Answer to Question #234950 in Statistics and Probability for talie

Question #234950

Hospital records show that of patients suffering from a certain disease, 25% die of it. (i) What is the probability that of 6 randomly selected patients, 4 will recover? [3 marks] (ii) What is the most probable number of recoveries out of 6 randomly selected patients? [2 marks] Page 3 of 5 (iii) Compute and plot a bar graph of the probability distribution of recoveries out of 6 randomly selected patients. [5 marks] (iv) Plot the distribution function of the number of recoveries out of 6 randomly selected patients.


1
Expert's answer
2021-09-09T14:48:38-0400

Let "X=" the number of recovered patients: "X\\sim Bin(n, p)."

Given "n=6, q=0.25, p=1-q=1-0.25=0.75."

(i)


"P(X=4)=\\dbinom{6}{4}(0.75)^4(0.25)^{6-4}"

"=0.2966"



(ii)

The mode is an integer "M" that satisfies


"(n+1)p-1\\leq M<(n+1)p"

"(6+1)\\cdot0.75-1\\leq M<(6+1)\\cdot0.75"

"4.25\\leq M<5.25"

"M=5"

"P(X=5)=\\dbinom{6}{5}(0.75)^5(0.25)^{6-5}"

"=0.35595703125"

The most probable number of recoveries out of 6 randomly selected patients is "5" patients.


(iii)


"P(X=0)=\\dbinom{6}{0}(0.75)^0(0.25)^{6-0}"

"=0.0002"




"P(X=1)=\\dbinom{6}{1}(0.75)^1(0.25)^{6-1}"

"=0.0044"




"P(X=2)=\\dbinom{6}{2}(0.75)^2(0.25)^{6-2}"

"=0.0330"




"P(X=3)=\\dbinom{6}{3}(0.75)^3(0.25)^{6-3}"

"=0.1318"




"P(X=4)=\\dbinom{6}{4}(0.75)^4(0.25)^{6-4}"

"=0.2966"




"P(X=5)=\\dbinom{6}{5}(0.75)^5(0.25)^{6-5}"

"=0.3560"




"P(X=6)=\\dbinom{6}{6}(0.75)^6(0.25)^{6-6}"

"=0.1780"






(iv)






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