Question

1. A survey of recent business college graduates revealed that their annual incomes were normally distributed with a mean income of $52,000 and a standard deviation of $18,000. Answer the following questions. (3.5 pts.) Note: Show your z score rounded to two decimal places (i.e., 2.85) (Showing calculations may help you earn partial credit.) Show your probability rounded to four decimal places (i.e., .6567). a) What is the probability of a random student earning less than $35,000? b) What is the probability of a random student earning more than $65,000? c) What is the probability of a random student earning between $35,000 and $65,000?

Answer 1

Solution :

Given that ,

mean = = 52000

standard deviation = = 18000

a) P(x < 35000) = P[(x - ) / < (35000 - 52000) / 18000]

= P(z < -0.94)

Using z table,

= 0.1736

b) P(x > 65000) = 1 - p( x< 65000)

=1- p P[(x - ) / < (65000 - 52000) / 18000]

=1- P(z < 0.72)

Using z table,

= 1 - 0.7642

= 0.2358

c) P(35000 < x < 65000) = P[(35000 - 52000)/ 18000) < (x - ) /  < (65000 - 52000) / 18000) ]

= P(-0.94 < z < 0.72)

= P(z < 0.72) - P(z < -0.94)

Using z table,

= 0.7642 - 0.1736

= 0.5906