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.
send. help.
You have a good start, but you have not quite finished the problem. Here I am reviewing, extending, and formatting your work:
The number who read e-books is $X \sim \mathsf{Binom}(n = 400, p=.07).$ $\mu = np = 600(.07) = 42;\; \sigma = \sqrt{np(1-p)} = \sqrt{39.06} = 6.2498 .$
Using the continuity correction, you seek $$P(X = 45) = P(44.5 < X < 45.5) = P\left(\frac{44.5 - 42}{6.2498} < \frac{X-\mu}{\sigma} < \frac{45.5 -42}{6.2498}\right)\\ \approx P(0.40 < Z < 0.56) = ??\,,$$ where $Z$ is standard normal and the answer can be obtained from printed standard normal tables. (Of course, you should verify all of my computations.)
Using software, you could find the exact value of $P(X = 45) = 0.0551.$ The result from R statistical software is shown below. It is typical for normal approximations to the binomial to be correct to about two decimal places.
The figure shows the part of the binomial distribution with significant probability along with the 'best fitting' normal curve. The exact binomial probability is the height of the solid black bar at $x = 45;$ the normal approximation is the area beneath the curve and between the vertical dotted lines.