Question

Assume that a procedure yields a binomial distribution with a trial repeated n=15n=15 times. Find the probability of x≥9x≥9 successes given the probability p=0.6p=0.6 of success on a single trial.

(Report answer accurate to 3 decimal places.)

P(x≥9)=

Answer 1

Here, n = 15, p = 0.6, (1 - p) = 0.4 and x = 9
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 8).
P(X <= 8) = (15C0 * 0.6^0 * 0.4^15) + (15C1 * 0.6^1 * 0.4^14) + (15C2 * 0.6^2 * 0.4^13) + (15C3 * 0.6^3 * 0.4^12) + (15C4 * 0.6^4 * 0.4^11) + (15C5 * 0.6^5 * 0.4^10) + (15C6 * 0.6^6 * 0.4^9) + (15C7 * 0.6^7 * 0.4^8) + (15C8 * 0.6^8 * 0.4^7)
P(X <= 8) = 0 + 0 + 0 + 0.002 + 0.007 + 0.024 + 0.061 + 0.118 + 0.177
P(X <= 8) = 0.389

P(X>= 9) = 1 - P(x < =8)
= 1 - 0.389
= 0.611