Question

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.

Answer 1

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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) We have to find $x_0$ such that

P(X < x)  = 0.1

$P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.1$  

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

$P(z < -1.282) = 0.1$

$\displaystyle\frac{x_0 - 56}{3} = -1.282\\\\x_0 = 52.15$

b) We have to find $x_0$ such that

P(X < x)  = 0.8

$P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.8$  

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

$P(z < 0.842) = 0.8$

$\displaystyle\frac{x_0 - 56}{3} = 0.842\\\\x_0 = 58.53$

c) We have to find $x_0$ such that

P(X > x)  = 0.01

$P( X > x) = P( z > \displaystyle\frac{x - 56}{3})=0.01$  

$= 1 -P( z \leq \displaystyle\frac{x - 56}{3})=0.01$  

$=P( z \leq \displaystyle\frac{x - 56}{3})=0.99$  

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

$P(z < 2.326) = 0.99$

$\displaystyle\frac{x_0 - 56}{3} = 2.326\\\\x_0 = 62.98$