Question

a ballet instructor is interested in knowing what percent of each years class will continue on to the next, so that she can plan what classes to offer. over the years, she has established the following probability distribution.
let ( x ) = the number of years a student will study ballet with the teacher
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:

Step1: Calculate the sum of probabilities

For a probability distribution, the sum of all \(P(x)\) values must equal \(1\).

$$
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$$

Since \(\sum P(x)=1\), then \(P(4)=1 - 0.85=0.15\)

Step2: Calculate \(x\cdot P(x)\) for each \(x\)

• When \(x = 1\), \(x\cdot P(x)=1\times0.10 = 0.10\)
• When \(x = 2\), \(x\cdot P(x)=2\times0.05 = 0.10\)
• When \(x = 3\), \(x\cdot P(x)=3\times0.10 = 0.30\)
• When \(x = 4\), \(x\cdot P(x)=4\times0.15 = 0.60\)
• When \(x = 5\), \(x\cdot P(x)=5\times0.30 = 1.50\)
• When \(x = 6\), \(x\cdot P(x)=6\times0.20 = 1.20\)
• When \(x = 7\), \(x\cdot P(x)=7\times0.10 = 0.70\)

Answer:

The completed table should have \(P(4) = 0.15\) and the \(x\cdot P(x)\) values as \(0.10\) (for \(x = 1\)), \(0.10\) (for \(x = 2\)), \(0.30\) (for \(x = 3\)), \(0.60\) (for \(x = 4\)), \(1.50\) (for \(x = 5\)), \(1.20\) (for \(x = 6\)), \(0.70\) (for \(x = 7\)). So the original table (with missing \(P(4)\) and \(x\cdot P(x)\) values) is incorrect.