Question

A recent study reported that 73% of Americans could only converse in one language. A random sample of 130 Americans was randomly selected. What is the probability that 100 or fewer of these Americans could only converse in one language?

Answer 1

Probability that 100 or fewer of these Americans could only converse in one language is 0.8599.

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
`\hat p` = sample proportion of Americans who could only converse in one language.

The z score probability distribution for sample proportion is given by;

Z =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where, p = population proportion of Americans who could only converse in one language = 73%

`\hat p` = sample proportion =
`(100)/(130)` = 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(
`\hat p`
`\leq` 0.77)

P(
`\hat p`
`\leq` 0.77) = P(
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }`
`\leq`
`\frac{0.77-0.73}{\sqrt{(0.77(1-0.77))/(130) } }` ) = P(Z
`\leq` 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.