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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

For cheerleading practice, Kiara must be able to lift her legs so that they are parallel to her outstrached arms. For each side

podryga [215]

If this is the dance position you are describing, then the arm and the leg of the dancer both make "right" angles (90º angles) to the floor.

5 0

3 years ago

What two values the following number 250 is between

polet [3.4K]

Answer:

B

Step-by-step explanation:

3 0

2 years ago

Leona purchased a $1,000 bond having a quoted price of 99.875. She had to pay a 5.5% brokerage fee (of the selling price). What

iogann1982 [59]

Answer:

None of these choices are correct.

Step-by-step explanation:

If a bond is quoted at 99.875, it means that it is sold at 99.875% of the face value;

Face value = 1000

Therefore, Price = 0.99875 * 1000

Price = $998.75

If Leona pays 5.5% of the selling price, it means that she is paying additional cost which will make the total cost more than the quoted price;

5.5% of 998.75 is;

0.055*998.75 = $54.93

The total cost = $998.75 +$54.93

= $1053.68

Therefore, none of the given choices is correct.

4 0

3 years ago

Read 2 more answers

The population of a town is increasing by 300 inhabitants each year. If its population at the beginning of 1990 was 21,152, what

babunello [35]

Answer:

Step-by-step explanation:

You take the starting year of 1990 and subtract it from 1999 to get the year span of 9,

You then take the amount per year the population goes up by which is 300, so 300 multiplied by 9 is 2,700

You add 2,700 to 21,152 to get B-23,852

8 0

3 years ago

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