Find the mean of the distribution above
Mean = \(\frac{(0 \times 1) + (1 \times 2) + (2 \times 2) + (3 \times 1) + (4 \times 9)}{1 + 2 + 2 + 1 + 9}\)
= \(\frac{(0 + 2 + 4 + 3 + 36)}{15}\)
= \(\frac{45}{15}\) =3
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