Question

Question 6 Type numbers in the boxes A certain medical test is known to detect 48% of the m people who are afflicted with the disease Y. If 10 ッPart 23 pants people with the disease are administered the test, P points what is the probability that the test will show that: io points All 10 have the disease, rounded to four decimal places? At least 8 have the disease, rounded to four decimal places?O At most 4 have the disease, rounded to four decimal places? C 87 Part 1:3 points 9

Answer 1

Probability of detecting the disease, p = 0.48

q = 1-0.48 = 0.52

n= 10

Binomial distribution: _{$\small P(X)=\ _nC_x*p^x*q^{n-x}$}

a) Probability of all 10 having the disease, P(X=10) =

$\small P(10)=\ _{10}C_{10}*0.48^{10}*0.52^{0} =\textbf{0.0006}$

b) Probability that atleast 8 have the disease, P(X>= 8) = P(8) + P(9) + P(10)

$\small =\ _{10}C_{8}*0.48^{8}*0.52^{2} +\ _{10}C_{9}*0.48^{9}*0.52^{1} +\ _{10}C_{10}*0.48^{10}*0.52^{0}$

$\small = 0.0343+ 0.0070+0.0006 = \textbf{0.0419}$

c)  Probability that at most 4 have the disease, P(X<= 4) = P(0) + P(1) + P(2) +P(3) + P(4)

Using binomial distribution calculator

$\small P(X \leq 4 ) = \textbf{0.4270}$