Answer to Question #178315 in Statistics and Probability for anis

Question #178315

5.According to a local study, 80 per cent of workers in Malaysia work in an environment that could endanger their visual. If 20 Malaysia workers selected randomly as a sample, what is the probability


a) that exactly 16 workers work in such environment?


b) that more than 17 workers work in such environment?


c) that exactly 4 workers not work in such environment?


d) fewer than 3 workers the number not work in such environment?


1
Expert's answer
2021-04-13T15:50:40-0400

"P(x) = \\frac{n!}{x!(n-x)!} \\times p^x \\times (1-p)^{n-x}"

a) P = 80 % = 0.8

n = 20

Evaluate P(x) if x = 16

"P(x = 16) = \\frac{20!}{16!(20-16)!} \\times 0.8^{16} \\times (1 - 0.8)^{20-16} \\\\\n\n= 4845 \\times 0.8^{16} \\times 0.2^{4} \\\\\n\n= 0.218199"

b) P(x > 17) = P(18) + P(19) + P(20)

"= \\frac{20!}{18!(20-18)!} \\times 0.8^{18} \\times (1 \u2013 0.8)^{20-18} + \\frac{20!}{19!(20-19)!} \\times 0.8^{19} \\times (1 - 0.8)^{20-19} + \\frac{20!}{20!(20-20)!} \\times 0.8^{20} \\times (1 \u2013 0.8)^{20-20} \\\\\n\n= 0.1369 + 0.0576 + 0.0115 \\\\\n\n= 0.2060"

c) P = 20 % = 0.2 (not work in such environment)

n = 20

"P(x = 4) = \\frac{20!}{4!(20-4)!} \\times 0.2^4 \\times (1-0.2)^{20-4} \\\\\n\n= 4845 \\times 0.0016 \\times 0.02814 \\\\\n\n= 0.21814"

d) P(x<3) = P(x=0) + P(x=1) +P(x=2)

"P(x=0) = \\frac{20!}{0!(20-0)!} \\times 0.2^0 \\times (1-0.2)^{20-0} \\\\\n\n= 0.011529 \\\\\n\nP(x=1) = \\frac{20!}{1!(20-1)!} \\times 0.2^1 \\times (1-0.2)^{20-1} \\\\\n\n= 20 \\times 0.2 \\times 0.01441 \\\\\n\n= 0.05764 \\\\\n\nP(x=2) = \\frac{20!}{2!(20-2)!} \\times 0.2^2 \\times (1-0.2)^{20-2} \\\\\n\n= 190 \\times 0.04 \\times 0.01801 \\\\\n\n= 0.136876 \\\\\n\nP(x<3) = 0.011529 + 0.05764 + 0.136876 = 0.206045"


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