Question

Suppose that 12% of a certain large population of people are left-handed. A simple random sample of 9 people is selected from this population. Find the following probabilities.

a) What is the probability exactly 3 people in the sample are left-handed? Round to four decimal places. Answer
b) What is the probability that 2 or less people in the sample are left-handed? Round to four decimal places. Answer

Answer 1

Sample size , n =    9
Probability of an event of interest, p =   0.12

P ( X = 3) = C (9,3) * 0.12^3 * ( 1 - 0.12)^6=      0.0674

.................

	X	P(X)
P ( X = 0) = C (9,0) * 0.12^0 * ( 1 - 0.12)^9=	0	0.3165
P ( X = 1) = C (9,1) * 0.12^1 * ( 1 - 0.12)^8=	1	0.3884
P ( X = 2) = C (9,2) * 0.12^2 * ( 1 - 0.12)^7=	2	0.2119

p(x<=2) = p(0) +p(1)+p(2) = 0.9167

.................

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