Answer to Question #304042 in Statistics and Probability for Vimukthi

Question #304042

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?

1
Expert's answer
2022-03-04T04:39:51-0500

If "X" is a binomial random variable with mean "\u03bc = np" and variance "\u03c3^2 = npq," then the limiting form of the distribution of


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

as "n \u2192 \u221e," 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"


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