Answer to Question #235013 in Statistics and Probability for Light

Question

Hospitalrecordsshowthatofpatientssufferingfromacertaindisease,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.
[2

Answer 1

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)