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.
<em>Let </em><em> = sample proportion of Americans who can order a meal in a foreign language</em>
The z-score probability distribution for sample proportion is given by;
Z = ~ N(0,1)
where, = 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( > 0.50)
P( > 0.06) = P(
>
) = P(Z > 0.85) = 1 - P(Z
0.85)
= 1 - 0.80234 = <u>0.19766</u>
<em>Now, in the z table the P(Z </em> <em>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.</em>
Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.