Question

A graduate student majoring in linguistics is interested in studying the number of students in her college who are bilingual. Of the 1,320 students at the college, 466 of them are bilingual. If the graduate student conducts a study and samples 50 students at the college, use a calculator to determine the probability that 17 or fewer of them are bilingual.

Answer 1

Answer:

48.41% probability that 17 or fewer of them are bilingual.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 50, p = \frac{466}{1320} = 0.3530$

So

$\mu = E(X) = np = 50*0.3530 = 17.65$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.3530*0.6470} = 3.38$

Probability that 17 or fewer of them are bilingual.

Using continuity correction, this is $P(X \leq 17 + 0.5) = P(X \leq 17.5)$, which is the pvalue of Z when X = 17.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17.5 - 17.65}{3.38}$

$Z = -0.04$

$Z = -0.04$ has a pvalue of 0.4841

48.41% probability that 17 or fewer of them are bilingual.