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modeling real life the heart rates (in beats per minute) for a random s…

Question

modeling real life the heart rates (in beats per minute) for a random sample of 36 blue whales are shown in the table. blue whale heart rates 11 10 12 12 8 10 13 11 13 9 11 8 12 6 8 13 5 8 9 8 12 6 9 7 6 8 7 9 9 5 12 9 11 7 8 9 estimate the population mean \\( \mu \\). round your answer to the nearest hundredth. the population mean \\( \mu \\) is about \\( \square \\).

Explanation:

Step1: Sum all heart rates

First, we list all the heart rates and sum them up. Let's count each value:

5 appears 2 times: \(5\times2 = 10\)

6 appears 3 times: \(6\times3 = 18\)

7 appears 3 times: \(7\times3 = 21\)

8 appears 7 times: \(8\times7 = 56\)

9 appears 7 times: \(9\times7 = 63\)

10 appears 2 times: \(10\times2 = 20\)

11 appears 3 times: \(11\times3 = 33\)

12 appears 4 times: \(12\times4 = 48\)

13 appears 3 times: \(13\times3 = 39\)

Now sum all these products: \(10 + 18 + 21 + 56 + 63 + 20 + 33 + 48 + 39\)

Calculate step by step:

\(10+18 = 28\); \(28+21 = 49\); \(49+56 = 105\); \(105+63 = 168\); \(168+20 = 188\); \(188+33 = 221\); \(221+48 = 269\); \(269+39 = 308\)

Wait, wait, maybe I miscounted the number of each value. Let's count again by looking at the table:

First row: 11, 10, 12, 12, 8, 10 (values: 11,10,12,12,8,10)

Second row:13,11,13,9,11,8 (13,11,13,9,11,8)

Third row:12,6,8,13,5,8 (12,6,8,13,5,8)

Fourth row:9,8,12,6,9,7 (9,8,12,6,9,7)

Fifth row:6,8,7,9,9,5 (6,8,7,9,9,5)

Sixth row:12,9,11,7,8,9 (12,9,11,7,8,9)

Now let's list all 36 values:

11,10,12,12,8,10,

13,11,13,9,11,8,

12,6,8,13,5,8,

9,8,12,6,9,7,

6,8,7,9,9,5,

12,9,11,7,8,9

Now count each number:

5: positions (3rd row, 5th column), (5th row, 6th column) → 2 times.

6: (3rd row, 2nd column), (4th row, 4th column), (5th row, 1st column) → 3 times? Wait 3rd row: 12,6,8,13,5,8 → 6 (1), 4th row:9,8,12,6,9,7 → 6 (2), 5th row:6,8,7,9,9,5 → 6 (3). So 3 times.

7: (4th row, 6th column), (5th row, 3rd column), (6th row, 4th column) → 3 times? 4th row:9,8,12,6,9,7 →7 (1), 5th row:6,8,7,9,9,5 →7 (2), 6th row:12,9,11,7,8,9 →7 (3). So 3 times.

8: (1st row,5th), (2nd row,6th), (3rd row,3rd), (3rd row,6th), (4th row,2nd), (5th row,2nd), (6th row,5th) → let's count:

1st row:8 (1)

2nd row:8 (2)

3rd row:8 (3), 8 (4)

4th row:8 (5)

5th row:8 (6)

6th row:8 (7) → 7 times. Correct.

9: (2nd row,4th), (4th row,1st), (4th row,5th), (5th row,4th), (5th row,5th), (6th row,2nd), (6th row,6th) →

2nd row:9 (1)

4th row:9 (2), 9 (3)

5th row:9 (4), 9 (5)

6th row:9 (6), 9 (7) → 7 times. Correct.

10: (1st row,2nd), (1st row,6th) → 2 times.

11: (1st row,1st), (2nd row,2nd), (2nd row,5th), (6th row,3rd) →

1st row:11 (1)

2nd row:11 (2), 11 (3)

6th row:11 (4) → Wait, 1st row:11, 2nd row:11,11, 6th row:11 → that's 4 times? Wait 1st row:11 (1), 2nd row:13,11,13,9,11,8 → 11 (2), 11 (3), 6th row:12,9,11,7,8,9 → 11 (4). So 4 times? Wait my initial count was wrong. Oh no, I made a mistake earlier. Let's recount 11:

1st row:11 (1)

2nd row:11 (2), 11 (3)

6th row:11 (4) → 4 times.

12: (1st row,3rd), (1st row,4th), (3rd row,1st), (4th row,3rd), (6th row,1st) →

1st row:12 (1), 12 (2)

3rd row:12 (3)

4th row:12 (4)

6th row:12 (5) → Wait 1st row:12,12; 3rd row:12; 4th row:12; 6th row:12 → 5 times? Wait 1st row:12 (1), 12 (2); 3rd row:12 (3); 4th row:12 (4); 6th row:12 (5) → 5 times? Wait 1st row:11,10,12,12,8,10 → two 12s. 3rd row:12,6,8,13,5,8 → one 12. 4th row:9,8,12,6,9,7 → one 12. 6th row:12,9,11,7,8,9 → one 12. So total 2+1+1+1=5? Wait no, 1st row:2, 3rd:1, 4th:1, 6th:1 → 5. But earlier I thought 4. So my mistake.

13: (2nd row,1st), (2nd row,3rd), (3rd row,4th) →

2nd row:13 (1), 13 (2)

3rd row:13 (3) → 3 times. Correct.

Now let's recalculate the sum with correct counts:

5: 2 → \(5\times2 = 10\)

6: 3 → \(6\times3 = 18\)

7: 3 → \(7\times3 = 21\)

8: 7 → \(8\times7 = 56\)

9: 7 → \(9\times7 = 63\)

10: 2 → \(10\times2 = 20\)

11: 4 → \(11\times4 = 44\) (because 1st row:1, 2nd row:2, 6th row:1 → total 4)

12: 5 → \(12\times5 = 60\) (1st row:2, 3rd row:1,…

Answer:

\(9.19\)