# 1. The ages​ (in years) and weights​ (in pounds) of all wide receivers for a football team are listed. Find the coefficient of variation for each of the two data sets. Then compare the results. Weights:214 185 201 213 211 197 188 220Ages:25 28 23 26 29 26 26 27 2. Compare the results. What can you​ conclude?A. Ages are more variable than weights for all wide receivers on this team.B. Weights are more variable than ages for all wide receivers on this team.C. Ages and weights for all wide receivers on this team have about the same amount of variability.

2 months ago

## Solution 1

Guest #2391071
2 months ago

1)

Weights: =0.06321578

Ages: =0.06980003

2)  A) Ages are more variable than weights for all wide receivers on this team

Step-by-step explanation:

The coefficient of variation is calculated as follows

where s is the sample standard deviation and x bar is the sample mean,

where

∑(xi-x bar)^2/(n-1)

xi=i observation, n is the sample size

and,  x bar = ∑(xi) / n

## 📚 Related Questions

Question
1. Write a differential equation describing the following situation: The rate at which people become involved in a corporate bribing scheme is jointly proportional to the number of people already involved and the number of people who are not yet involved. Suppose there are a total of 6000 people in the company. Use k for the constant, P for the number of people who are involved in the scheme, and t for time.
Solution 1

The differential equation describing represent the given equation is

### Important information:

Total number of people in company N = 6000 and Total people involved in bribery P

### Equation:

Here Number of people not affected should be N - P

Now

The differential equation should be

Now we input the value of N

So,

Solution 2

dP/dt = k*(600 - P)*P

Step-by-step explanation:

Given:

- Total number of people in company N = 6000

- Total people involved in bribery P

The rate at which people become involved in a corporate bribing scheme is jointly proportional to the number of people already involved and the number of people who are not yet involved.

Find:

Write a differential equation

Solution:

- The following differential equation can be used to describe the situation above:

Number of people not affected = N - P

- Hence we can write a differential equation:

dP/ dt = k *( N - P )*P

- input the value of N:

dP/dt = k*(600 - P)*P

Question
In order to determine whether or not a driver's education course improves the scores on a driving exam, a sample of 6 students were given the exam before and after taking the course. The results are shown below. Let d = Score After - Score Before. Score Score Student Before the Course After the Course 1 83 87 2 89 88 3 93 91 4 77 77 5 86 93 6 79 83 a. Compute the test statistic. b. At 95% confidence using the p-value approach, test to see if taking the course actually increased scores on the driving exam.
Solution 1

a)

b)

The significance level would be

So the p value is higher than the significance level given 0.05, so then we can conclude that we FAIL to reject the null hypothesis that the mean after minus the mean before is lower or equal than 0.

So we don't have enoug evidence to conclude that the course actually increased scores on the driving exam

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.

Let put some notation :

x=test value before , y = test value after

x: 83 89 93 77 86 79

y: 87 88 91 77 93 83

Part a

The system of hypothesis for this case are:

Null hypothesis:

Alternative hypothesis:

The first step is calculate the difference and we obtain this:

d: 4, -1, -2, 0, 7, 4

The second step is calculate the mean difference

The third step would be calculate the standard deviation for the differences, and we got:

The 4 step is calculate the statistic given by :

Part b

The next step is calculate the degrees of freedom given by:

Now we can calculate the p value, since we have a left tailed test the p value is given by:

The significance level would be

So the p value is higher than the significance level given 0.05, so then we can conclude that we FAIL to reject the null hypothesis that the mean after minus the mean before is lower or equal than 0.

So we don't have enoug evidence to conclude that the course actually increased scores on the driving exam

Question
A set of final examination grades in a calculus course wasfound to be normally distributed with a mean of 69 and a standarddeviation of 9. a. what is the probality of getting a grade of 91or less on this exam? b. What percentage of students scored between 65 and89? c. What percentage of students scored between 81 and89? d. Only 5% of the students taking the test scored higherthan what grade?
Solution 1

a) P ( X < 91 ) = 0.9927

b) P ( 65 < X < 91 ) = 0.6585

c) P(81 < X < 89 ) =0.0781

d) X = 83.8

Step-by-step explanation:

Given:

- Mean of the distribution u = 69

- standard deviation sigma = 9

Find:

a. what is the probability of getting a grade of 91 or less on this exam?

b. What percentage of students scored between 65 and 89?

c. What percentage of students scored between 81 and 89?

d. Only 5% of the students taking the test scored higher than what grade?

Solution:

- We will declare a random variable X denoting the score that a student gets on a final exam. So,

X ~ N ( 69 , 9 )

- After defining our variable X follows a normal distribution. We can compute the probabilities as follows:

a) P ( X < 91 ) ?

- Compute the Z-score value as follows:

Z = (91 - 69) / 9 = 2.4444

- Now use the Z-score tables and look for z = 2.444:

P( X < 91 ) = P ( Z < 2.4444) = 0.9927

b) P ( 65 < X < 89 ) ?

- Compute the Z-score values as follows:

Z = (89 - 69) / 9 = 2.2.222

Z = (65 - 69) / 9 = -0.4444

- Now use the Z-score tables and look for z = 2.222 and Z = -0.4444:

P(65 < X < 89 ) = P ( -0.444< Z < 2.2222) = 0.6585

b) P ( 81 < X < 89 ) ?

- Compute the Z-score values as follows:

Z = (89 - 69) / 9 = 2.2.222

Z = (81 - 69) / 9 = 1.3333

- Now use the Z-score tables and look for z = 2.222 and Z = 1.333:

P(81 < X < 89 ) = P ( 1.333< Z < 2.2222) = 0.0781

c) P ( X > a ) = 0.05 , a?

- Compute the Z-score values as follows:

Z = (a - 69) / 9 = q

- Now use the Z-score tables and look for z value that corresponds to:

P( X > a ) = P ( Z > q ) = 0.05

- The corresponding Z-value is: q = 1.6444

Hence,

Z = (a - 69) / 9 = 1.644

a = 83.8

Question
Interpret each of the measures of central tendency. Interpret the mean. Choose the correct answer below. A. The mean is the average driving performance index value. B. The mean is the driving performance index value that occurs most often in the data set. C. The mean is the driving performance index value such that half of the values in the data set are higher than it.
Solution 1

A

Step-by-step explanation:

The answer is A The mean is the average driving performance index value, B is the definition of mode, C is the definition of the median or  percentile 50.

Solution 2

A. The mean is the average driving performance index value.

Step-by-step explanation:

Mean is a measure of central tendency, it is the average of a set of data. It can be derived by summing the data and dividing the sum by the number of data.

A. The mean is the average driving performance index value. - mean

B. The mean is the driving performance index value that occurs most often in the data set. - Mode

C. The mean is the driving performance index value such that half of the values in the data set are higher than it. - median

Therefore, the mean is the average driving performance index value(A)

Question
According to your model, as the population of San Francisco was revitalizing, for what value of the independent variable t did it reach 750 thousand
Solution 1

t =1.453 decades from 1980 i.e in year 1994.5

Step-by-step explanation:

Given:

- Population @ year 1980 P = 678.97 thousands.

- Population @ year 2000 P = 776.73 thousands.

- The rate of increase is linear - constant rate.

Find:

As the population of San Francisco was revitalizing, for what value of the independent variable t did it reach 750 thousand ?

Solution:

- Develop an expression of population P as a function of time t in decades elapsed from year 1980 on-wards

- The linear expression can take a form of :

P(t) = m*t + C

- Where, m is the rate of increase.

C is the initial population.

- Formulate m:

m = (776.73 - 678.97) / 2 = 48.88

- Formulate C:

C = P (@ 1980) = 678.97

- Evaluate P(t) = 750:

P(t) = 48.8*t + 678.97

750 = 48.8*t + 678.97

t = 71.03/48.8

Question
A research hypothesis proposes that consuming low carbohydrate diets results in increased weight loss. One group of participants follows a low-carb diet for 4 weeks, whereas a second group follows a high-carb diet containing the same number of calories for 4 weeks. The average number of pounds lost for each group is then is compared. What is the dependent variable?
Solution 1

The dependent variable is pounds lost for each group.

Step-by-step explanation:

We are given the following in the question:

A hypothesis is done to measure the average number of pounds lost based on the fact whether a person follows a low carb diet or high carb diet.

Dependent and independent variable:

• The dependent variable depends on the independent variable.
• A change in dependent variable is due to change in independent variable.
• Independent is the free variable in the research hypothesis.

For the hypothesis:

Independent variable:

Participants of a group follows a low-carb or high carb diet for 4 weeks.

Dependent Variable:

average number of pounds lost for each group

Thus, the dependent variable is pounds lost for each group.

Question
The accompanying data represent the miles per gallon of a random sample of cars with a​ three-cylinder, 1.0 liter engine. ​(a) Compute the​ z-score corresponding to the individual who obtained 37.837.8 miles per gallon. Interpret this result. ​(b) Determine the quartiles. ​(c) Compute and interpret the interquartile​ range, IQR. ​(d) Determine the lower and upper fences. Are there any​ outliers? LOADING... Click the icon to view the data. ​(a) Compute the​ z-score corresponding to the individual who obtained 37.837.8 miles per gallon. Interpret this result. The​ z-score corresponding to the individual is nothing and indicates that the data value is nothing standard​ deviation(s) ▼ below above the ▼ mean. median. ​(Type integers or decimals rounded to two decimal places as​ needed.)
Solution 1

The question is not complete because we were not given the data that represent the miles per gallon of a random sample of smart cars with a three cylinder, 1.0-liter engine.

However, we are going to assume some data below such that whatever data is latter provided, the solution to the question follows the sama procedure and pattern.

Sample Data = 31.5, 36.0, 37.8, 38.5, 40.1, 42.2, 34.2, 36.2, 38.1, 38.7, 40.6, 42.5, 34.7, 37.3, 38.2, 39.5, 41.4, 43.4, 35.6, 37.6, 38.4, 39.6, 41.7, 49.3

Now to solve for

(a) which is to compute the Z-score corresponding to the individual who obtained 37.837.8 miles per gallon.

Therefore, the value of z-score is calculated as follows,

μ = Σ lin (n) lin (i=1) X₁ / n → 933.1 /24 → 38.88

σ = √ Σ lin (n) lin (i=1) X₁² - nX⁻²/ n - 1  → 3.61

z = X - μ / σ  → 37.8 - 38.88/3.61 = −  0.299

Therefore, the value of z-score is -0.299.

(b) Considering the fact that the values are in ascending order is given as,

Recalling the values: 31.5, 36, 37.8 ,38.5, 40.1 ,42.2, 34.2 ,36.2, 38.1, 38.7, 40.6 ,42.5, 34.7, 37.3 ,38.2, 39.5 ,41.4, 43.4, 35.6, 37.6 ,38.4 ,39.6 ,41.7, 49.3

Q₁ = ( n+1/4)th value → (24+1/4)th value → 6.25th value  →  36.2 + 0.25 ∗ ( 37.3 −  36.2 )  =>  36.47

M = ( n+1/2)th value → (24+1/2)th value = (24+1/4)th value =>  12.5 t h  v a l u e  => 38.4 + 38.5/2 = 38.45

Q₃ = 3( n+1/4)th value  → 3(24+1/4)th value =>  18.75 t h  v a l u e  =>  40.6 + 0.75 ∗  ( 41.4  −  40.6 )  =>  41.2

(c) I n t e r q u a r t i l e r a n g e  =  Q ₃ −  Q ₁  =>  41.2  −  36.47  =>  4.25

Therefore, the IQR value is measure of spread of data, and here the value of IQR = 4.25.

(d) L o w e r f o u r t h  =  Q₁− 1.5 ( I Q R )  =>  36.75 −  1.5  ( 4.25 )  =>  30.375  U p p e r f o u r t h  =  Q ₃ +  1.5  ( I Q R )  =>  41  +  1.5  ( 4.25 )  =>  47.375

Hence, there is outliers since observations are not less than 30.375 and greater than 47.375.

The table for the entire problem is given as,

Z- Score     -0.299

Q₁                  36.47

Q₂                  38.45

Q₃                  41.2

IQR               4.25

Lower fourth   30.375

Upper fourth       47.375

Question
How to graph y= 3/2x
Solution 1

Step-by-step explanation:

There is no y intercept so start from the (0,0) and count 3 up and 2 to the left and just make a straight line with those points

Question
How do you Solve 4x-3-2x=-3x+4
Solution 1

Step-by-step explanation:

Collecting all the x terms together on the left side,

Simplifying the terms,

Moving all the x terms to one side of the equation,

The required solution of the equation is

Solution 2

6x-3=-3x+4

9x-3=4

9x=4

x=2.25

Step-by-step explanation:

Question
Of the last 100 customers entering a computer shop, 25 have purchased a computer. If the classical method for computing probability is used, the probability that the next customer will purchase a computer is .75 1.00 .25 .50 ___ The answer is apparently .50 but i am not understanding why that is so, wouldn't it be .25? Or is it that there is a option for the customer to buy or not buy so it makes it 50% so 50/100? I'm not sure...
Solution 1

0.50

Step-by-step explanation:

The question asks you to apply the classical method of computing probability. In this method, prior events do not interfere in the likelihood of an event happening in the future, instead it states that every possible outcome is equally likely to happen.

In this case there are only two possible outcomes: purchase or not purchase a computer. Therefore, the likelihood that the next customer will purchase a computer is 50% or 0.50.

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