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LUCKY_DIMON [66]

2 years ago

7

Use the linear regression feature on a graphing calculator to find the correlation coefficient for the data. Round your answer t

o three decimal places.

1 answer:

Lerok [7]2 years ago

7 0

Answer:

0.9769

Step-by-step explanation:

Given the data :

X : 7.5, 9.9, 4.8, 9, 8.1, 1.7, 3.1

Y : 35, 6, 45, 18, 20, 89, 84

The Correlation Coefficient obtained by using a graphing calculator to model the data above is : 0.9769, this means a strong positive correlation exists between the two variables.

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A researcher drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 209 times. The observed freque

ddd [48]

Answer:

Step-by-step explanation:

Given that the observed frequencies for the outcomes as follows:

To check this we can use chi square goodness of fit test.

$H_0: equally likely\\H_a: not equally likely$

(Two tailed test at 5% significance level)

Assuming equally likely expected observations are found out and then chi square is calculated as (0-E)^2/E

Df = 6-1 =5

Outcome Frequency Expected frequency (Obs-exp)^2/Exp

1 36 34.83333333 0.03907496

2 30 34.83333333 0.670653907

3 41 34.83333333 1.091706539

4 40 34.83333333 0.766347687

5 23 34.83333333 4.019936204

6 39 34.83333333 0.498405104

209 209 7.086124402

p value =0.214

Since p >alpha, we accept null hypothesis

It appears that the loaded die does not behave differently than a fair die at 5% level of significance

5 0

3 years ago

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

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

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

2 years ago

PLEASE HELP ME!!!!!!!!

iren [92.7K]

SAS

there is an included angle on each triangle so if you look carefully there are two sides in which are given

5 0

2 years ago

Write the number 63 4 ways

taurus [48]

Sixty-three, 60 + 3, 9 x 7, 63

6 0

3 years ago

Read 2 more answers

Mr. Malone is contemplating which chauffeured car service to take to the airport. The first costs $28 upfront and $2 per mile. T

Ksju [112]

Answer:

The distance is 8 miles.

Step-by-step explanation:

First, we need to write an equation for each of chauffeured car services.

The first one = 28 + 2x = ?

The second one = 13 + 5x = ?

Next, we need to figure out what x is. Usually I just plug in random numbers, but I'm sure there's a better way to do it.

I first plugged in 2 as x.

28 + 4 = 32

13 + 10 = 23

That didn't work, so I then plugged in 5.

28 + 10 = 38

13 + 25 = 38

So 5 would be the correct answer for what is the distance.

8 0

2 years ago