Question

(a) Find a z₀ that has area 0.9525 to its left. (Round your answer to two decimal places.)

z₀ =

(b) Find a z₀ that has area 0.05 to its left. (Round your answer to two decimal places.)

z₀ =

Find the following percentiles for the standard normal random variable z. (Round your answers to two decimal places.)

(a)    90th percentile
z =  

(b)    95th percentile
z =  

(c)    97th percentile
z =  

(d)    99th percentile
z =

A normal random variable x has mean μ = 1.2 and standard deviation σ = 0.11. Find the probabilities of these X-values. (Round your answers to four decimal places.)

Answer 1

Answer:- X is normal random variable .

Mean = 1.2

Standard deviation =0.11

1)

A) value of Z for area 0.9525 to its left

= P(Z < ? ) = 0.9525

= Z = 1.6696 ( from normal table)

B) value of Z for area 0.05 to its left

= P( Z < ? ) = 0.05

= Z = -1.65 (from normal table)

2)

these value is calculated from normal (table Z value table)

(a)    90th percentile
z = 1.282 ( 90% of area under standard normal curve is below 1.282)

(b)    95th percentile
z = 1.645 ( 95% of area under standard normal curve is below 1.645)

(c)    97th percentile
z = 1.881 ( 97% of area under standard normal curve is below 1.881)

(d)    99th percentile
z =2.326 ( 99% of area under standard normal curve is below 2.326)