Question

Need to find A, B, and C!

Answer 1

Answer:

A. Mean = $41900

B. Median = $37000

C. Mode = $37000

Step-by-step explanation:

A. Mean

Here

n=40

Mean = Sum of values/n

$Mean = \frac{(3)(18000)+(3)(22000)+(3)(25000)+(5)(34000)+(17)(37000)+(2)(45000)+52000+(5)(80000)+140000}{40}\\=\frac{54000+66000+75000+170000+629000+90000+52000+400000+140000}{40} \\=\frac{1676000}{40}\\=41900$

Mean = $41900

B. Median:

As the number of salaries is even,

the median will be mean of middle two terms

$Median= \frac{1}{2}(\frac{n}{2}th\ term+ \frac{n+2}{2}th\ term)\\= \frac{1}{2}(\frac{40}{2}th\ term+ \frac{40+2}{2}th\ term)\\=\frac{1}{2} (20th + 21st)}$

The 20th and 21st term will be 37000

So their mean will be same

So,

Median = $37000

C. Mode

Mode is the value which occurs most of the time in data.

the occurrence of 37000 is highest in the given data.

So,

Mode = $37000 ..

Answer 2

Answer:

Mean: 4.44 add up every number and divide it by how many there is.

Median: 3 put from least to greatest and count till the middle.

Mode: 3 because it appears the most