Suppose x is a normally distributed random variable with µ = 56 and σ = 3. Find a value x0 of the random variable x that satisfies the following equations or statements.a.​ 10% of the values of x are less than x0.b.​ 80% of the values of x are less than x0.c.​1% of the values of x are greater than x0.

Respuesta :

Answer:

a) 52.15

b) 58.53

c) 62.98        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 56

Standard Deviation, σ = 3

We are given that the distribution of x is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) We have to find [tex]x_0[/tex] such that

P(X < x)  = 0.1

[tex]P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.1[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < -1.282) = 0.1[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = -1.282\\\\x_0 = 52.15[/tex]

b) We have to find [tex]x_0[/tex] such that

P(X < x)  = 0.8

[tex]P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.8[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < 0.842) = 0.8[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = 0.842\\\\x_0 = 58.53[/tex]

c) We have to find [tex]x_0[/tex] such that

P(X > x)  = 0.01

[tex]P( X > x) = P( z > \displaystyle\frac{x - 56}{3})=0.01[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 56}{3})=0.01 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 56}{3})=0.99 [/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < 2.326) = 0.99[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = 2.326\\\\x_0 = 62.98[/tex]