Find the outlier of the data set 3.1, 5.3, 4.3, 0.7, 5.6, 5.2, 5.8, 5.4

Mathematics

2 answers:

Blizzard [7]3 years ago

8 0

Answer:

0.7

Step-by-step explanation:

An outlier of the data set is the data point which is quiet different from the rest of the data points or we can say that they lie outside the the other points.

Since in the given data, the similar data points ranges from 3.1 to 5.8, all points are nearly in between in this range but 0.7 is the only point which does not lie in this range. Hence, 0.7 is the outlier of the given data set.

One can also find the outlier by arranging the given data set from lowest to greatest that is if we arrange 3.1, 5.3, 4.3, 0.7, 5.6, 5.2, 5.8, 5.4 from least value to the greatest value that is 0.7, 3.1, 4.3, 5.2, 5.3, 5.4, 5.6, 5.8, we see that there is a large change in the value of the median from 0.7 to 3.1 as compared to rest of the values.

Median of 0.7 and 3.1 is $\frac{0.7+3.1}{2}$ = $1.9$ , while median of rest of the data points are nearly same.

Thus, the outlier will be 0.7.

NNADVOKAT [17]3 years ago

6 0

An outlier is a value that is much higher or lower than the rest of the data.
The answer here is .7 because all other values range from 3.1-5.8.

You might be interested in

Solve for x in the equation x squared + 2 x + 1 = 17.

Papessa [141]

Answer:

$x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\$

Step-by-step explanation:

given equation

$x^2 +2x +1 = 17$

subtracting 17 from both sides

$x^2 +2x +1 = 17\\x^2 +2x +1 -17= 17-17\\x^2 +2x - 16 = 0\\$

the solution for quadratic equation

$ax^2 + bx + c = 0$ is given by

x = $x = -b + \sqrt{b^2 - 4ac} /2a \\\\and \ \\-b - \sqrt{b^2 - 4ac} /2a$

________________________________

in our problem

a = 1

b = 2

c = -16

$x =( -2 + \sqrt{2^2 - 4*1*-16}) /2*1 \\x =( -2 + \sqrt{4 + 64}) /2\\x =( -2 + \sqrt{68} )/2\\x = ( -2 + \sqrt{4*17} )/2\\x = ( -2 + 2\sqrt{17} )/2\\x = - 1 + \sqrt{17}\\and\\\\x = - 1 - \sqrt{17}\\$