Question

Suppose x is a normally distributed random variable with

muμequals=3434

and

sigmaσequals=44.

Find a value

x 0x0

of the random variable x.

a.

​P(xgreater than or equals≥x 0x0​)equals=.5

b.

​P(xless than<x 0x0​)equals=.025

c.

​P(xgreater than>x 0x0​)equals=.10

d.

​P(xgreater than>x 0x0​)equals=.95

Answer 1

Solution:-

Given that,

mean = = 34

standard deviation = = 4

a) Using standard normal table,

P(Z z) = 0.5

= 1 - P(Z z) = 0.5  

= P(Z z) = 1 - 0.5

= P(Z z ) = 0.5

= P(Z 0) = 0.5  

z = 0

Using z-score formula,

x₀ = z * +

x₀ = 0 * 4 + 34

x₀ = 34

b) Using standard normal table,

P(Z < z) = 0.025

= P(Z < z) = 0.025  

= P(Z < -1.96 ) = 0.025

z = -1.96

Using z-score formula,

x₀ = z * +

x₀ = -1.96 * 4 + 34

x₀ = 26.16

c) Using standard normal table,

P(Z > z) = 0.10

= 1 - P(Z < z) = 0.10  

= P(Z < z) = 1 - 0.10

= P(Z < z ) = 0.90

= P(Z < 1.28 ) = 0.90  

z = 1.28

Using z-score formula,

x₀ = z * +

x₀ = 1.28 * 4 + 34

x₀ = 39.12

d) Using standard normal table,

P(Z > z) = 0.95

= 1 - P(Z < z) = 0.95

= P(Z < z) = 1 - 0.95

= P(Z < z ) = 0.05

= P(Z < -1.65 ) = 0.05

z = -1.65

Using z-score formula,

x₀ = z * +

x₀ = -1.65* 4 + 34

x₀ = 27.40