Question

1. A certain population has an income distribution given by a normal random variable whose mean is $35000 and whose standard deviation is $7500.
2. A. What is the probability that a randomly selected person has an income of at least $35000?

Answer 1

Solution :

Given that ,

mean = = $ 35000

standard deviation = = $ 7500

P(x $ 35000 ) = 1 - P(x   $ 35000 )

= 1 - P[(x - ) / ( 35000 - 35000 ) / 7500 ]

= 1 -  P(z 0.00 )  

Using z table

= 1 - 0.5000

= 0.5000