Normal Standard distribution: Binomial Probability?

Question

Given a 79% chance of rain forecast for the next 9 days. Would it be a usual event to rain exactly 3 days?

Answer 1

depends on your definition of "usual"  It is more likely to rain on more than 3 days than not.

Answer 2

This question only deals with the binomial distribution, not the normal.

Let X be the number of days there is rain.  X has the binomial distribution with n = 9 trials and success probability p = 0.79



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 , p )

the mean of the binomial distribution is n * p = 7.11

the variance of the binomial distribution is n * p * (1 - p) = 1.4931

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.221925

The Probability Mass Function, PMF,

f(X) = P(X = x) is:

P( X =  0 ) =  7.9428e-07

P( X =  1 ) =  2.689205e-05

P( X =  2 ) =  0.0004046614

P( X =  3 ) =  0.003552028 ← answer

P( X =  4 ) =  0.02004358

P( X =  5 ) =  0.07540205

P( X =  6 ) =  0.1891036

P( X =  7 ) =  0.3048813

P( X =  8 ) =  0.2867336

P( X =  9 ) =  0.1198516