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

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

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:

The standard deviation of the binomial distribution is:

Normal probability distribution

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

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

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 , .

In this problem, we have that:

So

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when . So

has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

answer: no

step-by-step explanation:

you are only given the cost not the amount of pounds.