a. X be the number of stored radios with two slots has hypergeometric distribution
P ( X = x ) = h ( x ; n , M , N ) = ( M x ) ( N − M n − x ) ( N n ) P(X=x)=h(x;n, M, N)={\dbinom{M}{x}\dbinom{N-M}{n-x} \over \dbinom{N}{n}} P ( X = x ) = h ( x ; n , M , N ) = ( n N ) ( x M ) ( n − x N − M ) Given that N = 20 , M = 12 , n = 6. N=20, M=12, n=6. N = 20 , M = 12 , n = 6.
b.
P ( X ≤ 3 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) P(X\leq3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) P ( X ≤ 3 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 )
P ( X = 0 ) = ( 12 0 ) ( 20 − 12 6 − 0 ) ( 20 6 ) = 1 ( 28 ) 38760 = 28 38760 P(X=0)={\dbinom{12}{0}\dbinom{20-12}{6-0} \over \dbinom{20}{6}}={1(28) \over 38760}={28 \over 38760} P ( X = 0 ) = ( 6 20 ) ( 0 12 ) ( 6 − 0 20 − 12 ) = 38760 1 ( 28 ) = 38760 28
P ( X = 1 ) = ( 12 1 ) ( 20 − 12 6 − 1 ) ( 20 6 ) = 12 ( 56 ) 38760 = 672 38760 P(X=1)={\dbinom{12}{1}\dbinom{20-12}{6-1} \over \dbinom{20}{6}}={12(56) \over 38760}={672 \over 38760} P ( X = 1 ) = ( 6 20 ) ( 1 12 ) ( 6 − 1 20 − 12 ) = 38760 12 ( 56 ) = 38760 672
P ( X = 2 ) = ( 12 2 ) ( 20 − 12 6 − 2 ) ( 20 6 ) = 66 ( 70 ) 38760 = 4620 38760 P(X=2)={\dbinom{12}{2}\dbinom{20-12}{6-2} \over \dbinom{20}{6}}={66(70) \over 38760}={4620 \over 38760} P ( X = 2 ) = ( 6 20 ) ( 2 12 ) ( 6 − 2 20 − 12 ) = 38760 66 ( 70 ) = 38760 4620
P ( X = 3 ) = ( 12 3 ) ( 20 − 12 6 − 3 ) ( 20 6 ) = 220 ( 56 ) 38760 = 12320 38760 P(X=3)={\dbinom{12}{3}\dbinom{20-12}{6-3} \over \dbinom{20}{6}}={220(56) \over 38760}={12320\over 38760} P ( X = 3 ) = ( 6 20 ) ( 3 12 ) ( 6 − 3 20 − 12 ) = 38760 220 ( 56 ) = 38760 12320
P ( X ≤ 3 ) = 28 38760 + 672 38760 + 4620 38760 + 12320 38760 = P(X\leq3)={28 \over 38760}+{672 \over 38760}+{4620 \over 38760}+{12320 \over 38760}= P ( X ≤ 3 ) = 38760 28 + 38760 672 + 38760 4620 + 38760 12320 =
= 17640 38760 = 147 323 ≈ 0.4551 ={17640 \over 38760}={147\over 323}\approx0.4551 = 38760 17640 = 323 147 ≈ 0.4551
c. Calculate the mean and standard deviation of X.
μ = E ( X ) = n ⋅ M N = 6 ⋅ 12 20 = 3.6 \mu=E(X)=n\cdot{M \over N}=6\cdot{12 \over 20}=3.6 μ = E ( X ) = n ⋅ N M = 6 ⋅ 20 12 = 3.6
V a r ( X ) = σ 2 = ( N − n N − 1 ) ⋅ n ⋅ M N ( 1 − M N ) = Var(X)=\sigma^2=({N-n \over N-1})\cdot n\cdot{M \over N}(1-{M \over N})= Va r ( X ) = σ 2 = ( N − 1 N − n ) ⋅ n ⋅ N M ( 1 − N M ) =
= ( 20 − 6 20 − 1 ) ⋅ 6 ⋅ 12 20 ( 1 − 12 20 ) = 504 475 =({20-6\over 20-1})\cdot 6\cdot{12 \over 20}(1-{12 \over 20})={504 \over 475} = ( 20 − 1 20 − 6 ) ⋅ 6 ⋅ 20 12 ( 1 − 20 12 ) = 475 504
σ = 504 475 ≈ 1.03 \sigma=\sqrt{{504 \over 475}}\approx1.03 σ = 475 504 ≈ 1.03 d.
μ + σ ≈ 3.6 + 1.03 = 4.63 \mu+\sigma\approx3.6+1.03=4.63 μ + σ ≈ 3.6 + 1.03 = 4.63
P ( 3.6 < X ≤ 4.63 ) = P ( X = 4 ) = P(3.6<X\leq4.63)=P(X=4)= P ( 3.6 < X ≤ 4.63 ) = P ( X = 4 ) =
= ( 12 4 ) ( 20 − 12 6 − 4 ) ( 20 6 ) = 495 ( 28 ) 38760 = 13860 38760 = ={\dbinom{12}{4}\dbinom{20-12}{6-4} \over \dbinom{20}{6}}={495(28) \over 38760}={13860\over 38760}= = ( 6 20 ) ( 4 12 ) ( 6 − 4 20 − 12 ) = 38760 495 ( 28 ) = 38760 13860 =
= 231 646 ≈ 0.3576 ={231\over 646}\approx0.3576 = 646 231 ≈ 0.3576 e.
Moderately negative skewed.