Answer to Question #315643 in Statistics and Probability for Silling

Question

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.

Answer 1

$1a:\\\int_1^3{\frac{1}{2}dx}=\frac{1}{2}\left( 3-1 \right) =1\\1b:\\P\left( 2<X<2.5 \right) =\int_2^{2.5}{\frac{1}{2}dx}=\frac{1}{4}\\1c:\\P\left( X\leqslant 1.6 \right) =\int_1^{1.6}{\frac{1}{2}dx}=0.3\\2a:\\P\left( X<4 \right) =\int_2^4{\frac{2\left( 1+x \right)}{27}}dx=\frac{5^2-3^2}{27}=\frac{16}{27}\\2b:\\P\left( 3\leqslant X<4 \right) =\int_3^4{\frac{2\left( 1+x \right)}{27}dx}=\frac{5^2-4^2}{27}=\frac{1}{3}\\3: T\,\,-\,\,number\,\,of\,\,dimes\\P\left( T=1 \right) =\frac{C_{4}^{1}}{C_{6}^{3}}=0.2\\P\left( T=2 \right) =\frac{C_{2}^{1}C_{4}^{2}}{C_{6}^{3}}=0.6\\P\left( T=3 \right) =\frac{C_{4}^{3}}{C_{6}^{3}}=0.2$

$4:\\P\left( X=0 \right) =\left( \frac{4}{6} \right) ^3=\frac{8}{27}\\P\left( X=1 \right) =C_{3}^{1}\left( \frac{4}{6} \right) ^2\left( \frac{2}{6} \right) ^1=\frac{4}{9}\\P\left( X=2 \right) =C_{3}^{2}\left( \frac{4}{6} \right) ^1\left( \frac{2}{6} \right) ^2=\frac{2}{9}\\P\left( X=3 \right) =\left( \frac{2}{6} \right) ^3=\frac{1}{27}$