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Natalka [10]
3 years ago
9

Help me please like ASAP I need it now please

Mathematics
2 answers:
Len [333]3 years ago
8 0
80 cm becauwu uwueueueeiieidb. Zakinthos z Habq abs brush. Gujfbvhdjf did
lina2011 [118]3 years ago
5 0

Answer:

C. 600cm^{2}

<em>hope this helped :)</em>

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The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical
Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

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 = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

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

By the Central Limit Theorem

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

Z = \frac{36 - 38.4}{0.8538}

Z = -2.81

Z = -2.81 has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

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

Z = \frac{41 - 38.4}{0.8538}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08

So

a)

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

b)

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

Z = \frac{36 - 38.4}{1.08}

Z = -2.22

Z = -2.22 has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

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

Z = \frac{41 - 38.4}{1.08}

Z = 2.41

Z = 2.41 has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

4 0
3 years ago
Land once connecting asia and north america
Whitepunk [10]
The land bridge, It includes the Chukchi Sea, the Bering Sea, the Bering Strait, the Chukchi and Kamchatka Peninsulas in Russia as well as Alaska in the United States. The area includes land lying on the North American Plate and Siberian land east of the Chersky Range.

You could've just looked it up dude.
7 0
4 years ago
A bag contains 22 red marbles, 8 white marbles, and 3 gray marble. You randomly pick out a marble, record its color, and put it
frez [133]

Total marbles : 33

W + G marbles : 11

Probability of picking either a W or G : 11/33

Number of W or G marbles = 11/33 × 45 = 15

4 0
4 years ago
The area of a rectangle is 229.2 m2. If the length is 15 m, what is the perimeter of the rectangle
igor_vitrenko [27]

Answer:

\large\boxed{P=60.56\ m}

Step-by-step explanation:

The formula of an area of a rectangle:

A=lw

l - length

w - width

We have

A=229.2\ m^2\\\l=15\ m

Substitute:

15w=229.2          <em>divide both sides by 15</em>

w=15.28\ m

The formula of a perimeter of a rectangle:

P=2(l+w)

Substitute:

P=2(15+15.28)=2(30.28)=60.56\ m

8 0
4 years ago
If x=-4 what is the value of the following expression?
Gre4nikov [31]

Answer:

2x^2 +3x -2

18

12

-22

-46

Step-by-step explanation:

6 0
3 years ago
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