Jake plays basketball. Assume y= the number of basketball goals that Jake makes during a given season, and assume y can only take values from 0 to 6. The following table represents the probability distribution for the discrete random variable y.
y0123456 P(y) 0.09 0.11 0.24 0.25 ? 0.12 0.08
(a) Find the probability that Jake would score exactly 4 goals. (b) Find the probability that Jake would score at least 2 goals.
(c) Find the mean of the probability distribution.
(d) Find the standard deviation of the probability distribution.
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To find the missing value for y = 4 in the probability distribution table, we need to know that the sum of all probabilities in a probability distribution is equal to 1. So, we can find the missing value by subtracting the sum of all the given probabilities from 1.
The sum of the given probabilities is 0.09 + 0.11 + 0.24 + 0.25 + 0.12 + 0.08 = 0.89.
To find the missing probability for y = 4, we subtract the sum of the given probabilities (0.89) from 1:
1 - 0.89 = 0.11
Therefore, the missing probability for y = 4 is 0.11.
(a) The probability that Jake would score exactly 4 goals is 0.11.
(b) To find the probability that Jake would score at least 2 goals, we need to sum the probabilities for y = 2, 3, 4, 5, and 6.
P(y ≥ 2) = P(y = 2) + P(y = 3) + P(y = 4) + P(y = 5) + P(y = 6) = 0.11 + 0.24 + 0.25 + 0.12 + 0.08 = 0.8
Therefore, the probability that Jake would score at least 2 goals is 0.8.
(c) The mean of a probability distribution can be calculated by multiplying each value of y by its corresponding probability and summing them up.
Mean (μ) = Σ(y * P(y))
Mean = (0 * 0.09) + (1 * 0.11) + (2 * 0.24) + (3 * 0.25) + (4 * 0.11) + (5 * 0.12) + (6 * 0.08)
Mean = 0 + 0.11 + 0.48 + 0.75 + 0.44 + 0.6 + 0.48 = 2.96
Therefore, the mean of the probability distribution is 2.96.
(d) The standard deviation (σ) of a probability distribution can be calculated using the formula:
σ = √(Σ[(y - μ)^2 * P(y)])
Where μ is the mean of the distribution.
Standard Deviation = √((0 - 2.96)^2 * 0.09 + (1 - 2.96)^2 * 0.11 + (2 - 2.96)^2 * 0.24 + (3 - 2.96)^2 * 0.25 + (4 - 2.96)^2 * 0.11 + (5 - 2.96)^2 * 0.12 + (6 - 2.96)^2 * 0.08)
Standard Deviation = √((8.6016 * 0.09) + (3.7696 * 0.11) + (0.0904 * 0.24) + (0.0036 * 0.25) + (1.6304 * 0.11) + (5.8564 * 0.12) + (12.4416 * 0.08))
Standard Deviation ≈ √(0.77415 + 0.414656 + 0.021696 + 0.0009 + 0.179344 + 0.702768 + 0.995328)
Standard Deviation ≈ √3.08
Therefore, the standard deviation of the probability distribution is approximately 1.75.