Question

An article written for a magazine claims that

78
%
78%78, percent of the magazine's subscribers report eating healthily the previous day. Suppose we select a simple random sample of
675
675675 of the magazine's approximately
50
,
000
50,00050, comma, 000 subscribers to check the accuracy of this claim.
Assuming the article's
78
%
78%78, percent claim is correct, what is the approximate probability that less than
80
%
80%80, percent of the sample would report eating healthily the previous day?

Answer 1

0.0058 or 0.58%

Step-by-step explanation:

We can use a normal approximation to the binomial distribution to find the probability that less than 80% of the sample would report eating healthily the previous day.

First, we need to calculate the mean and standard deviation of the binomial distribution.

The mean is given by:

mean = (sample size) x (proportion of success) = 675 x 0.78 = 522.5

The standard deviation is given by:

standard deviation = sqrt((sample size) x (proportion of success) x (proportion of failure)) = sqrt(675 x 0.78 x 0.22) = 7.7

Then we can use the standard normal distribution (z-score) to find the probability of getting a sample proportion less than 0.8:

z = (80% - 78%) / (standard deviation / sqrt(675)) = (0.02 - 0.78) / (7.7/sqrt(675)) = -2.6

Using a standard normal table, we can find that the probability of getting a z-score less than -2.6 is about 0.0058 or 0.58%.

So the approximate probability of getting a sample proportion less than 80% is 0.0058 or 0.58%