Using statcruch: Create a set of about 11 points (use an odd number of points) t

Question

Using statcruch: Create a set of about 11 points (use an odd number of points) that are very close together that ehave a mean of about 4. Create the set of points by click in the white space above the number line. Record the mean, median, standard deviation and IQR of your distribution. Drag the farthest ost point on the left farther left. Don't drag it past the 1 or it will delete the point 1. The mean Select) 2. The median Select ] 3. The standard deviation ISelect ] 4. The IQR [Select] when the point was moved to the left. when the point was moved to the left. when the point was moved to the left. when the point was moved to the left Question 2 2 pts Click the Reset button. Create a set of about 11 points (use an odd number of points) that are very close together that have a mean of about 2. Create the set of points by click in the white space above the number line. Record the mean, median, standard deviation and IQR of your distribution. Drag the furthest most point on the right further right. Don't drag it past the 5 or it will delete the point 1. The mean ISelect ] 2. The median [Select) 3. The standard deviation ISelect] 4. The IQR I Select ] when the point was moved to the right when the point was moved to the right when the point was moved to the right. when the point was moved to the right. MacBook Pro

Explanation / Answer

(1)

Here we have to create a set of 11 points, such that when we arrange these points on the number scale they have a mean value close to 2.

Since we have 11 points, so the median will be the 6th point, and first quartile Q1 will be the 3rd point, and third quartile will be the 9th point.

Now, when the farthest left point is moved a bit further left, so the median will not change because the 6th point is left undistrubed. Also since Q1 and Q3 are also unchanged, so the Inter Quartile Range, IQR = Q3-Q1, will also be unchanged.

Only changes visible will be in the mean and standard deviation.

Since point is moved further left, its value is decreased.

Since mean depends on the sum of all values, so it will get decreased.

Since moving the point further extreme causes the data to be more spread out, so the standard deviation will increase.

So based on the above reasonings, we have the following answers:

The mean decreases when the point was moved to the left.

The median remains unchanged when the point was moved to the left.

The standard deviation increases when the point was moved to the left.

The IQR remains unchanged when the point was moved to the left.

(2)

In this case, we have the exact same scenario as above, except that the mean in this case is equal to 2, and the point which is farthest right is moved further right.

Using the exact same reasonings as above:

Since we have 11 points, so the median will be the 6th point, and first quartile Q1 will be the 3rd point, and third quartile will be the 9th point.

Now, when the farthest right point is moved a bit further right, so the median will not change because the 6th point is left undistrubed. Also since Q1 and Q3 are also unchanged, so the Inter Quartile Range, IQR = Q3-Q1, will also be unchanged.

Only changes visible will be in the mean and standard deviation.

Since point is moved further right, its value is increased.

Since mean depends on the sum of all values, so it will get increased.

Since moving the point further extreme causes the data to be more spread out, so the standard deviation will increase.

So based on the above reasonings, we have the following answers:

The mean increases when the point was moved to the right.

The median remains unchanged when the point was moved to the right.

The standard deviation increases when the point was moved to the right.

The IQR remains unchanged when the point was moved to the right.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.