# Using the data from the set included, what is the minimum number of 4's that must be added to make 6 an outlier?

Data Values:-3,-1,1,3,4,5,6,7,9
Frequency: 34,6,12,4,7+x,1,20,3,12
Cumulative Frequency: 34,40,52,56,63+x, 64+x,84+x,87+x,99+x

This is not for homework, it is for an exam review and I am really confused on this one concept.

## To determine the minimum number of 4's that must be added to make 6 an outlier, we need to understand the concept of outliers in a dataset.

An outlier is a value that is significantly different from other values in a dataset. In this case, 6 is considered an outlier if it is significantly lower or higher than the other values in the dataset.

To determine if 6 is an outlier, we can use a method called the "1.5 times the interquartile range" rule. This rule helps in identifying outliers beyond the upper and lower bounds.

The interquartile range (IQR) is the range between the first quartile (Q1) and the third quartile (Q3) of the dataset. It tells us the spread of the middle 50% of the data.

In your provided dataset, the cumulative frequency helps us to determine the quartiles. The cumulative frequency column with 63+x shows that 63 values are less than or equal to 4.

To calculate the quartiles, we can use the following steps:

Step 1: Find the median (Q2)
Since the dataset has an odd number of values, the median is the middle value. In this case, the median is 4.

Step 2: Find the first quartile (Q1)
The first quartile (Q1) is the median of the lower half of the dataset. Since the median is 4, we can see that Q1 is 3 (the value just before the median in the sorted list).

Step 3: Find the third quartile (Q3)
The third quartile (Q3) is the median of the upper half of the dataset. Looking at the cumulative frequency, we can see that the median is located between 63+x and 64+x. Therefore, 64+x is the third quartile (Q3).

Now that we have the quartiles, we can calculate the interquartile range (IQR):
IQR = Q3 - Q1 = (64+x) - 3 = 61+x.

According to the "1.5 times the interquartile range" rule, we compute the lower bound (Q1 - 1.5 * IQR) and the upper bound (Q3 + 1.5 * IQR).

Lower bound = 3 - 1.5 * (61 + x)
Upper bound = (64 + x) + 1.5 * (61 + x)

If 6 is considered an outlier, it should be beyond these bounds. In other words:

6 < Lower bound or 6 > Upper bound.

Let's first substitute 6 into the lower bound inequality:

6 < 3 - 1.5 * (61 + x)
3 > -1.5 * (61 + x)
2 > -1.5 * (61 + x) / 1.5
2 > -(61 + x)
-2 > 61 + x
-63 > x

Based on the above calculation, for 6 to be considered an outlier, the value of x must be greater than -63.

Next, let's substitute 6 into the upper bound inequality:

6 > (64 + x) + 1.5 * (61 + x)
6 > 64 + x + 1.5 * 61 + 1.5 * x
-118 > 2.5x + 97.5
-215.5 > 2.5x
-86.2 > x

Therefore, for 6 to be considered an outlier, the value of x must be less than -86.2.

Combining both conditions, we have:

-63 < x < -86.2.

So, the value of x should be less than -86.2 and greater than -63 for 6 to be considered an outlier.

Since you are looking for the minimum number of 4's that must be added to make 6 an outlier, you can choose the smallest whole number that satisfies the condition.

In this case, x = -87 satisfies the condition. Therefore, the minimum number of 4's that must be added to make 6 an outlier is 87.