Whcih describes how to calculate the range of this data set?

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

Cuales son los terminos que siguen en cada sucesion:a)2,4,8,16,32,64,....B)1,3,5,7,9,....C)1,3,6,10,15,21,28,36,45

Elanso [62]

Answer:

a) Número siguiente = Número anterior * 2

b) Número siguiente = Número anterior + 2

c) Número siguiente = Número anterior + (n + 1)

Step-by-step explanation:

a) Para la secuencia 2,4,8,16,32,64, ....

El siguiente número es dos veces el número anterior.

Por ejemplo, 4 = 2 * 2, 8 = 4 * 2, 16 = 8 * 2 etc.

b) Para la secuencia 1,3,5,7,9, ....

El siguiente número es igual al número anterior + 2

Por ejemplo, 3 = 1 + 2, 5 = 3 + 2, 7 = 5 + 2 etc.

c) Para la secuencia1,3, 6,10,15,21,28,36,45

La diferencia entre los números aumenta en +1.

3-1 = 2

6-3 = 3

10-6 = 4

15-10 = 5

21-15 = 6

28-21 = 7

3 0

3 years ago

Give a correct interpretation of a 99% confidence level of 0.328 < μ < 0.486.

stepladder [879]

Answer:

D

Step-by-step explanation:

Lower bound of the population mean is 0.328

Upper bound of the population mean is 0.486

Confidence level is 99%

Therefore, the correct interpolation of the confidence interval is that we are 99% confident that the true value of the population mean is between 0.328 and 0.486

5 0

4 years ago

A video rental store keeps a list of their top 15 movie rentals each week. This week the list includes 6 action, 4 comedies, 3 d

azamat

Answer:

Step-by-step explanation:

Probability is the likelihood or chance that an event will occur.

Probability = Expected outcome of event/Total outcome.

If a video rental store keeps a list of their top 15 movie rentals each week, the total outcome is 15.

If the list for the week includes 6 action, 4 comedies, 3 dramas, and 2 mysteries and the store manager removes a copy of each of the 15 movies from the shelf, then randomly selects 3 of the 15 to show on the display monitors in the store, the probability that she selected 2 comedies and 1 action movie will be calculated as shown;

Probability of selecting 2 comedies = 4/15*4/15 = 16/225 (Note that the expected outcome in this case is 4).

Probability of selecting 1 action movie = 6/15 = 2/5

<em>Hence, the probability that she selected 2 comedies and 1 action movie will be equivalent to 16/225*2/5 = 32/1125</em>

<em>Note that the rented movies will have to be returned hence reason for the replacement. </em>

8 0

3 years ago

Which expression uses the greatest common factor and distributive property to find 10+50?

Rudiy27

Ab+ac=a(b+c) where a is the greatest common factor
find greatest common factor of 10 and 50
10=1,2,5,10
50=1,2,5,10,25,50
greatest common is 10
a=10
10(1)+10(5)=10(1+5)=10(6)=60

3 0

3 years ago