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Sagot :
To find the standard deviation of the heights of a group of professional basketball players as summarized in the given frequency distribution, we will follow these steps:
1. Determine the midpoints of each height range:
- For the height range [tex]\(70-71\)[/tex], the midpoint is [tex]\(\frac{70 + 71}{2} = 70.5\)[/tex].
- For the height range [tex]\(72-73\)[/tex], the midpoint is [tex]\(\frac{72 + 73}{2} = 72.5\)[/tex].
- For the height range [tex]\(74-75\)[/tex], the midpoint is [tex]\(\frac{74 + 75}{2} = 74.5\)[/tex].
- For the height range [tex]\(76-77\)[/tex], the midpoint is [tex]\(\frac{76 + 77}{2} = 76.5\)[/tex].
- For the height range [tex]\(78-79\)[/tex], the midpoint is [tex]\(\frac{78 + 79}{2} = 78.5\)[/tex].
- For the height range [tex]\(80-81\)[/tex], the midpoint is [tex]\(\frac{80 + 81}{2} = 80.5\)[/tex].
- For the height range [tex]\(82-83\)[/tex], the midpoint is [tex]\(\frac{82 + 83}{2} = 82.5\)[/tex].
So, the midpoints are: [tex]\([70.5, 72.5, 74.5, 76.5, 78.5, 80.5, 82.5]\)[/tex].
2. Calculate the mean height:
- Frequency (f): [tex]\([3, 7, 16, 12, 10, 4, 1]\)[/tex]
- Midpoints (m): [tex]\([70.5, 72.5, 74.5, 76.5, 78.5, 80.5, 82.5]\)[/tex]
- Total number of players: [tex]\(3 + 7 + 16 + 12 + 10 + 4 + 1 = 53\)[/tex]
The mean height ([tex]\(\mu\)[/tex]) is calculated as:
[tex]\[ \mu = \frac{\sum (f_i \cdot m_i)}{\sum f_i} = \frac{3 \cdot 70.5 + 7 \cdot 72.5 + 16 \cdot 74.5 + 12 \cdot 76.5 + 10 \cdot 78.5 + 4 \cdot 80.5 + 1 \cdot 82.5}{53} \approx 75.82 \][/tex]
3. Calculate the variance:
- Variance ([tex]\(\sigma^2\)[/tex]) is calculated as:
[tex]\[ \sigma^2 = \frac{\sum f_i \cdot (m_i - \mu)^2}{\sum f_i} \][/tex]
Substituting the values:
[tex]\[ \sigma^2 = \frac{3 \cdot (70.5 - 75.82)^2 + 7 \cdot (72.5 - 75.82)^2 + 16 \cdot (74.5 - 75.82)^2 + 12 \cdot (76.5 - 75.82)^2 + 10 \cdot (78.5 - 75.82)^2 + 4 \cdot (80.5 - 75.82)^2 + 1 \cdot (82.5 - 75.82)^2}{53} \approx 7.54 \][/tex]
4. Calculate the standard deviation:
- The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} = \sqrt{7.54} \approx 2.75 \][/tex]
Thus, the standard deviation of the basketball players' heights is approximately [tex]\(2.75\)[/tex]. In the context of the options given:
- 3.2
- 2.8
- 3.0
- 3.3
The closest option to [tex]\(2.75\)[/tex] is [tex]\(2.8\)[/tex]. Therefore, the correct answer is:
[tex]\[ 2.8 \][/tex]
1. Determine the midpoints of each height range:
- For the height range [tex]\(70-71\)[/tex], the midpoint is [tex]\(\frac{70 + 71}{2} = 70.5\)[/tex].
- For the height range [tex]\(72-73\)[/tex], the midpoint is [tex]\(\frac{72 + 73}{2} = 72.5\)[/tex].
- For the height range [tex]\(74-75\)[/tex], the midpoint is [tex]\(\frac{74 + 75}{2} = 74.5\)[/tex].
- For the height range [tex]\(76-77\)[/tex], the midpoint is [tex]\(\frac{76 + 77}{2} = 76.5\)[/tex].
- For the height range [tex]\(78-79\)[/tex], the midpoint is [tex]\(\frac{78 + 79}{2} = 78.5\)[/tex].
- For the height range [tex]\(80-81\)[/tex], the midpoint is [tex]\(\frac{80 + 81}{2} = 80.5\)[/tex].
- For the height range [tex]\(82-83\)[/tex], the midpoint is [tex]\(\frac{82 + 83}{2} = 82.5\)[/tex].
So, the midpoints are: [tex]\([70.5, 72.5, 74.5, 76.5, 78.5, 80.5, 82.5]\)[/tex].
2. Calculate the mean height:
- Frequency (f): [tex]\([3, 7, 16, 12, 10, 4, 1]\)[/tex]
- Midpoints (m): [tex]\([70.5, 72.5, 74.5, 76.5, 78.5, 80.5, 82.5]\)[/tex]
- Total number of players: [tex]\(3 + 7 + 16 + 12 + 10 + 4 + 1 = 53\)[/tex]
The mean height ([tex]\(\mu\)[/tex]) is calculated as:
[tex]\[ \mu = \frac{\sum (f_i \cdot m_i)}{\sum f_i} = \frac{3 \cdot 70.5 + 7 \cdot 72.5 + 16 \cdot 74.5 + 12 \cdot 76.5 + 10 \cdot 78.5 + 4 \cdot 80.5 + 1 \cdot 82.5}{53} \approx 75.82 \][/tex]
3. Calculate the variance:
- Variance ([tex]\(\sigma^2\)[/tex]) is calculated as:
[tex]\[ \sigma^2 = \frac{\sum f_i \cdot (m_i - \mu)^2}{\sum f_i} \][/tex]
Substituting the values:
[tex]\[ \sigma^2 = \frac{3 \cdot (70.5 - 75.82)^2 + 7 \cdot (72.5 - 75.82)^2 + 16 \cdot (74.5 - 75.82)^2 + 12 \cdot (76.5 - 75.82)^2 + 10 \cdot (78.5 - 75.82)^2 + 4 \cdot (80.5 - 75.82)^2 + 1 \cdot (82.5 - 75.82)^2}{53} \approx 7.54 \][/tex]
4. Calculate the standard deviation:
- The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} = \sqrt{7.54} \approx 2.75 \][/tex]
Thus, the standard deviation of the basketball players' heights is approximately [tex]\(2.75\)[/tex]. In the context of the options given:
- 3.2
- 2.8
- 3.0
- 3.3
The closest option to [tex]\(2.75\)[/tex] is [tex]\(2.8\)[/tex]. Therefore, the correct answer is:
[tex]\[ 2.8 \][/tex]
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