Let X be the number of times an odd number is rolled. X has the binomial distribution with n = 9 trials and success probability p = 0.5
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.
The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.
X ~ Binomial( n = 9 , p = 0.5 )
the mean of the binomial distribution is n * p = 4.5
the variance of the binomial distribution is n * p * (1 - p) = 2.25
the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.5
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Let X be the number of times an odd number is rolled. X has the binomial distribution with n = 9 trials and success probability p = 0.5
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.
The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.
X ~ Binomial( n = 9 , p = 0.5 )
the mean of the binomial distribution is n * p = 4.5
the variance of the binomial distribution is n * p * (1 - p) = 2.25
the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.5
The Probability Mass Function, PMF,
f(X) = P(X = x) is:
P( X = 0 ) = 0.001953125
P( X = 1 ) = 0.01757813
P( X = 2 ) = 0.0703125
P( X = 3 ) = 0.1640625
P( X = 4 ) = 0.2460938
P( X = 5 ) = 0.2460938
P( X = 6 ) = 0.1640625
P( X = 7 ) = 0.0703125
P( X = 8 ) = 0.01757813
P( X = 9 ) = 0.001953125
The Cumulative Distribution Function, CDF,
F(X) = P(X ≤ x) is:
x
∑ P(X = t) =
t = 0
P( X ≤ 0 ) = 0.001953125
P( X ≤ 1 ) = 0.01953125
P( X ≤ 2 ) = 0.08984375 = P(X < 3) <<< ANSWER
P( X ≤ 3 ) = 0.2539063
P( X ≤ 4 ) = 0.5
P( X ≤ 5 ) = 0.7460938
P( X ≤ 6 ) = 0.9101563
P( X ≤ 7 ) = 0.9804688
P( X ≤ 8 ) = 0.9980469
P( X ≤ 9 ) = 1