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Bingel [31]
2 years ago
9

Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and ob

serving the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met
Mathematics
1 answer:
klio [65]2 years ago
6 0

Answer:

Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.

A.  The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.

C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.

E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake

The answer is E.

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A quality control manager at an auto plant measures the paint thickness on newly painted cars. A certain part that they paint ha
Scorpion4ik [409]

Answer:

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

Step-by-step explanation:

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

Normal probability distribution

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.

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 problem, we have that:

\mu = 2, \sigma = 0.8, n = 100, s = \frac{0.8}{\sqrt{100}} = 0.08

What is the probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value?

This is the pvalue of Z when X = 2 + 0.1 = 2.1 subtracted by the pvalue of Z when X = 2 - 0.1 = 1.9. So

X = 2.1

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

By the Central Limit Theorem

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

Z = \frac{2.1 - 2}{0.08}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

X = 1.9

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

Z = \frac{1.9 - 2}{0.08}

Z = -1.25

Z = -1.25 has a pvalue of 0.1056

0.8944 - 0.1056 = 0.7888

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

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(5x^3 + 4)(25x^6 - 20x3 +16)  remember factoring the cubes?
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Answer:

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Answer:

The average rate of change is 1.275

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The average rate of change of f(x) from x=a to x=b is given by:

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The money Terry invested is modeled by the function f(x)=0.01(2)^x where x represents number of days.

The average rate of change from day 2 to day 10 is given by:

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The summation indicates the sum from n = 1 to n = 3 of the expression 2(n+5).

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For n = 3, it equals 16

So the sum from n=1 to n=3 will be 12 + 14 + 16 = 42

Sum of an Arithmetic Series can also be written as:

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