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The given probability distribution table is:

Mathematics

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The given probability distribution table is:

xx325ttkk
P(X=x)P(X=x)325\frac{3}{25}120\frac{1}{20}14\frac{1}{4}kk

It is also given that E(X)=10150E(X) = \frac{101}{50}.

a) The value of the constants tt and kk

Step 1: Use the property that the sum of all probabilities in a distribution must equal 1.

∑P(X=x)=1\sum P(X=x) = 1
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