Hi,
The question is asking you to calculate the approximate probability. The 'exact' probability would be determined using the binomial distribution. This would've been the approach to use had you not been asked to calculate the approximate probability.
Let X~Bin(n,p). Then the approximate distribution (say) Y ~ N(np, npq)
A continuity correction is required when moving from a discrete to a continuous distribution.
The P(X > 100) ~ P(Y > 100.5).
Standardising the distribution of Y to the standard normal with mean 0, and variance 1 gives
P(Z > [100.5 - 120] / sqrt(72)) = P(Z> -2.298)
Graphically, this probability represents the area under the curve to the right of -2.298 of a N(0,1) distribution.
Now, the normal distribution is a symmetric distribution. This property is desrable and means that P(Z>-2.298) = P(Z<2.298) = phi(2.298) [the area under the curve to the left of 2.298.]
Hope that helps.
Last edited: May 14, 2020
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