Statistic approximation

SOLUTION: The number of cats below 2200 grams is:

$\text{\#Cats}_{<2200} = \text{\#Cats}_{1500 ≤ m < 2000} + \text{\#Cats}_{Partial 2000 ≤ m < 2500}$

In the assumption that "the distribution is unskewed between intervals", we can find the no of cats (inside this partial interval) that each space occupies and we will multiply this by the space of: $2000 - 2200 space = 200 space$ :

$\text{\#Cats}_{Partial 2000 ≤ m < 2500} = \frac{22 \text{ cats}}{500 \text{ space}} \times 200 \text{ space}$

$\text{\#Cats}_{Partial 2000 ≤ m < 2500} = 8.8 \text{ cats}$

We now can solve for $\text{\#Cats}_{<2200} = 8.8 \text{ cats} + 9 \text{ cats} = \boxed{17 \text{ cats}}$ (we rounded down)

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The same goes with the other but since we have that the largest weighting cat is $3745 \text{ grams}$ , we must change the interval from $3500 ≤ m < 4000$ to $3500 ≤ m < 3745$ (and these 2 intervals have the same cats in this case of ours).

$\text{\#Cats}_{>3600 gr} = \text{\#Cats}_{Partial 3500 ≤ m < 3745}$

In the assumption that "the distribution is unskewed between intervals", we can find the no of cats (inside the partial interval) that each space occupies and we will multiply this by the space of $3745 - 3600 space = 145 space$ :

$\text{\#Cats}_{Partial 3500 ≤ m < 3745} = \frac{12 \text{ cats}}{(3745-3500) \text{ space}} \times 145 \text{ space}$

$\text{\#Cats}_{Partial 3500 ≤ m < 3745} = \boxed{7 \text{ cats}}$ (we rounded down)

$17+7 = \boxed{24}$