Respuesta :
Answer:
Probability that 100 or fewer of these Americans could only converse in one language is 0.8599.
Step-by-step explanation:
We are given that a recent study reported that 73% of Americans could only converse in one language.
A random sample of 130 Americans was randomly selected.
Let [tex]\hat p[/tex] = sample proportion of Americans who could only converse in one language.
The z score probability distribution for sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, p = population proportion of Americans who could only converse in one language = 73%
[tex]\hat p[/tex] = sample proportion = [tex]\frac{100}{130}[/tex] = 0.77
n = sample of Americans = 130
Now, probability that 100 or fewer of these Americans could only converse in one language is given by = P( [tex]\hat p[/tex] [tex]\leq[/tex] 0.77)
P( [tex]\hat p[/tex] [tex]\leq[/tex] 0.77) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.77-0.73}{\sqrt{\frac{0.77(1-0.77)}{130} } }[/tex] ) = P(Z [tex]\leq[/tex] 1.08) = 0.8599
The above probability is calculated by looking at the value of x = 1.08 in the z table which has an area of 0.8599.
