Question

A survey reports that 67% of college students prefer to drink more coffee during the exams week. If we randomly select 80 college students and ask each whether they drink more coffee during exams week. Using Normal approximation to binomial distribution, with correction for continuity, what is the probability that at most 50 say that they drink coffee during exam week

Answer 1

Answer:

The probability that at most 50 say that they drink coffee during exam week is 0.166.

Step-by-step explanation:

The random variable X can be defined as the number of college students who prefer to drink more coffee during the exams week.

The probability of the random variable X is p = 0.67.

A random sample of n = 80 college students are selected.

The response of every students is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 80 and p = 0.67.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10
2. n(1 - p) ≥ 10

Check the conditions as follows:

$np=80\times 0.67=53.6>10\\n(1-p)=80\times (1-0.67)=26.4>10$

Thus, a Normal approximation to binomial can be applied.

$X\sim N(np, np(1-p))$

The mean of the distribution of X is:

$\mu=np=80\times 0.67=53.6$

The standard deviation of the distribution of X is:

$\sigma=\sqrt{np(1-p)}=\sqrt{80\times 0.67\times (1-0.67)}=4.206$

A Normal distribution is a continuous distribution. So, the probability at a point cannot be computed for the Normal distribution. To compute the probability at a point we need to apply the continuity correction.

Compute the probability that at most 50 say that they drink coffee during exam week as follows:

Apply continuity correction:

$P(X\leq 50)=P(X$

                 $=P(X$

*Use a z-table for the probability.

Thus, the probability that at most 50 say that they drink coffee during exam week is 0.166.