Suppose X is normally distributed with mean 20 and standard deviation 4, find the value x0 such that P(X ≥ x0) = 0.975.

Mathematics

1 answer:

abruzzese [7]2 years ago

7 0

Answer:

12.16

Step-by-step explanation:

From your question, we have the following information

Mean = u = 20

Standard deviation sd = 4

P(X>Xo) = 0.975

1-0.975 = 0.025

P(z<Xo-20/4) = 0.025

Xo-20/4 = -1.96

We cross multiply

Xo - 20 = -1.96 x 4

Xo - 20 = -7.84

Xo = -7.84+30

Xo = 12.16

12.16 is the value of Xo and therefore the answer to this question.

Thank you!

You might be interested in

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ

Deffense [45]

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$