Question

•The probability that a student fails the screening test for scoliosis (curvature of the spine) at a local high school is known to be 0.004. Of the next 1875 students who are screened for scoliosis, find the probability that (a) fewer than 5 fail the test.

Answer 1

$Let,\;X\;-\;random\;variable\;representing\;number\;of\\students,\;who\;failed\;test.\\Consider\;that\;it\;is\;success\;when\;student\;fails\;test.\\Trials\;are\;independent\Rightarrow X\;has\;binomial\;distribution\;with\\n\;=\;1875\;,\;p\;=\;0.004\\p\;is\;small\;and\;n\;is\;large,\;\mu\;=\;n\times p=7.5\;is\;finite.\\U\sin g\;poisson\;approximation,\;X\;has\;a\;Poisson\;distribution\\with\;parameter\;\mu=7.5\\Probability\;mass\;function:\;p(x;\;7.5)\;=\;\frac{e^{-7.5}{(7.5)}^x}{x!}\\P(X<5)\;=\;P(0)+P(1)+P(2)+P(3)+P(4)=\\=\frac{e^{-7.5}{(7.5)}^0}{0!}+...+\frac{e^{-7.5}{(7.5)}^4}{4!}=0.00055+0.00415+\\+0.01556+0.03889++0.7292=0.1321\\\\$