. what is the normal approximation for the binomial probability that x=4, n=12 and p=0.3? how does this compare to the value of P(x=4)
I need so much help in searching for z scores, area and figuring probabilities and learning the right tables to use. Thank you very much
I need so much help in searching for z scores, area and figuring probabilities and learning the right tables to use. Thank you very much
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P(success) = p = 0.3
q = 1-p = 1-0.3 = 0.7
n = 12
Binomial probability = P(x=4)
= 12C4*0.7^8*0.3^4
= 0.2311
Normal approximation to the Binomial distribution (applied when n is very large and continuity correction is applied) )
z = (X-np) / sqrt npq
P(x=4) becomes P(3.5 < 4 < 4.5)
z1 = (3.5 - 12*0.3) / sqrt 12*0.3*0.7
z1 = - 0.1 /1.59
z1 = - 0.06
z2 = (4.5 -12*0.3) / sqrt 12*0.3*0.7
z2 = 0.9 / 1.59
z2 = + 0.57
The area under the standard normal curve between z1 and z2 indicates the required probability.
Required probability = 0.0239 + 0.2157 = 0.2396
Since the z1 is -ve and z2 is +ve, the areas corresponding to the z-scores are added
Search in the google for z-scores chart
q = 1-p = 1-0.3 = 0.7
n = 12
Binomial probability = P(x=4)
= 12C4*0.7^8*0.3^4
= 0.2311
Normal approximation to the Binomial distribution (applied when n is very large and continuity correction is applied) )
z = (X-np) / sqrt npq
P(x=4) becomes P(3.5 < 4 < 4.5)
z1 = (3.5 - 12*0.3) / sqrt 12*0.3*0.7
z1 = - 0.1 /1.59
z1 = - 0.06
z2 = (4.5 -12*0.3) / sqrt 12*0.3*0.7
z2 = 0.9 / 1.59
z2 = + 0.57
The area under the standard normal curve between z1 and z2 indicates the required probability.
Required probability = 0.0239 + 0.2157 = 0.2396
Since the z1 is -ve and z2 is +ve, the areas corresponding to the z-scores are added
Search in the google for z-scores chart