all independent, there are only two outcomes- success or failure, the probability for success is the same for each event, and the goal is the find the probability of r successes out of n trials. In a probability distribution, (n) represents the number of trials in an event, (p) represents the probability of a certain event happening, and (q) is the probability that the same event won’t happen. The three methods for computing binomial probabilities are: with a formula: P(r)=n!r!(n-r)!prq(n-r),with the table which has calculations for 1-20 trails, the number of events for each trial, and the probabilities of .1-.95 of the event happening, and on a calculator: where you use (binompdf) of n,p,q. To calculate the mean for a probability distribution, you multiply the number of trials for the distribution (n) with the probability of the event happening (p). To find the standard deviation, you multiply the number of trials (n), the probability of the event happening (p), and the probability of the event not happening (q), then take the square root of that value.
Example:
Bob has to take a 10 question multiple-choice quiz and each question has five answers, but he hasn’t been in class lately. What is the probability that he will get all answers right? Wrong?
The probability that he will get all the answers correct is .000. The probability that he’ll get all of them wrong is .107.
For this problem, the (n) would be 10, which is the number of questions, (p) is the .20, the probability that he’ll get the answer for one question correct, and (q) is .80, the probability that he will get the answer for one question wrong.
To compute the binomial probability, we used the binomial distribution table which was a quick way to find the probabilities for the answer to each question.
This is an example of a binomial distribution because there are a fixed number of trials: the 10 questions on the quiz, all the trials are independent: the answer you give for one question has no effect on the other questions, there are only two outcomes-- success or failure: he either gets the question right or wrong, and the probability of success is the same for each event: there is a 50% chance for each answer being correct.
Part 2: You struggle to get to work on time in the morning, and only make it about 75% of the time. What is the probability you make it to work on time at least 8 of the next 10 days? What’s the probability you’re not late at all? The probability you make it to work 8 of the next 10 days is 0.526. The probability you’re not late is 0.000.