If anyone has knowledge in statistics area, do you know how to do these 2?

Replacement times for new Ford Taurus GL automobiles are normally distributed with a mean of 7.1 years and a standard deviation of 1.4 years. Find the probability that seven randomly selected Taurus GL automobiles will have a mean replacement time greater than 7.0 years.


Among U.S. households, 24% have telephone answering machines. If a telemarketing campaign involves 2,500 households, find the probability that more than 650 have answering machines.

Find the z score for 7 years. (7 - 7.1)/(1.4)

Then look at your z score table, find the area, and subtract from 1. The subtraction is necessary because the table will give you the "less than" probabilities, and you want greater than.

24% of 2500 will be 600. The procedure will be the same as for the car, but instead of a standard deviation you'll use an expression that uses the proportion of households (24%)

That expression will be square root(p(1-p)/n) where p is your percent (change it to a decimal) and n is your sample size.

hmm P(T>7) = P(Z> (7-7.1)/1.4) = P(Z > -0.0714)
use your normal table.....

2nd one
use normal approximation
mean will be np = 2500 * 0.24 =600
deviation = root of 2500*0.24*0.76 = 21.254
P(X>650) = P( Z> (650.5 - 600)/21.254) = P ( Z> 2.376)
use 650.5 in stead of 650 because of the correction thingy from approximation.
use your normal table....