Question

Anasia is a basketball player who regularly shoots sets of

\[2\] free-throws. Suppose that each shot has probability
\[0.7\] of being made, and the results of the shots are independent.
The table below displays the probability distribution of
\[X\], the number of shots that Anasia makes in a set of
\[2\] attempts.
\[X= \# \text{ of makes}\]
\[0\]
\[1\]
\[2\]
\[P(X)\]
\[0.09\]
\[0.42\]
\[0.49\]
Given that
\[\mu_X=1.4\] makes, find the standard deviation of the number of shots that Anasia makes.
Round your answer to two decimal places.

Answer 1

To find the standard deviation of the number of shots Anasia makes, calculate the variance using the given mean (1.4) and probabilities for each outcome, then take the square root of the variance. The result is a standard deviation of approximately 0.51.

Step-by-step explanation:

The student is asking about the standard deviation of the number of shots made by Anasia, a basketball player, in sets of 2 free-throws.

To find the standard deviation, we can use the formula for the standard deviation of a probability distribution, which is √(σ^2), where σ^2 is the variance.

The variance is calculated by σ^2 = Σ[(x - μ_X)^2 ⋅ P(x)], where μ_X is the mean of the distribution and P(x) is the probability of each outcome x.

The variance of X can be found using the given probabilities and the values of X:

1. √(σ^2) = √(Σ[(x - μ_X)^2 ⋅ P(x)])
2. σ^2 = Σ[(x - 1.4)^2 ⋅ P(x)]
3. σ^2 = (0 - 1.4)^2 ⋅ 0.09 + (1 - 1.4)^2 ⋅ 0.42 + (2 - 1.4)^2 ⋅ 0.49
4. σ^2 = 0.196 ⋅ 0.09 + 0.16 ⋅ 0.42 + 0.36 ⋅ 0.49
5. σ^2 = 0.01764 + 0.0672 + 0.1764
6. σ^2 = 0.26124
7. Standard deviation, σ = √(0.26124) ≈ 0.51

So the standard deviation of the number of shots Anasia makes in a set of 2 attempts is approximately 0.51, rounded to two decimal places.