Mathematics, 13.03.2020 01:29 dondre54

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?

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Answer from: TTHOR1935

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

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

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

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

Answer from: Quest

step-by-step explanation:

f(x)=2x^2 -1

translation is a function transformation that transform the function in a way of translating. by translating the function its graph is moved up an down. moving the function f(x) 2 units to the left will be the new function g(x) with the equation: g(x)=f(x+2)=2(x+2)^2 -1

Answer from: Quest

18.75 gallons

step-by-step explanation:

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