Discrete random variable practice problems

A table is a valid probability distribution for a discrete random variable only if two rules both hold: every probability is between 0 and 1 inclusive, and the probabilities across all possible values add to exactly 1.

Check Model A against the second rule: $0.45 + 0.30 = 0.75$, then $0.75 + 0.15 = 0.90$, then $0.90 + 0.05 = 0.95$. The total is $0.95$, not $1$, so Model A is not valid.

Check Model B against the first rule: $P(3) = -0.10$ is negative, and no probability can be below 0, so Model B is not valid.

Model B is worth a second look, because its total is fine: $0.50 + 0.35 = 0.85$, then $0.85 + 0.25 = 1.10$, then $1.10 + (-0.10) = 1.00$. Adding to 1 is not enough on its own when one entry sits outside the interval from 0 to 1.

Check Model C against both rules. Every value ($0.40$, $0.35$, $0.20$, $0.05$) lies between 0 and 1, and $0.40 + 0.35 = 0.75$, then $0.75 + 0.20 = 0.95$, then $0.95 + 0.05 = 1.00$.

Model C passes both rules, so it is the only valid distribution of the three.

For part (a), use the rule that the probabilities of a discrete distribution add to 1. Add the three probabilities you can read: $0.15 + 0.30 = 0.45$, then $0.45 + 0.20 = 0.65$.

Subtract that total from 1 to recover the missing entry: $P(1) = 1 - 0.65 = 0.35$.

Confirm the result is legal: $0.35$ lies between 0 and 1, so the completed table is a valid probability distribution.

For part (b), "at least two solos" means $X = 2$ or $X = 3$. Those two outcomes cannot both happen in one performance, so add their probabilities: $P(X \geq 2) = 0.30 + 0.20$.

Add: $0.30 + 0.20 = 0.50$.

Write the formula for the mean of a discrete random variable, which weights every value by its probability: $\mu_X = \sum x_i P(x_i)$, where $x_i$ is a possible value and $P(x_i)$ is its probability.

Check the table first: $0.35 + 0.30 = 0.65$, then $0.65 + 0.20 = 0.85$, then $0.85 + 0.10 = 0.95$, then $0.95 + 0.05 = 1.00$. The probabilities add to 1, so the distribution is valid.

Multiply each package count by its probability: $1(0.35) = 0.35$, $2(0.30) = 0.60$, $3(0.20) = 0.60$, $4(0.10) = 0.40$, and $5(0.05) = 0.25$.

Add the five products in order: $0.35 + 0.60 = 0.95$, then $0.95 + 0.60 = 1.55$, then $1.55 + 0.40 = 1.95$, then $1.95 + 0.25 = 2.20$.

The mean is $2.20$ packages. Notice it is not one of the listed values, which is normal for a mean.

Find the mean first, because the standard deviation formula needs it. Apply $\mu_X = \sum x_i P(x_i)$: $2(0.2) = 0.4$, $3(0.4) = 1.2$, $4(0.3) = 1.2$, and $5(0.1) = 0.5$.

Add those products: $0.4 + 1.2 = 1.6$, then $1.6 + 1.2 = 2.8$, then $2.8 + 0.5 = 3.3$. So $\mu_X = 3.3$ routes.

Set up the standard deviation, which is the square root of the probability-weighted squared distances from the mean: $\sigma_X = \sqrt{\sum (x_i - \mu_X)^2 P(x_i)}$.

Build the full table one column at a time, using $\mu_X = 3.3$:

Add the last column to get the variance: $0.338 + 0.036 = 0.374$, then $0.374 + 0.147 = 0.521$, then $0.521 + 0.289 = 0.810$. So $\sigma_X^2 = 0.81$.

Take the square root: $\sigma_X = \sqrt{0.81} = 0.9$ routes. A typical visit lands about $0.9$ routes away from the mean of $3.3$.

For part (a), confirm the table is valid: $0.15 + 0.40 = 0.55$, then $0.55 + 0.30 = 0.85$, then $0.85 + 0.15 = 1.00$.

Apply $\mu_X = \sum x_i P(x_i)$: $2(0.15) = 0.30$, $3(0.40) = 1.20$, $4(0.30) = 1.20$, and $5(0.15) = 0.75$.

Add the products: $0.30 + 1.20 = 1.50$, then $1.50 + 1.20 = 2.70$, then $2.70 + 0.75 = 3.45$. So $\mu_X = 3.45$ kittens.

For part (b), no single litter can hold 3.45 kittens, since $X$ only takes the values 2, 3, 4, and 5. The mean of a random variable is a long-run average, not a value any one outcome has to equal.

A correct interpretation names the variable, the context, and the long run: over many surrendered litters, this shelter averages about 3.45 kittens per litter.

For part (c), the expected total is the per-litter mean times the number of litters: $60 \times 3.45 = 207$ kittens.

Check both tables. Route A: $0.5 + 0.3 + 0.2 = 1.0$. Route B: $0.4 + 0.5 + 0.1 = 1.0$. Both are valid distributions.

Apply $\mu_X = \sum x_i P(x_i)$ to Route A: $22(0.5) = 11.0$, $26(0.3) = 7.8$, and $40(0.2) = 8.0$.

Add: $11.0 + 7.8 = 18.8$, then $18.8 + 8.0 = 26.8$ minutes, so $E(A) = 26.8$ minutes.

Apply the same formula to Route B: $24(0.4) = 9.6$, $27(0.5) = 13.5$, and $30(0.1) = 3.0$.

Add: $9.6 + 13.5 = 23.1$, then $23.1 + 3.0 = 26.1$ minutes, so $E(B) = 26.1$ minutes.

Compare: $26.1 < 26.8$, so Route B is faster on average by $26.8 - 26.1 = 0.7$ minutes per delivery. Route A has the single fastest outcome (22 minutes), but its 40-minute delay happens often enough to push its average above B's.

For part (b), multiply the per-delivery saving by the number of deliveries: $50 \times 0.7 = 35$ minutes. As a check, $50 \times 26.8 = 1340$ minutes and $50 \times 26.1 = 1305$ minutes, and $1340 - 1305 = 35$.

Count the equally likely outcomes first. Choosing 2 dryers out of 5 without regard to order gives $\binom{5}{2} = \frac{5 \times 4}{2} = 10$ possible pairs, and every pair is equally likely because the choice is random.

Label the working dryers $W_1$, $W_2$, $W_3$ and the broken ones $B_1$, $B_2$. For $X = 0$ both picks must be broken, and the only such pair is $\{B_1, B_2\}$, so that is 1 pair out of 10 and $P(0) = \frac{1}{10} = 0.1$.

For $X = 1$ the student picks one working and one broken dryer: 3 choices of working times 2 choices of broken gives $3 \times 2 = 6$ pairs, so $P(1) = \frac{6}{10} = 0.6$.

For $X = 2$ both picks work, and the pairs are $\{W_1, W_2\}$, $\{W_1, W_3\}$, and $\{W_2, W_3\}$, which is 3 pairs, so $P(2) = \frac{3}{10} = 0.3$. Check the total: $0.1 + 0.6 + 0.3 = 1.0$, and the counts check too, since $1 + 6 + 3 = 10$.

For part (b), apply $\mu_X = \sum x_i P(x_i)$: $0(0.1) = 0$, $1(0.6) = 0.6$, and $2(0.3) = 0.6$, so $\mu_X = 0 + 0.6 + 0.6 = 1.2$ working dryers.

Now use $\sigma_X = \sqrt{\sum (x_i - \mu_X)^2 P(x_i)}$. Squared deviations from $1.2$: $(0 - 1.2)^2 = 1.44$, $(1 - 1.2)^2 = 0.04$, and $(2 - 1.2)^2 = 0.64$.

Weight each by its probability: $1.44(0.1) = 0.144$, $0.04(0.6) = 0.024$, and $0.64(0.3) = 0.192$. Add for the variance: $0.144 + 0.024 = 0.168$, then $0.168 + 0.192 = 0.360$.

Take the square root: $\sigma_X = \sqrt{0.36} = 0.6$ working dryers.

For part (a), the probabilities must add to 1. Add the known entries: $0.30 + 0.20 = 0.50$, then $0.50 + 0.10 = 0.60$. So $P(2) = 1 - 0.60 = 0.40$.

For part (b), apply $\mu_X = \sum x_i P(x_i)$: $1(0.30) = 0.30$, $2(0.40) = 0.80$, $3(0.20) = 0.60$, and $4(0.10) = 0.40$.

Add: $0.30 + 0.80 = 1.10$, then $1.10 + 0.60 = 1.70$, then $1.70 + 0.40 = 2.10$. So $\mu_X = 2.1$ items per visitor.

For part (c), use $\sigma_X = \sqrt{\sum (x_i - \mu_X)^2 P(x_i)}$ with $\mu_X = 2.1$:

Add the last column for the variance: $0.363 + 0.004 = 0.367$, then $0.367 + 0.162 = 0.529$, then $0.529 + 0.361 = 0.890$. So $\sigma_X^2 = 0.89$.

Take the square root: $\sigma_X = \sqrt{0.89} \approx 0.943$ items.

For part (d), the expected number of items is the mean per visitor times the number of visitors: $40 \times 2.1 = 84$ items.

Convert items to time: $84 \times 20 = 1680$ volunteer minutes, and $\frac{1680}{60} = 28$ hours.