Respuesta :
Answer:
Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.
Step-by-step explanation:
We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.
Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.
Let [tex]\hat p[/tex] = sample proportion of Americans who can order a meal in a foreign 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, [tex]\hat p[/tex] = sample proportion
p = population proportion of Americans who can order a meal in a foreign language = 0.47
n = sample of Americans = 200
Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( [tex]\hat p[/tex] > 0.50)
P( [tex]\hat p[/tex] > 0.06) = P( [tex]\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] > [tex]\frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } }[/tex] ) = P(Z > 0.85) = 1 - P(Z [tex]\leq[/tex] 0.85)
= 1 - 0.80234 = 0.19766
Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.
Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.
