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

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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step-by-step explanation:

Answer from: Quest

starting with:

5 + x - 12 = 2x - 7

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)

5 - 12 = x - 7 (combine and cancel x's from both sides)

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)