Question

Please post all the work so I can understand the process! Thanks :)

Question 3:

The 9-month salaries at a daycare center are normally distributed with a mean of $18,000 and a standard deviation of $5,000.

a.	What is the probability that an employee will have a salary between $15,520 and $18,480?
b.	What is the probability that an employee will have a salary more than $19,880?
c.	What is the probability that an employee will have a salary less than $28,440?

Answer 1

Solution:

Given, X follows Normal distribution with,

   = 18,000

= 5,000

a)

P(15,520 < x< 18,480)

= P(X < 18,480) - P(X < 15,520 )

=  P[(X - )/ <  (18,480 - 18,000)/5,000] -   P[(X - )/ <  (15,520 - 18,000)/5,000]

= P[Z < 0.096] - P[Z < -0.496]

= 0.5382 - 0.3099 ..Use z table

= 0.2283

b)

P(X > 19880)

= P[(X - )/ >  (19880 - )/]

= P[Z > (19880 - 18,000)/5,000]

= P[Z > 0.376]

= 1 - P[Z < 0.376]

=    1 - 0.6465( use z table)

=    0.3535

c)

P(X < 28,440)

= P[(X - )/ <  (28,440 - )/]

= P[Z <  (28,440 - 18,000)/5,000]

= P[Z < 2.088]

= 0.9816 ... ( use z table