Question

A)

Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.

mean=101; standard deviation= 11

P(x > or = 120) = ?

B)

mean=2.6; standard deviation= 0.38

P(x > or = 2) = ?

C) Find z such that 2.6% of the standard normal curve lies to the left of z.

z= ?

D) Find z such that 4.2% of the standard normal curve lies to the right of z

z= ?

E) Find the z value such that 96% of the standard normal curve lies between -z and z

z = ?

Answer 1

a)

µ = 101     

σ = 11

  

P ( X ≥120.00) = P( (X-µ)/σ ≥ (120-101) / 11)

= P(Z ≥1.727) = P( Z <-1.727) = 0.0421(answer)

b)

µ = 2.6     

σ = 0.38

  

P ( X ≥2.00) = P( (X-µ)/σ ≥ (2-2.6) / 0.38)

= P(Z ≥-1.579) = P( Z <1.579) = 0.9428(answer)

c)

µ=0     

σ = 1

proportion=0.026

  

Z value at 0.026=-1.943 (answer)   

(excel formula =NORMSINV(0.026) )

d)

µ=0     

σ = 1

area lies to left= 1 - 0.042 =0.958

  

Z value at 0.958=1.728

(excel formula =NORMSINV(0.958) )

e)

µ = 0       

σ = 1

proportion=0.9600

proportion left =1-0.96 = 0.04    is equally distributed both left and right side of normal curve  

z value at 0.04/2 = 0.02    is ±2.054

(excel formula =NORMSINV(0.04/ 2 ) )

so, Z=2.054