Stats problem

[cliftonite]

Can someone please do these two problems for me I have an exam in 4:30 hrs and i dont feel like thinking anymore (yes a final on a SUNDAY morning)

The average amount of money that an 18 to 24 year old living with parents spends in a year is 4,190. Suppose this amount is normally distributed with a standard deviation of 1200. What proportion of 18 to 24 year olds living with parents spend over 6,000?

Market research indicates that 70% of US households own at least one personal computer. US households are selected at random. What is the probablity that fewer than 4 of these households own at least on PC?

 

[b0mbrman]

Enable your PMs or IM me on AIM...

 

[Ameesh]

do you have any idea what time it is?

 

[b0mbrman]

The average amount of money that an 18 to 24 year old living with parents spends in a year is 4,190. Suppose this amount is normally distributed with a standard deviation of 1200. What proportion of 18 to 24 year olds living with parents spend over 6,000?

6000 corresponds to a z-score of:
(6000-mean)/stdDev
=(6000 - 4190)/1200
=1.51

Check your normal distribution table...the area to the right of 1.51 is 0.0657 or 6.57%

Market research indicates that 70% of US households own at least one personal computer. US households are selected at random. What is the probablity that fewer than 4 of these households own at least on PC?

How many US households?

 

[cliftonite]

PMs are enabled and its 12 households. Thanks for the help guys, I got two more problems if u guys want to do them

 

[cliftonite]

bump