Answer:
a) 0.0139; b) 0.1809; c) 0.4278
Step-by-step explanation:
We use a normal approximation to a binomial distribution for these problems.
The sample size, n, for each is 30; p, the probability of success, is 0.808. This makes the mean, μ = np = 30(0.808) = 24.24. The standard deviation,
σ = √(npq) = √(30(0.808)(1-0.808)) = √(30(0.808)(0.192)) = √4.65408 = 2.1573
For part a,
We are asked for P(X < 20). Using continuity correction to account for the discrete variable, we find
P(X < 19.5)
z = (19.5-24.24)/(2.1573) = -4.74/2.1573 = -2.20
Using a z table, we see that the area under the curve to the left of this is 0.0139.
For part b,
We are asked for P(X = 24). Using continuity correction, we find
P(23.5 < X < 24.5)
z = (23.5-24.24)/2.1573 = -0.74/2.1573 = -0.34
z = (24.5-24.24)/2.1573 = 0.26/2.1573 = 0.12
Using a z table, we see that the area under the curve to the left of z = -0.34 is 0.3669. The area under the curve to the left of z = 0.12 is 0.5478. The area between them is then
0.5478-0.3669 = 0.1809.
For part c,
We are asked to find P(25 ≤ X ≤ 28). Using continuity correction, we find
P(24.5 < X < 28.5)
z = (24.5-24.24)/2.1573 = 0.26/2.1573 = 0.12
z = (28.5-24.24)/2.1573 = 4.26/2.1573 = 1.97
Using a z table, we see that the area under the curve to the left of z = 0.12 is 0.5478. The area under the curve to the left of z = 1.97 is 0.9756. The area between them is 0.9756 - 0.5478 = 0.4278.
