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Solved: A Fish Story Ethan and Drew went on a 10-day
Chapter 9, Problem 21AYU(choose chapter or problem)
Ethan and Drew went on a 10-day fishing trip. The number of smallmouth bass caught and released by the two boys each day was as follows:
Ethan |
9 |
24 |
8 |
9 |
5 |
8 |
9 |
10 |
8 |
10 |
Drew |
15 |
2 |
3 |
18 |
20 |
1 |
17 |
2 |
19 |
3 |
(a) Find the population mean and the range for the number of smallmouth bass caught per day by each fisherman. Do these values indicate any differences between the two fishermen’s catches per day? Explain.
(b) Find the population standard deviation for the number of smallmouth bass caught per day by each fisherman. Do these values present a different story about the two fishermen’s catches per day? Which fisherman has the more consistent record? Explain.
(c) Discuss limitations of the range as a measure of dispersion.
Questions & Answers
QUESTION:
Ethan and Drew went on a 10-day fishing trip. The number of smallmouth bass caught and released by the two boys each day was as follows:
Ethan |
9 |
24 |
8 |
9 |
5 |
8 |
9 |
10 |
8 |
10 |
Drew |
15 |
2 |
3 |
18 |
20 |
1 |
17 |
2 |
19 |
3 |
(a) Find the population mean and the range for the number of smallmouth bass caught per day by each fisherman. Do these values indicate any differences between the two fishermen’s catches per day? Explain.
(b) Find the population standard deviation for the number of smallmouth bass caught per day by each fisherman. Do these values present a different story about the two fishermen’s catches per day? Which fisherman has the more consistent record? Explain.
(c) Discuss limitations of the range as a measure of dispersion.
ANSWER:
Step 1 of 4
(a) The formula for population mean:
\(\bar{x}=\frac{\sum_{i=1}^{n} x_{i}}{n}\) ……….(1)
Where,
The population mean of Ethan is obtained by substituting the values in (1).
\(\bar{X}_{E} =\frac{9+24+8+9+5+8+9+10+8+10}{10} \)
\(=\frac{100}{10} \)
\(=10\)
The population mean of Drew is obtained by substituting the values in (1).
\(\bar{X}_{D}=\frac{15+2+3+18+20+1+17+2+19+3}{10}\)
\(=\frac{100}{10} \)
\(=10\)
Hence, the population mean of the number of smallmouth bass caught per day by each fisherman is 10.