Please help with this two thank you

Part 1

a) Greatest possible weight range of gorillas = 60 kg.

b) 20 gorillas weigh 80 kg or less.

c) Midpoint weight of the modal group = 105 kg.

d) The estimate of the mean gorilla weight = 99 kg.

Part 2

a) The greatest range of the lengths of snakes = 250 cm.

b) 40 snakes have lengths between 1.5 m and 2.5 m.

c) Midpoint length of the modal group = 175 cm.

d) The estimate of the mean gorilla length = 154 cm.

Step-by-step explanation:

Part 1 - The Gorilla part

a) Greatest possible weight range of gorillas = (Maximum weight of gorillas on the table) - (Minimum weight of gorillas on the table)

= 120 - 60 = 60 kg

b) How many gorillas weigh 80 kg or less

6 gorillas weigh between 60 < W ≤ 70

14 gorillas weigh between 70 < W ≤ 80

So, 6 + 14 = 20 gorillas weigh 80 kg or less.

c) Midpoint weight of the modal group

To find the modal class, we first use (n+1)/2 th

where N = number of variables = 160

Modal weight will be (160+1)/2 = 80.5 weight

The 80.5th weight is in the 100 to 110 class. This is how we know

6 + 14 + 22 + 34 = 76

Indicating that the 80.5th weight is in the next class (100 < W ≤ 110)

The midpoint weight of the modal class is then

(100+110)/2 = 105 kg

d) To calculate the mean weight, we use the midpoint theory where we replace all the groups with the midpoint weight of each weight class.

Midpoint weight is W, frequency is f

W | f

65 | 6

75 | 14

85 | 22

95 | 34

105 | 40

115 | 44

The mean is given as

Mean = (Σfx)/(Σf)

Σfx = (65×6) + (75×14) + (85×22) + (95×34) + (105×40) + (115×44) = 15800

Σf = 160

Mean = (15800/160) = 98.75 kg = 99 kg to the nearest whole number.

Part 2 - The Snake part

a) Greatest possible range of lengths of snakes = (Maximum length of snakes on the table) - (Minimum length of snakes on the table)

= 250 - 0 = 250 cm

b) How many snakes are between 1.5 m and 2.5 m in length?

1.5 m = 150 cm, 2.5 m = 250 cm

19 snakes have lengths between 150 < L ≤ 200

21 snakes have lengths between 200 < L ≤ 250

So, 19 + 21 = 40 snakes have lengths between 1.5 m and 2.5 m

c) Midpoint length of the modal group

To find the modal class, we first use (n+1)/2 th

where N = number of variables = 72

Modal length will be (72+1)/2 = 36.5th length

The 36.5th length is in the 150 to 200 class. This is how we know

4 + 11 + 17 = 32

Indicating that the 36.5th length is in the next class (150 < L ≤ 200)

The midpoint length of the modal class is then

(150+200)/2 = 175 cm

d) To calculate the mean length, we use the midpoint theory where we replace all the groups with the midpoint length of each length class.

Midpoint length is L, frequency is f

L | f

25 | 4

75 | 11

125 | 17

175 | 19

225 | 21

The mean is given as

Mean = (Σfx)/(Σf)

Σfx = (25×4) + (75×11) + (125×17) + (175×19) + (225×21) = 11100

Σf = 72

Mean = (11100/72) = 154.167 cm = 154 cm to the nearest whole number.

Hope this Helps!!!

y=mx+b is the final answer!   : )

what graph

step-by-step explanation: