Census data show that 4% of students in a certain country went to college or a t

Question

Census data show that 4% of students in a certain country went to college or a trade school within a year of graduating from high school. A random sample of 779 high school graduates is selected. Use the normal approximation to the binomial distribution to complete parts a through e below a. What are the mean and standard deviation of this distribution? (Round to two decimal places as needed.) b. What is the probability that 17 or fewer students will go to college? P(xs 17) (Round to four decimal places as needed)

Explanation / Answer

NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
a.
mean ( np ) = 779 * 0.04 = 31.16
standard deviation ( npq )= 779*0.04*0.96 = 5.4693
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
b.
P(X > 17) = (17-31.16)/5.4693
= -14.16/5.4693 = -2.589
= P ( Z >-2.589) From Standard Normal Table
= 0.9952
P(X < = 17) = (1 - P(X > 17))
= 1 - 0.0048 = 0.9952
c.
We calculate exactly value by using the interval precision,
P(X=17) = P(16.5 < X < 17.5)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 16.5) = (16.5-31.16)/5.4693
= -14.66/5.4693 = -2.6804
= P ( Z <-2.6804) From Standard Normal Table
= 0.00368
P(X < 17.5) = (17.5-31.16)/5.4693
= -13.66/5.4693 = -2.4976
= P ( Z <-2.4976) From Standard Normal Table
= 0.00625
P(16.5 < X < 17.5) = 0.00625-0.00368 = 0.0026
d.
P(X < 39) = (39-31.16)/5.4693
= 7.84/5.4693= 1.4335
= P ( Z <1.4335) From Standard NOrmal Table
= 0.9241
P(X > = 39) = (1 - P(X < 39))
= 1 - 0.9241 = 0.0759
e.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 17) = (17-31.16)/5.4693
= -14.16/5.4693 = -2.589
= P ( Z <-2.589) From Standard Normal Table
= 0.00481
P(X < 29) = (29-31.16)/5.4693
= -2.16/5.4693 = -0.3949
= P ( Z <-0.3949) From Standard Normal Table
= 0.34645
P(17 < X < 29) = 0.34645-0.00481 = 0.3416

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