Hi,

Expected value and SD				

n	p	np	n*n*p	

0	0.54	0	0	

1	0.16	0.16	0.16	

2	0.06	0.12	0.24	

3	0.04	0.12	0.36	

4	0.2	0.8	3.2	

		1.2	3.96	

variance = 3.96 - (1.2)^2 = 2.52

SD =  = 1.5875

mean = 1.2

P(x ≥ 3)

Using TI: The syntax is normalcdf(smaller, larger, µ, σ)

Note: 100 is used as the larger value to be at least 5 standard deviations from the mean.

P(x ≥ 3) = 1 - normalcdf(3,100,1.2, 1.5875) 

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