The average mark of candidates in an aptitude test was 128.5 with a standard deviation of 8.2. Three scores extracted from the test are 148, 102, 15zs2. What is the average of the extracted scores that are extreme values (outliers)

Solution:

The average marks of candidates in an aptitude test was 128.5 with a standard deviation of 8.2. Three scores extracted from the test are; 148, 102 and 152.

To find : average of the extracted scores that are the extreme values (outliers)

The average marks of candidates in an aptitude test was 128.5

=> Mean = 128.5

Standard Deviation = 8.2

outlier will be out sides   then range    Mean  ± 3* SD

128.5 ± 8.2 × 3

= 128.5 ± 24.6

Range is   ( 103.9  ,  153.1)

102 is outside this range

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form

tip $2.90

total $17.40