Find the mean, median and mode for the above set of data

a. Mean = 35.2 ≈ 35

b. Median = 35.6 ≈ 36

c. Mode = 36.6 ≈ 37

Step-by-step Explanation:

==>Given:

Class of ages in yrs

No. of cases of each class = f

Midpoint of each class = x

Product of midpoint and no. of cases of each class = fx

==>Required:

a. Mean

b. Median

c. Mode

==>SOLUTION:

a. Mean = (Σfx)/Σf

Σf = sum of no. of cases = 5+10+20+22+13+5 = 75

Σfx = 47.5+195+590+869+643.5+297.5 = 2,642.5

Mean = 2,642.5/75 = 35.2 ≈ 35

b. Median = Lm + [((Σf/2) - Cfb)/fm]Cw

Our median is between the 37th and the 38th term, which can be found in the class interval 35-44. This is our median class.

Lm = Lower class boundary of the median class = lower limit of the Medina class + upper limit of the class before the median class ÷ 2 = (35+34)/2 = 34.5

Σf/2 = 75/2 = 37.5

Cfb = Cumulative frequency of class before the median class = 5+10+20 = 35

fm = frequency of the Medina class = 22

Cw = Class width = upper class boundary - lower class boundary = 44.5-34.5 = 10

Median = 34.5 + [(37.5-35)/22] × 10

= 34.5 + [2.5/22] × 10

= 34.5 + [25/22]

= 34.5 + 1.1

= 35.6 ≈ 36

c. Mode = Lm + [∆¹/(∆¹+∆²)]Cw

Modal class = (35-44) [i.e. the class with the highest frequency, which is where our mode falls in]

Lm = lower class boundary of the modal class = lower limit of the modal class + upper limit of the class before the modal class ÷ 2 = (35+34)/2 = 34.5

∆¹ = difference between the frequency of the modal class & the frequency of the class before the modal class = 22 - 20 = 2

∆² = difference between the frequency of the modal class & the frequency of the class after the modal class = 22 - 13 = 9

Cw == Upper class boundary - Lower class boundary = 44.5 - 34.5 = 10

Mode = 34.5 + [2/(2+9)] × 10

= 34.5 + [2/11] × 10

= 34.5 + [20/11]

= 34.5 + 1.8

Mode = 36.6 ≈ 37

idk

explanation:

hi.

i'm not sure about the answer, but, i would say c. *when the recipient was not included on the original email.*

hope this , well if it's right. uwu

merry christmas! ^-^