Question #315643

Self Assessment/ student activity



1. A continuous random variable X that can assume values between x =



1 and x = 3 has a density function given by f (x) = 1/2. (a) Show that the



area under the curve is equal to 1. (b) Find P (2 <X < 2.5). (c) Find P



(X ≤ 1.6)



2. A continuous random variable X that can assume values between x = 2



and x = 5 has a density function given by f (x) = 2(1 +x)/27.



Find (a) P(X < 4); (b) P (3 ≤ X < 4)



3. From a box containing 4 dimes and 2 nickels, 3 coins are selected at



random without replacement. Find the probability distribution for the total



T of the 3 coins. Express the probability distribution graphi- cally as a



probability histogram.



4. From a box containing 4 black balls and 2 green balls, 3 balls are drawn in



succession, each ball being replaced in the box before the next draw is made.



Find the probability distribution for the number of green balls.

Expert's answer

1a:1312dx=12(31)=11b:P(2<X<2.5)=22.512dx=141c:P(X1.6)=11.612dx=0.32a:P(X<4)=242(1+x)27dx=523227=16272b:P(3X<4)=342(1+x)27dx=524227=133:TnumberofdimesP(T=1)=C41C63=0.2P(T=2)=C21C42C63=0.6P(T=3)=C43C63=0.21a:\\\int_1^3{\frac{1}{2}dx}=\frac{1}{2}\left( 3-1 \right) =1\\1b:\\P\left( 2<X<2.5 \right) =\int_2^{2.5}{\frac{1}{2}dx}=\frac{1}{4}\\1c:\\P\left( X\leqslant 1.6 \right) =\int_1^{1.6}{\frac{1}{2}dx}=0.3\\2a:\\P\left( X<4 \right) =\int_2^4{\frac{2\left( 1+x \right)}{27}}dx=\frac{5^2-3^2}{27}=\frac{16}{27}\\2b:\\P\left( 3\leqslant X<4 \right) =\int_3^4{\frac{2\left( 1+x \right)}{27}dx}=\frac{5^2-4^2}{27}=\frac{1}{3}\\3: T\,\,-\,\,number\,\,of\,\,dimes\\P\left( T=1 \right) =\frac{C_{4}^{1}}{C_{6}^{3}}=0.2\\P\left( T=2 \right) =\frac{C_{2}^{1}C_{4}^{2}}{C_{6}^{3}}=0.6\\P\left( T=3 \right) =\frac{C_{4}^{3}}{C_{6}^{3}}=0.2


4:P(X=0)=(46)3=827P(X=1)=C31(46)2(26)1=49P(X=2)=C32(46)1(26)2=29P(X=3)=(26)3=1274:\\P\left( X=0 \right) =\left( \frac{4}{6} \right) ^3=\frac{8}{27}\\P\left( X=1 \right) =C_{3}^{1}\left( \frac{4}{6} \right) ^2\left( \frac{2}{6} \right) ^1=\frac{4}{9}\\P\left( X=2 \right) =C_{3}^{2}\left( \frac{4}{6} \right) ^1\left( \frac{2}{6} \right) ^2=\frac{2}{9}\\P\left( X=3 \right) =\left( \frac{2}{6} \right) ^3=\frac{1}{27}


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