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Digiron [165]
3 years ago
13

The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is

generated in the city. The study included 129 residents whose mean number of pounds of trash generated per person per week was 34.2 pounds and the standard deviation was 8.2 pounds.
a. The sampling distribution follows a ______ distribution.
b. With 95% confidence the population mean number of pounds per person per week is between_____ and_____ pounds.
c. If many groups of 120 randomly selected people in the city are studied, then a different confidence interval would be produced from each group. About_____ percent of these confidence intervals will contain the true population mean number of pounds of trash generated per person per week and about________ percent will not contain the true population mean number of pounds of trash generated per person per week.
Mathematics
1 answer:
saul85 [17]3 years ago
4 0

Solution :

Given :

Sample mean, $\overline X = 34.2$

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

   α = 0.05, α/2 = 0.025

  From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

  t =T.INV.2T(0.05, 128) = 2.34

  Taking the positive value of t, we get  

  Confidence interval is ,

  $\overline X \pm t \times \frac{s}{\sqrt n}$

 $34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

 (32.52, 35.8)

 95% confidence interval is  (32.52, 35.8)

So with $95 \%$ confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About $95 \%$ of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about $5 \%$ that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.  

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Answer:

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Step-by-step explanation:

<u>To find:</u>

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<u>Given:</u>

  • Use the formula of a rectangular prism.

<u>Note:</u>

  • \Longrightarrow: \sf{V=L*W*H}

<u>Solutions:</u>

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<u>Multiply the numbers from left to right.</u>

\Longrightarrow:\sf{6*4*8=\boxed{\sf{192}}

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Suppose the heights of 18-year-old men are approximately normally distributed, with mean 69 inches and standard deviation 2 inch
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Answer:

a) 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

b) 95.44% probability that the mean height x is between 68 and 70 inches.

Step-by-step explanation:

To solve this question, it is important to know the normal probability distribution and the Central Limit Theorem.

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:

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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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 69, \sigma = 2

(a) What is the probability that an 18-year-old man selected at random is between 68 and 70 inches tall?

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68. So

X = 70

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

Z = \frac{70 - 69}{2}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

X = 68

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

Z = \frac{68 - 69}{2}

Z = -0.5

Z = -0.5 has a pvalue of 0.3085.

So there is a 0.6915 - 0.3085 = 0.383 = 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

(b) If a random sample of sixteen 18-year-old men is selected, what is the probability that the mean height x is between 68 and 70 inches?

Now we use the Central Limit Theorem, with n = 16, s = \frac{2}{\sqrt{16}} = 0.5

The probability is also the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68, but with s as the standard deviation. So

X = 70

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

Z = \frac{70 - 69}{0.5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 68

Z = \frac{X - \mu}{0.5}

Z = \frac{68 - 69}{0.5}

Z = -2

Z = -2 has a pvalue of 0.0228.

So there is a 0.9772 - 0.0228 = 0.9544 = 95.44% probability that the mean height x is between 68 and 70 inches.

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