According to a study done by Wakefield 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. What is the probability the proportion of Americans who can order a meal in a foreign language is greater than 0.5

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 = 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 =  ~ 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 = 0.19766

Now, in the z table the P(Z  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.

c: lkllmn.

hope it's you

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week