Answer:
23.27%
Step-by-step explanation:
From the statement we know that random sample n is 110 and that p is 75% and x the percentage to evaluate is 78%
We have that the probability would be equal:
P (x > 0.78) = [tex]P(z <\frac{x-p}{\sqrt{\frac{p*(1-p)}{n} }})[/tex]
Replacing we have:
[tex]P(z <\frac{0.78-0.75}{\sqrt{\frac{0.75*(1-0.75)}{110} }})[/tex]
P ( z < 0.73) = 1 - P ( z => 0.73)
= 1 - 0.7673
= 0.2327
Therefore the probability is 23.27%
