Example 1
The probability of tossing a head with a weighted coin is 0.4, if the coin is tossed 60 times, what is the probability that: a) 10 or fewer heads are tossed b) exactly 20 heads are tossed
Solution a):
n = 60 p = 0.4 q = 0.6
Can we use normal approximation? np = 24 > 5. YES P(x⇐10) = P(x < 10.5) =
now convert the x score to the z score P(x<10.5) P(z<) P(z←3.55) = 0.00019262 This is almost zero.
Solution b):
P(x=20) = P(19.5<x<20.5) z-score area: P(z<zscore of 19.5) = P(z←1.18) = 0.119 P(z<zscore of 20.5) = P(z←0.92) = 0.17879 P(19.5<x<20.5) = 0.17879 - 0.119 = 0.05979
When you use BP P(x=20), you get 0.0616. not too bad. almost 0.02 off, very damn close
Example 2
A student is writing a true-false test. if he is answering 10 questions, what is the probability that he will answer 3 or less questions correctly
Solution
Can we use normal distributions? n = 10 p = 0.5 q = 0.5 np = 5. 5 is not > 5. WE CANNOT USE NORMAL DISTRIBUTION. WE MUST USE BINOMIAL DISTRIBUTION use Binomial Distribution to solve it. BD: P(x<3) = P(x=0) + P(x=1) + P(x=2) + P(x=3) …
Example 3
Eric rolls a dice 100 times. what is pribability that he will roll a sum of 7 between 10 to 15 times? n = 100 p = 1/6 q = 5/6 np = 100/6 > 5 P(10⇐x⇐15) turn to continuous interval P(9.5⇐x⇐15.5) find z scores z score of 9.5 = -1.92 z score of 15.5 = -0.31 Always the bigger number first. P(z←0.31) - P(z←1.92) = 0.6274 - 0.3783 = 0.3509
Example 4
What is the probability of rolling more than 10 “3”s when rolling a dice 100 times? p=1/6 q = 5/6 n=100 100/6 > 5. we can use continuous P(x>=10) to continuous interval is: P(x>9.5) z-score of 9.5 is: … 0.7967 79.67%