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Sagot :
The normal approximation is not suitable as np<5.
Given,
mean = np
It turns out that if np and nq are both at least 5, the normal distribution can be used to approximate the binomial distribution. Additionally, keep in mind that a binomial distribution has a mean of np and variance of npq.
Hence apply the mean formula and get the value for np
n= 12 and p=0.
therefore mean = np
np = n×p
np = 12×0.3×
np = 3.6
np is less than 5.
nq=?
q=1-p
q=1-0.3
q=0.7
∴nq = 12 × 0.7
nq = 8.4
nq is not less than 5
It's remarkable that the normal distribution may accurately represent the binomial distribition when n, np and nq are large.
since here only nq is greater than five and np is less than five, therefore we conclude that normal approximation is not suitable.
Hence np<5 so normal approximation is not suitable.
Learn more about "normal distribution" here-
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