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navik [9.2K]
4 years ago
13

Assume that trees are subjected to different levels of carbon dioxide atmosphere with 7% of the trees in a minimal growth condit

ion at 370 parts per million (ppm), 10% at 440 ppm (slow growth), 49% at 550 ppm (moderate growth), and 34% at 670 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees
Mathematics
1 answer:
taurus [48]4 years ago
3 0

Answer: The mean and standard deviation are 567.2 and 89.88 resp.

Step-by-step explanation:

Since we have given that

For 370 parts per million = 7% = 0.07

For 440 parts per million = 10% = 0.10

For 550 parts per million = 49% = 0.49

For 670 parts per million = 34% = 0.34

So, Mean of the carbon dioxide atmosphere for these trees would be

E[x]=370\times 0.07+440\times 0.1+550\times 0.49+670\times 0.34=567.2

And

E[x^2]=370^2\times 0.07+440^2\times 0.1+550^2\times 0.49+670^2\times 0.34=329794

So, Variance would be

Var\ x=E[x^2]-E[x]^2=329794-567.2^2=8078.16

So, the standard deviation would be

\sigma=\sqrt{8078.16}=89.88

Hence, the mean and standard deviation are 567.2 and 89.88 resp.

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A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m
love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is c = 20.6153.

Step-by-step explanation:

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}}.

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.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that \mu = 20.4, \sigma = 1.23

Consider the sample mean weight of 100 watermelons of this variety. This means that n = 100.

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when X = 20.27.

Since we are working with the sample mean, we use s instead of \sigma in the Z score formula

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

Z = \frac{20.27 - 20.4}{0.123}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of X = c that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use Z = 1.75

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

1.75 = \frac{c - 20.4}{0.123}

c - 20.4 = 0.123*1.75

c = 20.6153

This value is c = 20.6153.

6 0
3 years ago
State which of the following sets of ordered pairs represent a function.
anastassius [24]

Answer:

d. Set A and Set B

Step-by-step explanation:

We have been given three sets:

Set A: (5, 2), (4, 3), (3, 4), (2, 5)

Set B: (-1, -6), (0, 2), (1, 2), (3, 6)

Set C: (2, 1), (4, 2), (2, 3), (8, 4)

Now we need to state about which of the following sets of ordered pairs represent a function.

We know that a function can't have repeated values in domain that is x-value can't repeat.

we see that set C has repeated x-value "2".

Then set C is not a function.

Hence correct choice is:

d. Set A and Set B

8 0
3 years ago
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Arisa [49]

Answer:

38 degrees

Step-by-step explanation:

By the Law of Cosines,

5² = 3² + 7² - 2(3)(7) cos A

5² - 3² - 7² = -2(3)(7) cos A

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If sec 50 = cosec 70, prove that 0 = 7.5°​
LiRa [457]

Answer:

The proofing is given below:

Step-by-step explanation:

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Hence, it is proved

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3 years ago
Select all the expressions that are equivalent to 4x + 8
RoseWind [281]

Answer:

2(2x+4)  4(x+2)  2(2x)+2(4)

Step-by-step explanation:

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