Suppose the following set of random numbers is being used to simulate the event of a basketball player making three free throws
2 answers:
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Answer:
(a) The probability distribution is shown in the attachment.
(b) The value of E (<em>Y</em>) is 7.85.
(c) The value of E (X) and E (X²) are 1.45 and 3.25 respectively.
(d) The value of P (Y ≤ 2) is 0.60.
(e) Verified that the value of E (Y) is 7.85.
Step-by-step explanation:
(a)
The random variable <em>Y</em> is defined as:
For <em>X</em> = {0, 1, 2, 3} the value of <em>Y</em> are:
The probability of <em>Y</em> for different values are as follows:
P (Y = 1) = P (X = 0) = 0.20
P (Y = 2) = P (X = 1) = 0.40
P (Y = 9) = P (X = 2) = 0.15
P (Y = 22) = P (X = 3) = 0.25
The probability distribution of <em>Y</em> is shown below.
(b)
The expected value of a random variable using the probability distribution table is:
Compute the expected value of <em>Y</em> as follows:
Thus, the value of E (<em>Y</em>) is 7.85.
(c)
Compute the expected value of <em>X</em> as follows:
Compute the expected value of <em>X</em>² as follows:
Thus, the value of E (X) and E (X²) are 1.45 and 3.25 respectively.
(d)
Compute the value of P (Y ≤ 2) as follows:
Thus, the value of P (Y ≤ 2) is 0.60.
(e)
The value of E (Y) is 7.85.
Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).
Hence verified.
