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iragen [17]
3 years ago
12

An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos

e 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?
Mathematics
1 answer:
Schach [20]3 years ago
8 0

Answer:

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:

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:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

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 X = 0.8*675 = 540. So

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

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

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

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