PLEASE HELP Given f(x)= (x-5)/3 SOLVE FOR f^-1(3)

1 answer:

7 0

As you already know, to get the inverse of any expression, we start off by doing a quick switcharoo on the variables, and then solve for "y", so let's do so,

$\bf \stackrel{f(x)}{y}=\cfrac{x-5}{3}\qquad inverse\implies \boxed{x}=\cfrac{\boxed{y}-5}{3}
\\\\\\
3x=y-5\implies 3x+5=\stackrel{f^{-1}(x)}{y}\\\\
-------------------------------\\\\
3(3)+5=f^{-1}(3)\implies 14=f^{-1}(3)$

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In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$