The problem asks us to determine if there are any outliers in the given data set: $51, 56, 70, 74, 76, 76, 77, 80, 81, 83$ using the Interquartile Range (IQR) method.

1. Problem Description

The problem asks us to determine if there are any outliers in the given data set: $51, 56, 70, 74, 76, 76, 77, 80, 81, 83$ using the Interquartile Range (IQR) method.

2. Solution Steps

First, we need to find the first quartile (Q1), the third quartile (Q3), and then calculate the IQR.

The given data set is already sorted in ascending order: $51, 56, 70, 74, 76, 76, 77, 80, 81, 83$.

There are $n = 10$ data points.

Q1 is the median of the lower half of the data.

The lower half of the data is $51, 56, 70, 74, 76$.

Q1 is the median of this lower half, which is $70$.

Q3 is the median of the upper half of the data.

The upper half of the data is $77, 80, 81, 83$.

Q3 is the median of this upper half, which is $\frac{77 + 80}{2} = \frac{157}{2} = 78.5$.

The IQR is the difference between Q3 and Q1:

$IQR = Q3 - Q1 = 78.5 - 70 = 8.5$

Now, we need to calculate the lower and upper bounds for outliers.

Lower bound $= Q1 - 1.5 \times IQR = 70 - 1.5 \times 8.5 = 70 - 12.75 = 57.25$

Upper bound $= Q3 + 1.5 \times IQR = 78.5 + 1.5 \times 8.5 = 78.5 + 12.75 = 91.25$

Any data points below the lower bound or above the upper bound are considered outliers.

The data points are $51, 56, 70, 74, 76, 76, 77, 80, 81, 83$.

The minimum value is $51$ and the maximum value is $83$.

Since $51 < 57.25$, $51$ is not an outlier because it must be less than 57.25 to be an outlier.

Since $83 < 91.25$, $83$ is not an outlier.

Therefore, there are no outliers.

3. Final Answer

D. There are no outliers.