A small class has 9 students, 3 of whom are girls and 6 of whom are boys. The teacher is going to choose two of the students at
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Answer:
(a) <u>Sampling distribution </u>
P(25) = 0,04
P(35) = 0.1 + 0.1 = 0,2
P(42,5) = 0.06 + 0.06 = 0,12
P(45) = 0,25
P(52,5) = 0.15 + 0.15 = 0,3
P(60) = 0,09
(b) E(X) = 45.5 oz
(c) E(X) = μ
Step-by-step explanation:
The variable we want to compute is
For this we need to know all the possible combinations of X1 and X2 and the probability associated with them.
(a) <u>Sampling distribution </u>
Calculating all the 9 combinations (3 repeated, so we end up with 6 unique combinations):
P(25) = P(X1=25) * P(X2=25) = p25*p25 = 0.2 * 0.2 = 0,04
P(35) = p25*p45+p45*p25 = 0.2*0.5 + 0.5*0.2 = 0.1 + 0.1 = 0,2
P(42,5) = p25*p60 + p60*p25 = 0.2*0.3 + 0.3*0.2 = 0.06 + 0.06 = 0,12
P(45) = p45*p45 = 0.5 * 0.5 = 0,25
P(52,5) = p45*p60 + p60*p45 = 0.5*0.3 + 0.3*0.5 = 0.15 + 0.15 = 0,3
P(60) = p60*p60 = 0.3*0.3 = 0,09
(b) Using the sample distribution, E(X) can be expressed as:
The value of E(X) is 45.5 oz.
(c) The value of μ can be calculated as
We can conclude that E(X)=μ
We could have arrived to this conclusion by applying