Question

let ( p(x) ) = the probability that a student will study ballet ( x ) years. complete table 4.28 using the data provided. is the following table correct?

Explanation:

Step-by-Step Format:

Step1: Calculate \(x*P(x)\) for \(x = 1\)

\(1\times0.10=0.10\)

Step2: Calculate \(x*P(x)\) for \(x = 2\)

\(2\times0.05 = 0.10\)

Step3: Calculate \(x*P(x)\) for \(x = 3\)

\(3\times0.10=0.30\)

Step4: Calculate \(P(x)\) for \(x = 4\)

Since \(\sum_{x = 1}^{7}P(x)=1\), let \(P(4)=p\). Then \(0.10 + 0.05+0.10 + p+0.30 + 0.20+0.10 = 1\).
\(p=1-(0.10 + 0.05+0.10 + 0.30 + 0.20+0.10)=0.15\)

Step5: Calculate \(x*P(x)\) for \(x = 4\)

\(4\times0.15 = 0.60\)

Step6: Calculate \(x*P(x)\) for \(x = 5\)

\(5\times0.30=1.50\)

Step7: Calculate \(x*P(x)\) for \(x = 6\)

\(6\times0.20 = 1.20\)

Step8: Calculate \(x*P(x)\) for \(x = 7\)

\(7\times0.10=0.70\)

Answer:

For \(x = 1\), \(xP(x)=0.10\); for \(x = 2\), \(xP(x)=0.10\); for \(x = 3\), \(xP(x)=0.30\); \(P(4) = 0.15\), \(xP(x)=0.60\) (already shown in the second - table for \(x = 4\)); for \(x = 5\), \(xP(x)=1.50\); for \(x = 6\), \(xP(x)=1.20\); for \(x = 7\), \(xP(x)=0.70\). The second - table is incomplete (missing rows for \(x = 5\), \(x = 6\), \(x = 7\)), but for the rows \(x = 1\), \(x = 2\), \(x = 3\), \(x = 4\) that are present in both tables, the values of \(P(x)\) and \(xP(x)\) are correct.