Let X be number of cars with airbags among the next 4 cars sold by the agency.
Therefore
X ∽ B i n o m i a l ( n = 4 , p = 0.4 ) X\backsim Binomial (n=4,p=0.4)\\ X ∽ B in o mia l ( n = 4 , p = 0.4 )
Probability mass function f(x)
f ( x ) = ( 4 x ) 0. 5 x ( 1 − 0.5 ) 4 − x = ( 4 x ) 0. 5 x ( 1 − 0.5 ) 4 − x = ( 4 x ) 0. 5 4 = 1 16 ( 4 x ) x = 0 , 1 , 2 , 3 , 4 f(x)=\dbinom{4}{x}0.5^x(1-0.5)^{4-x}=\\
\dbinom{4}{x}0.5^x(1-0.5)^{4-x}=\dbinom{4}{x}0.5^4=\dfrac{1}{16}\dbinom{4}{x}\\
x=0,1,2,3,4 f ( x ) = ( x 4 ) 0. 5 x ( 1 − 0.5 ) 4 − x = ( x 4 ) 0. 5 x ( 1 − 0.5 ) 4 − x = ( x 4 ) 0. 5 4 = 16 1 ( x 4 ) x = 0 , 1 , 2 , 3 , 4
Cumulative distribution function
F ( x ) = ∑ t ≤ x f ( t ) f ( 0 ) = 1 16 ( 4 0 ) = 1 16 f ( 1 ) = 1 16 ( 4 1 ) = 1 4 f ( 2 ) = 1 16 ( 4 2 ) = 3 8 f ( 3 ) = 1 16 ( 4 3 ) = 1 4 f ( 4 ) = 1 16 ( 4 4 ) = 1 16 F ( 0 ) = 1 16 F ( 1 ) = f ( 0 ) + f ( 1 ) = 5 16 F ( 2 ) = f ( 0 ) + f ( 1 ) + f ( 2 ) = 11 16 F ( 3 ) = f ( 0 ) + f ( 1 ) + f ( 2 ) + f ( 3 ) = 15 16 F ( 4 ) = f ( 0 ) + f ( 1 ) + f ( 2 ) + f ( 3 ) + f ( 4 ) = 1 F ( X ) = { 1 16 for 0 ≤ x > 1 5 16 for 1 ≤ x > 2 5 16 for 2 ≤ x > 3 15 16 for 3 ≤ x > 4 1 for x ≥ 4 F(x)=\displaystyle\sum_{t\leq{x}}f(t)\\
f(0)=\dfrac{1}{16}\dbinom{4}{0}=\dfrac{1}{16}\\
f(1)=\dfrac{1}{16}\dbinom{4}{1}=\dfrac{1}{4}\\
f(2)=\dfrac{1}{16}\dbinom{4}{2}=\dfrac{3}{8}\\
f(3)=\dfrac{1}{16}\dbinom{4}{3}=\dfrac{1}{4}\\
f(4)=\dfrac{1}{16}\dbinom{4}{4}=\dfrac{1}{16}\\
F(0)=\dfrac{1}{16}\\
F(1)=f(0)+f(1)=\dfrac{5}{16}\\
F(2)=f(0)+f(1)+f(2)=\dfrac{11}{16}\\
F(3)=f(0)+f(1)+f(2)+f(3)=\dfrac{15}{16}\\
F(4)=f(0)+f(1)+f(2)+f(3)+f(4)=1\\
F(X)=\begin{cases}
\dfrac{1}{16} &\text{for } 0\leq{x}\gt1\\
\\
\dfrac{5}{16} &\text{for } 1\leq{x}\gt2\\
\\
\dfrac{5}{16} &\text{for } 2\leq{x}\gt3\\
\\
\dfrac{15}{16} &\text{for } 3\leq{x}\gt4\\
\\
1 &\text{for }x\geq4
\end{cases} F ( x ) = t ≤ x ∑ f ( t ) f ( 0 ) = 16 1 ( 0 4 ) = 16 1 f ( 1 ) = 16 1 ( 1 4 ) = 4 1 f ( 2 ) = 16 1 ( 2 4 ) = 8 3 f ( 3 ) = 16 1 ( 3 4 ) = 4 1 f ( 4 ) = 16 1 ( 4 4 ) = 16 1 F ( 0 ) = 16 1 F ( 1 ) = f ( 0 ) + f ( 1 ) = 16 5 F ( 2 ) = f ( 0 ) + f ( 1 ) + f ( 2 ) = 16 11 F ( 3 ) = f ( 0 ) + f ( 1 ) + f ( 2 ) + f ( 3 ) = 16 15 F ( 4 ) = f ( 0 ) + f ( 1 ) + f ( 2 ) + f ( 3 ) + f ( 4 ) = 1 F ( X ) = ⎩ ⎨ ⎧ 16 1 16 5 16 5 16 15 1 for 0 ≤ x > 1 for 1 ≤ x > 2 for 2 ≤ x > 3 for 3 ≤ x > 4 for x ≥ 4
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