Slot Machine Binomial Distribution

The probability of winning on a slot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution. Find the area under the normal curve over the interval (29.5,30.5). Binomial Distribution Slot Machine, highest rtp slot on betfred, joker poker free slots, les meilleures mains au poker Gambling online for real money is highly popular and great fun because you can play at any time of the day or night, on Binomial Distribution Slot Machine your pc, tablet or mobile. A month or so ago I answered the Question of the Day for the Las Vegas Advisor and I answered it in terms of the Binomial Distribution. There were some comments posted after that QOD indicating that people wanted to know more about it. Slot machines have three drums with the same number of positions on each drum. Each position contains a symbol. When you pull the lever on the machine, the drums spin. Assume that each drum is equally likely to stop in any position, and that the positions in which the three drums stop are independent of each other and independent from spin to spin. Suppose a particular slot machine has 8.

QSCI 482/Dr. ConquestTAs Kennedy, Malinick

Slot Machine Binomial Distribution Definition

HW1--DUE *THIS* FRIDAY, OCTOBER 4, 2002. ALSO, IF YOU DID NOT FILL OUT A

CLASS SURVEY SHEET ON DAY 1, PLEASE DO SO BY THE END OF THE WEEK.

1. Attach a copy of a recent photo of yourself; it will help me learn

everyone's names. I will return it if you wish. You don't have to be

smiling ['mug shots' are OK] but photo must be in good taste [no pictures

of your last skinny-dipping party, please. No 'Full Monty's'.]. Thanks!

2. In a 1995 the Puget Sound Gilnetters Association conducted a field test

to test the effects of different kinds of fishing gear on alcid bird

entanglement (diving seabirds like rhinocerous auklet and common murre,

covered under the Migratory Bird Treaty Act).First, a test was done to

see if the total bird entanglements followed a Poisson distribution with

mean mu=.09, since this is what was found from previous field seasons.

(That is, the overall entanglement rate is about .09 birds per net set.)

Out of 449 net sets (a fishing net set in the water and fishing for the

same specified period of time), 418 of them had 0 birds caught, 30 sets

had 1 bird caught, and 1 of the sets experienced 2-or-more birds caught.

At the .05 level of significance, do a goodness-of-fit test to test

whether the observed entanglements follow a Poisson distribution with mean

mu=.09, against the alternative hypothesis that they do not follow such a

distribution. What do you conclude?[NOTE. The Poisson probability for a

particular count, X, is on p. 571 in Zar (in the 3rd ed. it's p. 569) and

is: exp(-mu)*(mu^X)/X!.]

[Why do we care about whether the data follow this distribution or not? In

order to plan experimental designs for future field tests, it was

Slot Machine Binomial Distribution Calculator

necessary to check out some specific distributional assumptions first. The

eventual outcome of field studies done in 1995 and 1996 was a new

Washington state law that mandated the use of a particular kind of net to

minimize accidental bird entanglement.]

3. This exercise involves computing probabilities for the binomial

distribution (in Zar, the chapter titled 'More on Dichotomous Variables',

or look under 'binomial distribution' in any elementary statistics book).

According to a mathematician who works there, The Anchor Gaming Company is

the largest supplier of slot machines in the world. Let's consider a

simple slot machine and compute some EXPECTED VALUES OF OUTCOMES FOR A

BINOMIAL PROBABILITY DISTIRIBUTION.Suppose a slot machine has 4 slots,

each of which shows a picture of either a cherry or a pear, and the slots

work independently. For a given slot, the probability that a cherry shows

up is 0.4, and the probability that a pear shows up is 0.6. So for a given

'pull' of the machine, one can observe either: 4 cherries; 3 cherries and

1 pear; 2 cherries and 2 pears; 1 cherry and 3 pears; or 4 pears. Suppose

a given machine is 'pulled' 10,000 times. How many times would we expect

to see:

a. 4 cherries

b. 3 cherries and 1 pear

c. 2 cherries and 2 pears

d. 1 cherry and 3 pears

e. 4 pears

The following problems (4-6, on the back side of this sheet) fall under the heading, 'algebra review'. If you have not done algebra for awhile, this will give you needed practice. If this stuff comes easily to you, give a classmate some help--we're all in this together.

4. You may recall that the formula for the standard error for a sample

mean X-bar is: Standard Error of X-bar = sqrt(sample variance/n), where n

Slot machine binomial distribution calculator

is the sample size, and 'sqrt' stands for 'square root of'. You are

reading a journal paper and trying to find out what the original sample

size was (which the authors failed to state, and the editors did not catch

the omission).In a table, the authors state that 'the standard error for

the data was 10.3 kg.' Elsewhere in the paper, you find that the sample

variance was 1,697.44 kg^2. Now, solve for the original sample size, n.

5. We have a random sample of n = 5 weights (kg) of a particular kind of

animal: 3.1, 3.4, 3.6, 3.7, 4.0. The sample mean, X-bar, is 3.56 kg.

Now compute the sample variance of these data, using the standard

'machine formula' calculation for the sample variance, s^2:

s^2 = {Sum([Xi^2]) - [(Sum[Xi])^2]/n}/(n-1)

6. The following data (X) are known to come from a lognormal distribution;

that is, the natural logarithms of the data, Y = ln(X), will follow a

normal (bell-shaped) distribution. Here are the data in the original

units:3.67, 4.01, 3.85, 3.92, 3.71, 3.88, 3.74, 3.82 ml.

a. Compute the sample mean of the log-transformed data.

b. Compute the sample variance of the log-transformed data. Take the sqrt

to get the sample standard deviation.

c. It turns out that the 95% confidence interval (CI) for the mean of the

log-transformed data is xbar +- 2.365*(std. deviation/sqrt(n)). Compute the 95%

CI for the log-transformed data; then transform each of the endpoints back

to get the 95% CI in the original units [ml]. Comment upon the symmetry of

the CI for the log-transformed units [ln(ml)], and the back-transformed

Slot Machine Binomial Distribution Method

units [ml].