a company pays its employees and average hourly wage of 36.25 with a standard deviation of 6.50 more than 23.22 of 36 employees are randomly selected, what is the probability that their mean wage is more than 23.22

P [ X > 23,22 ]  = 0,8798      or  P [ X > 23,22 ]  =  87,98 %

Step-by-step explanation:

P [ X > 23,22 ]  =  1  -  P [X ≤ 23,22 ]

P [X ≤ 23,22 ] =  ( X - μ₀ ) / σ /√n

P [X ≤ 23,22 ] = ( 23,22 - 36,25 )/ 6,50/√36

P [X ≤ 23,22 ] =  - 13,03 *6 / 6,50

P [X ≤ 23,22 ] = - 12,02        or       P [X ≤ 23,22 ] = 12,02 %

(note the - sign only mean that values on the left tail of the bell shape curve)

Then

P [ X > 23,22 ]  =  1  -  0,1202

P [ X > 23,22 ]  = 0,8798      or  P [ X > 23,22 ]  =  87,98 %

109 trust me

step-by-step explanation:

this doesn't make sense lol