Math / Data & Statistics

QuestionApply Jefferson's method to the following sets of states with the given populations. Assume that 100 delegates are to be apportioned. State whether the quota criterion is satisfied.
A: 115; B: 675; C: 205 (modified divisor 9.78)
Determine the number of representatives for each state using Jefferson's method with a modified divisor of 9.78 . A: 11 B: 69 C. 20
State whether the quota criterion is satisfied. The quota criterion is $\square$ because the number of seats assigned to at least one of the states $\square$ the same as its $\square$ quota either rounded up or down to the nearest integer

Studdy Solution

STEP 1

What is this asking? We need to divide up 100 delegates between three states based on their populations using Jefferson's method and a given modified divisor, then check if the result follows the quota rule. Watch out! Jefferson's method can be tricky!
Remember, we're looking for a modified divisor that gives us the correct total number of delegates when we round the modified quotas down.

STEP 2

1. Calculate the modified quotas.
2. Check the sum of the apportionments.
3. Check the quota criterion.

STEP 3

Alright, let's dive in!
We're given a modified divisor of $9.78$ , so let's use that to calculate the modified quota for each state.
Remember, the modified quota is just the state's population divided by the modified divisor.

STEP 4

For state A, the population is $115$ , so the modified quota is $\frac{115}{9.78} \approx 11.7587$ .

STEP 5

For state B, the population is $675$ , so the modified quota is $\frac{675}{9.78} \approx 68.9980$ .

STEP 6

And for state C, the population is $205$ , so the modified quota is $\frac{205}{9.78} \approx 20.9612$ .

STEP 7

Now, with Jefferson's method, we round each modified quota down to get the apportionment.
So, state A gets $\lfloor 11.7587 \rfloor = 11$ delegates.
State B gets $\lfloor 68.9980 \rfloor = 68$ delegates.
And state C gets $\lfloor 20.9612 \rfloor = 20$ delegates.

STEP 8

Let's add those up: $11 + 68 + 20 = 99$ .
Oops, we're one short!
This isn't the correct apportionment using Jefferson's method with the given divisor.
The problem provided incorrect apportionments.
With the given divisor of $9.78$ , the sum should be $99$ , not $100$ .

STEP 9

The quota criterion states that each state's apportionment should be either its lower quota (standard quota rounded down) or its upper quota (standard quota rounded up).
Let's calculate the standard quotas first.
The total population is $115 + 675 + 205 = 1000$ .
With $100$ delegates, the standard divisor is $\frac{1000}{100} = 10$ .

STEP 10

State A's standard quota is $\frac{115}{10} = 11.5$ .
Its lower quota is $11$ and its upper quota is $12$ .

STEP 11

State B's standard quota is $\frac{675}{10} = 67.5$ .
Its lower quota is $67$ and its upper quota is $68$ .

STEP 12

State C's standard quota is $\frac{205}{10} = 20.5$ .
Its lower quota is $20$ and its upper quota is $21$ .

STEP 13

Using the modified divisor of $9.78$ , state A gets $11$ delegates (within the quota), state B gets $68$ delegates (within the quota), and state C gets $20$ delegates (within the quota).
So, even though the total number of delegates is incorrect, the quota criterion is satisfied for each state individually.

STEP 14

Using the modified divisor of $9.78$ , State A receives 11 delegates, State B receives 68 delegates, and State C receives 20 delegates, totaling 99 delegates.
The quota criterion is satisfied for each state individually, even though the total apportionment is incorrect.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the modified quotas.
2. Check the sum of the apportionments.
3. Check the quota criterion.

STEP 3

Alright, let's dive in!
We're given a modified divisor of $9.78$ , so let's use that to calculate the modified quota for each state.
Remember, the modified quota is just the state's population divided by the modified divisor.

STEP 4

For state A, the population is $115$ , so the modified quota is $\frac{115}{9.78} \approx 11.7587$ .

STEP 5

For state B, the population is $675$ , so the modified quota is $\frac{675}{9.78} \approx 68.9980$ .

STEP 6

And for state C, the population is $205$ , so the modified quota is $\frac{205}{9.78} \approx 20.9612$ .

STEP 7

Now, with Jefferson's method, we round each modified quota down to get the apportionment.
So, state A gets $\lfloor 11.7587 \rfloor = 11$ delegates.
State B gets $\lfloor 68.9980 \rfloor = 68$ delegates.
And state C gets $\lfloor 20.9612 \rfloor = 20$ delegates.

STEP 8

Let's add those up: $11 + 68 + 20 = 99$ .
Oops, we're one short!
This isn't the correct apportionment using Jefferson's method with the given divisor.
The problem provided incorrect apportionments.
With the given divisor of $9.78$ , the sum should be $99$ , not $100$ .

STEP 9

STEP 10

State A's standard quota is $\frac{115}{10} = 11.5$ .
Its lower quota is $11$ and its upper quota is $12$ .

STEP 11

State B's standard quota is $\frac{675}{10} = 67.5$ .
Its lower quota is $67$ and its upper quota is $68$ .

STEP 12

State C's standard quota is $\frac{205}{10} = 20.5$ .
Its lower quota is $20$ and its upper quota is $21$ .

STEP 13

STEP 14

Using the modified divisor of $9.78$ , State A receives 11 delegates, State B receives 68 delegates, and State C receives 20 delegates, totaling 99 delegates.
The quota criterion is satisfied for each state individually, even though the total apportionment is incorrect.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the modified quotas.2. Check the sum of the apportionments.3. Check the quota criterion.

STEP 3

Alright, let's dive in!We're given a modified divisor of 9.789.789.78, so let's use that to calculate the modified quota for each state.Remember, the modified quota is just the state's population divided by the modified divisor.

STEP 4

For state A, the population is 115115115, so the modified quota is 1159.78≈11.7587\frac{115}{9.78} \approx 11.75879.78115​≈11.7587.

STEP 5

For state B, the population is 675675675, so the modified quota is 6759.78≈68.9980\frac{675}{9.78} \approx 68.99809.78675​≈68.9980.

STEP 6

And for state C, the population is 205205205, so the modified quota is 2059.78≈20.9612\frac{205}{9.78} \approx 20.96129.78205​≈20.9612.

STEP 7

STEP 8

Let's add those up: 11+68+20=9911 + 68 + 20 = 9911+68+20=99.Oops, we're one short!This isn't the correct apportionment using Jefferson's method with the given divisor.The problem provided incorrect apportionments.With the given divisor of 9.789.789.78, the sum should be 999999, not 100100100.

STEP 9

STEP 10

State A's standard quota is 11510=11.5\frac{115}{10} = 11.510115​=11.5.Its lower quota is 111111 and its upper quota is 121212.

STEP 11

State B's standard quota is 67510=67.5\frac{675}{10} = 67.510675​=67.5.Its lower quota is 676767 and its upper quota is 686868.

STEP 12

State C's standard quota is 20510=20.5\frac{205}{10} = 20.510205​=20.5.Its lower quota is 202020 and its upper quota is 212121.

STEP 13

STEP 14

Using the modified divisor of 9.789.789.78, State A receives **11** delegates, State B receives **68** delegates, and State C receives **20** delegates, totaling **99** delegates.The quota criterion *is* satisfied for each state individually, even though the total apportionment is incorrect.

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1. Calculate the modified quotas.
2. Check the sum of the apportionments.
3. Check the quota criterion.

Alright, let's dive in!
We're given a modified divisor of $9.78$ , so let's use that to calculate the modified quota for each state.
Remember, the modified quota is just the state's population divided by the modified divisor.

For state A, the population is $115$ , so the modified quota is $\frac{115}{9.78} \approx 11.7587$ .

For state B, the population is $675$ , so the modified quota is $\frac{675}{9.78} \approx 68.9980$ .

And for state C, the population is $205$ , so the modified quota is $\frac{205}{9.78} \approx 20.9612$ .

Let's add those up: $11 + 68 + 20 = 99$ .
Oops, we're one short!
This isn't the correct apportionment using Jefferson's method with the given divisor.
The problem provided incorrect apportionments.
With the given divisor of $9.78$ , the sum should be $99$ , not $100$ .

State A's standard quota is $\frac{115}{10} = 11.5$ .
Its lower quota is $11$ and its upper quota is $12$ .

State B's standard quota is $\frac{675}{10} = 67.5$ .
Its lower quota is $67$ and its upper quota is $68$ .

State C's standard quota is $\frac{205}{10} = 20.5$ .
Its lower quota is $20$ and its upper quota is $21$ .

Using the modified divisor of $9.78$ , State A receives 11 delegates, State B receives 68 delegates, and State C receives 20 delegates, totaling 99 delegates.
The quota criterion is satisfied for each state individually, even though the total apportionment is incorrect.