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scoundrel [369]
3 years ago
7

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

e μ. (Round your answers to four decimal places.)
Required:
a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?
b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?
Mathematics
1 answer:
zmey [24]3 years ago
7 0

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

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

X = 205

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

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

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

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

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

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

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

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

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3 years ago
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A cereal box filling machine is designed to release an amount of 16 ounces of cereal into each box, and the machine’s manufactur
Paraphin [41]

Using the z-distribution, it is found that since the p-value is less than 0.05, there is evidence that the mean amount of cereal in each box is different from 16 ounces at 0.05 significance.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if the mean is not different to 16 ounces, that is:

H_0: \mu = 16

At the alternative hypothesis, it is tested if the mean is different, hence:

H_1: \mu \neq 16

<h3>What is the test statistic?</h3>

The test statistic is:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

In which:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • \sigma is the standard deviation of the population.
  • n is the sample size.

The parameters for this problem are:

\overline{x} = 15.75, \mu = 16, \sigma = 1.46, n = 150.

Hence:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{15.75 - 16}{\frac{1.46}{\sqrt{150}}}

z = -2.1

<h3>What is the p-value and the conclusion?</h3>

Using a z-distribution calculator, for a two-tailed test, as we are testing if the mean is different of a value, with z = -2.1, the p-value is of 0.0357.

Since the p-value is less than 0.05, there is evidence that the mean amount of cereal in each box is different from 16 ounces at 0.05 significance.

More can be learned about the z-distribution at brainly.com/question/16313918

#SPJ1

3 0
2 years ago
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Mrrafil [7]
I assume you need to solve for g?
12g + 6 = 78
- 6
12g = 72
÷ 12
g = 6
I hope this helps! Let me know if you want me to explain anything :)
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