Worked solution P43-Easy-1. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-2. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-3. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-4. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-5. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-6. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-7. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-8. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-9. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-10. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-11. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-12. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-13. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Easy-14. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-1. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-2. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-3. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-4. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-5. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-6. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-7. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-8. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-9. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-10. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-11. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-12. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-13. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Tough-14. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-1. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-2. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-3. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-4. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-5. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-6. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-7. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-8. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-9. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-10. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-11. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-12. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-13. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.

Worked solution P43-Toughest-14. The distribution is valid because all probabilities are nonnegative and sum to 1; $P(X\ge 2)=0.45$. $$\sum _{x}p(x)=0.20+0.35+0.30+0.15=1.00,P(X\ge 2)=0.30+0.15=0.45.$$ Interpretation: The random variable maps each outcome of the process to a numerical value; its distribution assigns long-run probability to each possible value. Validity: List the support completely and distinguish a probability mass function for discrete X from area under a density for continuous X. Error to reject: A continuous random variable has probability zero at any single exact value even though intervals can have positive probability.