Answer:
a. 0.0000
b. 0.1841
c. 0.9992
Explanation:
a. n = 285
p = 5% = 0.05
μ = np = 285 x 0.05
= 14.25
we fnd the standard deviation
sd = √np(1-p)
= [tex]\sqrt{285*0.05*0.95}[/tex]
= 3.6793
we find the z score
x = 30-0.5 = 29.5
[tex]z=\frac{29.5-14.25}{3.6793} \\= 4.14[/tex]
using the microsoft excel function
1-NORMSDIST(4.14)
probability = 1 -0.999982
= 0.0000
b.
n = 502
p = 0.05
np = 502x0.05
= 25.1
sd = [tex]\sqrt{np(1-p)}[/tex]
= [tex]\sqrt{502*0.05*0.95} \\= 4.8831[/tex]
x = 29.5
[tex]z = \frac{29.5-25.1}{4.8831} \\= 0.90[/tex]
1 - NORMSDIST(0.90)
= 1 - 0.815939875
PROB = 0.1841
c. n = 1033
p = 0.05
np = 1033*0.05
= 51.65
sd [tex]\sqrt{np(1-p)} \\= \sqrt{1033*0.05*0.95}[/tex]
= 7.0048
x = 29.5
[tex]z=\frac{29.5-51.65}{7.0048} \\= -3.16[/tex]
probability =
1 - normsdist(-3.16)
= 1 - 0.000788846
= 0.9992