ResEcon 8

What is the probability of getting at least 4 heads in 10 tosses of a fair coin?

0.828

the expected value of the Bernoulli random variable is just the sum of the n expect values of the Bernoulli trials

sum of n i=1 pi = npi

excel function to compute the normal distribution (CDF)

"=NORM.DIS (b,u,o,1 (CDF),0 (PDF) - NORM.DIST (a,u,o,1)"

A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Prices for 100 rats follow the following distribution:Price: $10.00 $12.50 $15.00Probability: 0.35 0.40 0.25How much should the lab budget for next year's rat orders be, assuming this distribution does not change?

$637 (Use the formula for the expected value of a discrete random variable. That result is the expected weekly cost of rats. Then, remember to multiply by the number of weeks to get the annual budget.)

There are 100 balls in a hat. 4 of them are RED, and 96 are BLACK. 3 are drawn at random with replacement. Calculate the expected number of red balls drawn in 3 pulls. (Round to two decimal places.)

0.12 (This baby takes a bit of work, at least until we introduce the binomial. Complete the following steps: 1. Create a probability distribution for R - the number of red balls in 3 pulls. The possible values are 0,1,2, and 3. 2. You also need probabilities for these - use a tree diagram to determine the probabilities for each value for the discrete random variable, R. 3. Multiply the values for the discrete random variable R, i.e. the values 0,1,2, and 3 by their respective probabilities. You'll have a third column in your table: R*P(R).4. Now, sum that column and you'll have the expected value for the discrete random variable, R.)

Suppose you get a cash prize of $100 if you win a game. The probability of winning the game is 0.6. What is the probability of winning at least $800 if you play the game 10 times, assuming that each game is independent?

0.167 (P(X=x) = n!/x!(n-x)! * pi^x(1-pi)^(n-x)

As reported in Trends in Television, the proportion of US households who have at least one VCR is 0.496. If 12 households are selected at random, without replacement, from all US households, what is the (approximate) probability that the number of households having at least one VCR is exactly 6. Be sure to use many decimal places in your calculations (at least 4), but report your answer to three decimal places

0.225

As reported in Trends in Television, the proportion of US households who have at least one VCR is 0.54. If 12 households are selected at random, without replacement, from all US households, what is the (approximate probability that the number of households having at least one VCR is no more than 5 but more than 3. Be sure to use many decimal places in your calculations (at least 4), but report your answer to three decimal places.

0.243

There are 100 balls in a hat. 13 of them are RED, and 87 are BLACK.3 are drawn at random with replacementCalculate the expected number of red balls drawn in 3 pulls. (Round to two decimal places.)

0.39

There are 100 balls in a hat. 17 of them are RED, and 83 are BLACK. 3 are drawn at random with replacement. Calculate the expected number of red balls drawn in 3 pulls. (Round to two decimal places.)

0.51

There are 100 balls in a hat. 22 of them are RED, and 78 are BLACK. 3 balls are drawn at random with replacement.The following is the discrete probability distribution where R is the number of red balls drawn from the hat described above. What is the standard deviation for this probability distribution?

0.717 (1. Compute the expected value using the information in the probability distribution. 2. Determine deviations from the expected value for each value of the discrete random variable. 3. Determine the squared deviations and weight them by their respective probabilities. 4. Apply the formula for the standard deviation for a discrete random variable. or you can use the binomial probability formula)

Choose the answer below that represents 5C3 (or equivalently, C53)

10

How many ways can 9 employees be selected from a group of 20 applicants?

167960 (20!/9!(20-9)!)

According to data collected from ResEc 211, 212A and 212B, [all combined together] the probability of a randomly selected student having a stereo is 0.494. Suppose a random sample of 25 students is selected from the 600 students in these classes. How many different combinations of exactly 11 students who have a stereo could be selected from these selected 25 students?

4457400 (The answer represents the binomial coefficient. To calculate, you must determine the value for the binomial coefficient. To consider this a binomial coefficient we would think in terms of having 11 successes (you select a student with a stereo) in a sample of 25 students drawn from the population. To calculate, use the formula: C1125 = 25! / 11! (25 - 11)!)

If n = 10 and p = 0.70, then the mean of the binomial distribution is

7.00. 10*0.70 = 7.00

If you flip a coin three times, the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. What is the probability of getting at least one head?

7/8 (because 7 out of 8 of the possible outcomes have at least one)

How many ways can 7 employees be selected from a group of 20 applicants?

77520

binomial probability distribution

A probability distribution showing the probability of x successes in n trials of a binomial experiment.

Bernoulli Experiment

A random experiment that has only two outcomes (success or failure) (collectively exhaustive that are mutually exclusive)

expected value formula

E = ∑ [n • P(x)]

Each trial is a Bernoulli trial with Expected Value

E(x) = 1*pi + 0*(1-pi) = pi

Mean or Expected value of a binomial distribution

E(x) = μx = npi

If p remains constant in a binomial distribution, an increase in n will not change the mean. (T or F)

F

Keeping the number of trials (n) constant, an increase in p will always increase the standard deviation. (T or F)

F

A survey of UMass students finds that the expected value for the discrete random variable "number of credit cards" is 1.76. Interpret this value.

If UMass students were repeatedly sampled, we would expect to find that they held 1.76 cards, on average.

if a random variable is distributed Binomial with _________ _____ of n and ________________ ___ ____________ of pi then.

sample size, probability of success

Binomial distribution formula

P(X=x) = p(x) = n!/x!(n-x)! * pi^x(1-pi)^(n-x)

binomial probability formula

P(x)= (nCx) (p^x) (q^n-x)

If p remains constant in a binomial distribution, an increase in n will increase the variance. (T or F)

T

The expected value of an unbiased estimator is equal to the parameter whose value is being estimated. (T or F)

T

What is true of a binomial distribution?

The probability of success must be constant from trial to trial. Each outcome is independent of the other. Each outcome may be classified as either "success" or "failure."

random variables

a numerical value associated with each possible outcome of a random experiment - the rule that assigns numerical values to the sample space, is denoted by X, while a value associated with a specific outcome in the sample space denoted by x

Binomial Experiment

a random experiment which consist of n independent and identical Bernoulli trials

discrete random variable

a variable that can take on a countable number of possible values along a specified interval. The values can be counted

random variable

a variable whose value is a numerical outcome of a random phenomenon

Identical

each trial is performed exactly in the same way

Combination counting rule

nCr = n!/r!(n-r)!

binomial coefficient

nCx = n!/x!(n-x)!

A binomial probability distribution for 1,000 trials with .65 as the probability of a success will have a distribution with a shape best described by:

normal (The key here is that there is a large number of trials. While a binomial probability distribution with a success probability (i.e., p) more than 0.5 is skewed to the left for a small number of trials, with 1,000 trials, the distribution will look normal. For any value of p not equal to 0.5, the level of skewness becomes less pronounced as n rises. At 1,000 trials, there will be no skewness observed.)

The general shape of a binomial random variable that has a probability of success of .5 and a large n will be:

normal or bell shaped (When the probability of success is close to 0.5, the binomial is symmetric. The larger n, the closer the binomial comes to the normal distribution)

variance of bernoulli experiment

probability of success times probability of failure

The general shape of a binomial random variable that has a probability of success of .25 will be:

right skewed or reverse j-shaped

Independence

the outcome of one trial doesn't influence or change the outcome of another

discrete probability distribution

the probability distribution of a discrete random variable X list of each possible outcome, x of X together with its corresponding probabilities, P(X=x)

In a binomial distribution:

the probability of success p is stable from trial to trial

Suppose the continuous random variable, X, is uniformly distributed between 1 and 9. What is the standard deviation of X?

σ = √(((9 - 1)2) / 12) = 2.309

standard deviation for a discrete random variable formula

σofx = √(x-μ)^2P(X=x)

standard deviation of a binomial distribution

√(np(1-p))