Question 1

- Find the sample correlation coefficient, r, between x and y and use it to describe the relationship between the two variables.
- Find the mean of all the
*x*-values - Find the standard deviation of all the
*x*-values (call it*s*) and the standard deviation of all the_{x}*y*-values (call it*s*)._{y}

For example, to find *s _{x}*, you would use the following equation:

- For each of the
*n*pairs (*x*,*y*) in the data set, take - Add up the
*n*results from Step 3. - Divide the sum by
*s*∗_{x}*s*._{y} - Divide the result by
*n*– 1, where*n*is the number of (*x*,*y*) pairs. (It’s the same as multiplying by 1 over*n*– 1.)

This gives you the correlation, *r*

- Find the best fitting line

Equations of straight lines are in the form **y** = **mx** + **c** (m and **c** are numbers). m is the gradient of the line and **c** is the **y**-intercept (where the graph crosses the **y**-axis).

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Trying to plot a straight line graph from the x and y in the table c is 2.1 and m is 1.3 as shown below

- What is the estimated amount of weight-loss if the dosage is 3.5? 6.65 lbs

using the formula of y=1.3x+2.1, replace x with 3.5

- What is the estimated amount of weight-loss if the dosage is 15? Use same formula

y=1.3(15)+2.1

y=21.6

- A recent census of employees at a large automotive company showed that 30% of the assembly line workers were extremely bored with their job. The management of the company arranged for a consulting firm to come and offer a program to reduce on-the-job boredom. After two months of the program, a random sample of 20 workers was selected to help evaluate its impact.
- If the program was not effective, how many workers in the sample would you expect to be bored with their job

30/100 of 20 =6 workers

- If the program was not effective, what is the probability that exactly 8 workers in the sample would report being bored with their job?

8/20 X100=40%

- If the program was not effective, what is the probability that at least 2 workers in the sample would report being bored with their job?

2/20 X 100=10%

- If the program was not effective, what is the probability that more than 1 but less than 19 workers in the sample would report being bored with their job?

19/20 X100=95%

- The mayor of the city of Chicago was informed that household water usage was a normally distributed random variable with mean of 25 gallons and a standard deviation of 5 gallons per day.
- If a household were to be randomly selected, what would be the expected amount of water usage per day? 25
- What is the probability that a randomly chosen household uses more than 20 and less than 28 gallons per day?

More than 20

Less than 28

Subtract the mean (μ in Step 1) from the greater than value 28-25=3

Divide the standard deviation (σ in Step 1) by the square root of your sample 5/root of 1=5

Divide your result from *a* by your result from *b 3/5=0.6 then*

Use the formula from Step 3 to find the z-values. This time, use Xbar2

Subtract the mean (μ in Step 1) from the greater than value 20-25=-5

Divide the standard deviation (σ in Step 1) by the square root of your sample 5/root of 1=5

Divide your result from *a* by your result from *b -5/5=-1*

Look up the value you calculated in Step 3 in the z-table

Z value of 0.6 corresponds to .2257

Z value of -1 corresponds to .3413 Note that the bell curve is symmetrical, so if you want to look up a negative value like -1, then just look up the positive counterpart. The area will be the same.

Add Step 5 and 6 together .2257+.3413=.567

- What is the probability that a randomly chosen household uses more than 20 gallons per day?

We already did for 20 gallons which is .3413 add 0.5 for the other half =.8413

- What is the probability that a randomly chosen household uses less than 20 gallons per day?

20-25/(5root1)=-1 Look up the z-score in the left-hand z-table. It is .8413 *However, this is not the answer,* as the question is asking for LESS THAN. Subtract from full 1-0.8413=.1587