According to a report on www.teleread.com (dated November 8, 2010), 7% of U.S. adults with online services currently read e-books. Assume that this percentage is true for the current population of U.S. adults with online services. Find the probability that in a random sample of 600 U.S. adults with online services, the number who read e-books is exactly 45. Round off to 4 decimal places. [CLUE: If it is exactly 45, use P open parentheses 44.50 less than x less than 45.50 close parentheses.] $P(44.5 < X < 45.5)$
What I did: mean = 42; standard deviation = 6.2498
$z = ((45-0.5)-42)/6.2498; z = 0.4000.$
Then I looked up the $z$ value in a $z$ table and got the probability of 0.6554; then I subtracted that from 1 : $1 - 0.6554 = 0.3446.$
my final answer is 0.3446.