18. Find the mean of the following frequency distribution Class-interval 0 - 50 50 - 100 100 - 150 150 - 200 200-250 250-300 Frequency 4 8 16 13 6 3

Step-by-step explanation:

Given

Class-interval ---- Frequency

0 - 50  ----------------- 4

50 - 100 ------------- 8

100 - 150  ------------ 16

150 - 200 ------------- 13

200-250 ------------- 6

250-300 ------------- 3

Required

Determine the Mean

We start by calculating the midpoint (M) of the class interval

Where:

Class-interval ---- ------Midpoint

0 - 50  --------------------0.5(0 + 50) = 25

50 - 100 ---------------- 0.5(50 + 100) = 75

100 - 150  ------------------ 0.5(100+150) = 125

150 - 200 -- ---------------- 0.5(150+200)=175

200-250 ------------------ 0.5(200+250) =225

250-300 ------------- 0.5 (250 + 300) = 275

Multiply the Midpoint by the corresponding frequency

Midpoint (M) ----- Frequency (F) ----- M* F

25 ------------------------ 4 ---------------------- 100

75 ------------------------- 8 --------------------- 600

125 ----------------------- 16 ---------------------2000

175 ----------------------- 13 --------------------- 2275

225 ----------------------- 6 --------------------- 1350

275 --------------------- 3 ------------------------- 825

Total -------------------50 ----------------------7150

Mean is then calculated as:

55? i think maybe

step-by-step explanation:

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40