gibesanna11p5nn28
22.10.2019 •
Mathematics
In a particular roulette game, there's a 1/36 chance of winning. in a single day, a gambler plays 100 rounds, and wins in 7 of them. what's the p-value for winning 7 or more out of 100 rounds? (hint: be sure you know which kind of probability distribution you're dealing with before you do the calculation.)
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Ответ:
0.021739
Step-by-step explanation:
This situation can be modeled with the Binomial Distribution which gives the probability of an event that occurs exactly k times out of n, and is given by
where
= combination of n elements taken k at a time.
p = probability that the event (“success”) occurs once
q = 1-p
In this case, the event “success” is winning the roulette game with probability 1/36 = 0.027777 and n=100 rounds.
The probability value of winning the roulette game 7 or more times out of 100, is
P(7;100)+P(8;100)+...+P(100;100) =
1 - P(0;100)+P(1;100)+...+P(6;100)
We can find the sum of these last 7 terms either by hand or computer-assisted and we would find
P(0;100)+P(1;100)+...+P(6;100) =
and the p-value (probability value) for winning 7 or more out of 100 rounds is
1-0.978261 = 0.021739
Ответ:
The slope of the line is m=−5138≈−1.34210526315789.
The y-intercept is (0,−14738)≈(0,−3.86842105263158).
The equation of the line in the slope-intercept form is y=−14738−51x38≈−3.86842105263158−1.34210526315789x.
Step-by-step explanation: