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To determine the standard deviation of the distribution for the number of free throws made out of 8 attempts, we proceed through the following steps:
1. Calculate the Mean (Expected Value) [tex]\( E(X) \)[/tex]:
The mean [tex]\(\mu\)[/tex] is given by the sum of each value of [tex]\(X\)[/tex] multiplied by its respective probability:
[tex]\[ \mu = \sum (x_i \cdot P(x_i)) \][/tex]
Where [tex]\( x_i \)[/tex] represents the number of free throws made and [tex]\( P(x_i) \)[/tex] represents the probability of making [tex]\( x_i \)[/tex] free throws.
Given:
[tex]\[ \{ X = 0, 1, 2, 3, 4, 5, 6, 7, 8 \} \][/tex]
and their respective probabilities:
[tex]\[ \{ P(X) = 0.002, 0.008, 0.04, 0.12, 0.23, 0.28, 0.21, 0.09, 0.02 \} \][/tex]
Calculate the mean:
[tex]\[ \mu = (0 \cdot 0.002) + (1 \cdot 0.008) + (2 \cdot 0.04) + (3 \cdot 0.12) + (4 \cdot 0.23) + (5 \cdot 0.28) + (6 \cdot 0.21) + (7 \cdot 0.09) + (8 \cdot 0.02) \approx 4.818 \][/tex]
2. Calculate the Variance [tex]\( \sigma^2 \)[/tex]:
The variance is given by:
[tex]\[ \sigma^2 = \sum ((x_i - \mu)^2 \cdot P(x_i)) \][/tex]
Using the mean [tex]\(\mu \approx 4.818\)[/tex], we compute each term [tex]\((x_i - \mu)^2\)[/tex], multiply it by the corresponding probability [tex]\(P(x_i)\)[/tex], and sum:
[tex]\[ \sigma^2 = ((0 - 4.818)^2 \cdot 0.002) + ((1 - 4.818)^2 \cdot 0.008) + ((2 - 4.818)^2 \cdot 0.04) + ... + ((8 - 4.818)^2 \cdot 0.02) \approx 1.965 \][/tex]
3. Calculate the Standard Deviation [tex]\( \sigma \)[/tex]:
The standard deviation is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} \approx \sqrt{1.965} \approx 1.401 \][/tex]
Given these computations:
- The mean of the distribution is approximately [tex]\(4.818\)[/tex],
- The variance of the distribution is approximately [tex]\(1.965\)[/tex],
- The standard deviation of the distribution is approximately [tex]\(1.401\)[/tex].
Therefore, the standard deviation of the distribution, closest to the options provided, is [tex]\(1.40\)[/tex]. Hence, the correct answer is:
[tex]\[ \boxed{1.40} \][/tex]
1. Calculate the Mean (Expected Value) [tex]\( E(X) \)[/tex]:
The mean [tex]\(\mu\)[/tex] is given by the sum of each value of [tex]\(X\)[/tex] multiplied by its respective probability:
[tex]\[ \mu = \sum (x_i \cdot P(x_i)) \][/tex]
Where [tex]\( x_i \)[/tex] represents the number of free throws made and [tex]\( P(x_i) \)[/tex] represents the probability of making [tex]\( x_i \)[/tex] free throws.
Given:
[tex]\[ \{ X = 0, 1, 2, 3, 4, 5, 6, 7, 8 \} \][/tex]
and their respective probabilities:
[tex]\[ \{ P(X) = 0.002, 0.008, 0.04, 0.12, 0.23, 0.28, 0.21, 0.09, 0.02 \} \][/tex]
Calculate the mean:
[tex]\[ \mu = (0 \cdot 0.002) + (1 \cdot 0.008) + (2 \cdot 0.04) + (3 \cdot 0.12) + (4 \cdot 0.23) + (5 \cdot 0.28) + (6 \cdot 0.21) + (7 \cdot 0.09) + (8 \cdot 0.02) \approx 4.818 \][/tex]
2. Calculate the Variance [tex]\( \sigma^2 \)[/tex]:
The variance is given by:
[tex]\[ \sigma^2 = \sum ((x_i - \mu)^2 \cdot P(x_i)) \][/tex]
Using the mean [tex]\(\mu \approx 4.818\)[/tex], we compute each term [tex]\((x_i - \mu)^2\)[/tex], multiply it by the corresponding probability [tex]\(P(x_i)\)[/tex], and sum:
[tex]\[ \sigma^2 = ((0 - 4.818)^2 \cdot 0.002) + ((1 - 4.818)^2 \cdot 0.008) + ((2 - 4.818)^2 \cdot 0.04) + ... + ((8 - 4.818)^2 \cdot 0.02) \approx 1.965 \][/tex]
3. Calculate the Standard Deviation [tex]\( \sigma \)[/tex]:
The standard deviation is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} \approx \sqrt{1.965} \approx 1.401 \][/tex]
Given these computations:
- The mean of the distribution is approximately [tex]\(4.818\)[/tex],
- The variance of the distribution is approximately [tex]\(1.965\)[/tex],
- The standard deviation of the distribution is approximately [tex]\(1.401\)[/tex].
Therefore, the standard deviation of the distribution, closest to the options provided, is [tex]\(1.40\)[/tex]. Hence, the correct answer is:
[tex]\[ \boxed{1.40} \][/tex]
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