The probability that a patient recovers from a delicate heart operation is 0.9. The next 100 patients having this operation, find the probability that (i) between 84 and 95 (inclusive values) survive? (ii) fewer than 86 survive?

If $X$ is a binomial random variable with mean $μ = np$ and variance $σ^2 = npq,$ then the limiting form of the distribution of

$Z=\dfrac{X-np}{\sqrt{npq}}$

as $n → ∞,$ is the standard normal distribution $N(0,1).$

Given $n=100, p=0.9, q=1-p=0.1$

$\mu=np=100(0.9)=90, npq=100(0.9)(0.1)=9$

Use the Continuity Correction Factor

(i)

$P(84\le X\le95)=P(X<95+0.5)-P(X<84-0.5)$

$=P(Z<\dfrac{95.5-90}{\sqrt{9}})-P(Z<\dfrac{83.5-90}{\sqrt{9}})$

$\approx P(Z<1.833333)-P(Z<-2.166667)$

$\approx0.96662-0.01513\approx0.9515$

(ii)

$P(X<86)=P(X<86-0.5)$

$=P(Z<\dfrac{85.5-90}{\sqrt{9}})\approx P(Z<-1.833333)$

$\approx0.0334$