Mathematics, 15.04.2020 22:16 Esme1995

In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weighs more than 180 pounds? Assume that we want 96% confidence that the error is no more than 2 percentage points.

A) 923
B) 1267
C) 2628
D) 2001

Answers: 2

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Answer from: lexipooh7894

B) 1267

Step-by-step explanation:

Percentage of people above 180 pounds.

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 152, \sigma = 26$

This percentage is 1 subtracted by the pvalue of Z when X = 26. So

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

$Z = \frac{180 - 152}{26}$

Z = 1.08

Z = 1.08 has a pvalue of 0.8599

1 - 0.8599 = 0.1401

14.01% of people above 180 pounds.

Confidence interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this question, we have that:

$\pi = 0.1401$

96% confidence level

So $\alpha = 0.04$ , z is the value of Z that has a pvalue of $1 - \frac{0.04}{2} = 0.98$ , so Z = 2.054 .

We need

A sample size of n.

n is found when M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.054\sqrt{\frac{0.1401*0.8599}{n}}$

$0.02\sqrt{n} = 2.054\sqrt{0.1401*0.8599}$

$\sqrt{n} = \frac{2.054\sqrt{0.1401*0.8599}}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.054\sqrt{0.1401*0.8599}}{0.02}}^{2}$

$n \cong 1267$

So the correct answer is:

B) 1267

Answer from: Quest

step-by-step explanation:

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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