Answer

Part (a)

\[ E[X] = \sum_{\text{all } x} x P[X=x] = 2 \cdot 0.1 + 4 \cdot 0.4 + 6 \cdot 0.3 + 8 \cdot 0.2 = 5.2 \]

Part (b)

\[ Var(X) = E[X-\mu]^{2} \] \[ = \sum_{\text{all } x} (x-5.2)^{2} P[X=x] \] \[ = (2-5.2)^{2} \cdot 0.1 + (4-5.2)^{2} \cdot 0.4 + (6-5.2)^{2} \cdot 0.3 + (8-5.2)^{2} \cdot 0.2 \] \[ = (-3.2)^{2} \cdot 0.1 + (-1.2)^{2} \cdot 0.4 + (0.8)^{2} \cdot 0.3 + (2.8)^{2} \cdot 0.2 \] \[ = 1.024 + 0.576 + 0.192 + 1.568 = 3.36 \]

Answer

Part (a)

\[ E[X] = \sum_{\text{all } x} x P[X=x] = 1 \cdot 0.1 + 2 \cdot 0.2 + 3 \cdot 0.4 + 4 \cdot 0.2 + 5 \cdot 0.1 = 3 \]

Part (b)

\[ Var(X) = E[X-\mu]^{2} \] \[ = \sum_{\text{all } x} (x-3)^{2} P[X=x] \] \[ = (1-3)^{2} \cdot 0.1 + (2-3)^{2} \cdot 0.2 + (3-3)^{2} \cdot 0.4 + (4-3)^{2} \cdot 0.2 + (5-3)^{2} \cdot 0.1 \] \[ = (-2)^{2} \cdot 0.1 + (-1)^{2} \cdot 0.2 + (0)^{2} \cdot 0.4 + (1)^{2} \cdot 0.2 + (2)^{2} \cdot 0.1 \] \[ = 0.4 + 0.2 + 0 + 0.2 + 0.4 = 1.2 \]

Answer

Part (a)

\[ E[X] = \sum_{\text{all } x} x P[X=x] = 10 \cdot 0.2 + 20 \cdot 0.3 + 30 \cdot 0.4 + 40 \cdot 0.1 = 24 \]

Part (b)

\[ Var(X) = E[X-\mu]^{2} \] \[ = \sum_{\text{all } x} (x-24)^{2} P[X=x] \] \[ = (10-24)^{2} \cdot 0.2 + (20-24)^{2} \cdot 0.3 + (30-24)^{2} \cdot 0.4 + (40-24)^{2} \cdot 0.1 \] \[ = (-14)^{2} \cdot 0.2 + (-4)^{2} \cdot 0.3 + (6)^{2} \cdot 0.4 + (16)^{2} \cdot 0.1 \] \[ = 39.2 + 4.8 + 14.4 + 25.6 = 84 \]

Answer

Part (a)

Let the random variable that measures profits be $X$ and so

Outcome	$P(X=x)$	$x$
Get contract	0.3	50000
Lose	0.4	0
Share	0.3	20000

Part (b)

The mean is

\[ EX = \sum_{\text{all } x} x P(X = x) = 0.3 \cdot 50000 + 0.3 \cdot 20000 = 15000 + 6000 = \$21000. \]

Answer

Part (a)

If the purchasing decisions of the customers are independent, then we have a Binomial $(5, 0.7)$ distribution for the number of customers who buy the extended warranty. Since none buying the warranty is an outcome of $s = 0$, we want $P[S = 0 \mid n = 5, \pi = 0.7]$. This is equal to: \[ (1 - 0.7)^{5} = 0.3^{5} = 0.00243 \] So it will happen $0.243\%$ of the time.

Part (b)

It is highly unlikely that the dealership's claim is correct (well below a 1 in 100 chance, or even a 5% significance threshold). Such an outcome is extremely rare if the true success probability were actually 70%, casting serious doubt on their claim.

Answer

Part (a)

The function $p(x) = \frac{1}{4}(x+1)$ is linear and strictly positive on the interval $0 < x < 2$, starting at $\frac{1}{4}$ and ending at $\frac{3}{4}$.

Checking that the total area under the density curve integrates to one:

\[ \int p(x)dx = \int_{0}^{2} \frac{1}{4}(x + 1)dx \] \[ = \frac{1}{4} \left\{ \int_{0}^{2} xdx + \int_{0}^{2} dx \right\} \] \[ = \frac{1}{4} \left[ \frac{x^2}{2} + x \right]_{0}^{2} \] \[ = \frac{1}{4} \left\{ \left(\frac{4}{2} + 2\right) - 0 \right\} = \frac{1}{4} \{ 2 + 2 \} = \frac{4}{4} = 1. \]

Part (b)

For the mean, we evaluate the expected value integral:

\[ \int xp(x)dx = \int_{0}^{2} x \cdot \frac{1}{4}(x + 1)dx = \int_{0}^{2} \frac{1}{4}(x^2 + x)dx \] \[ = \frac{1}{4} \left\{ \int_{0}^{2} x^2dx + \int_{0}^{2} xdx \right\} \] \[ = \frac{1}{4} \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_{0}^{2} \] \[ = \frac{1}{4} \left\{ \frac{8}{3} + \frac{4}{2} \right\} = \frac{1}{4} \left\{ \frac{8}{3} + 2 \right\} \] \[ = \frac{1}{4} \cdot \frac{14}{3} = \frac{14}{12} = \frac{7}{6}. \]

Answer

We have wages \( X \sim N(18.50, 1.80^2) \)

Part (a)

We need to work out here the wage such that the probability of being above that wage is 10%, i.e., solve for \( P[X > x] = 10\% \). We have that:

From our tables we can see that this means that \( (x - 18.50)/1.80 \) should equal 1.2816 (1.28 would have been fine). We solve for:

\( x = 18.50 + 1.80 \times 1.2816 = 20.81 \) dollars, i.e., about $20.81.

Part (b)

Very few earn this little.

Answer

Part (a)

We can work this out two ways:

(1) \(E[Y] = E[3X^2] = \sum_{\text{all } x} 3x^2 P[X = x] = 3(-1)^2 \times 0.1 + 3(0)^2 \times 0.8 + 3(1)^2 \times 0.1 = 0.3 + 0 + 0.3 = 0.6\).

(2) We could have alternatively noticed that the distribution of \(Y\) is determined by mapping values of \(x \to y\):

• For \(x = -1\) and \(x = 1\), \(y = 3(1) = 3\). Thus, \(P(Y = 3) = 0.1 + 0.1 = 0.2\).
• For \(x = 0\), \(y = 0\). Thus, \(P(Y = 0) = 0.8\).

\(y\)	\(P(Y = y)\)
0	0.8
3	0.2

so \(E[Y] = \sum_{\text{all } y} y P[Y = y] = 0 \times 0.8 + 3 \times 0.2 = 0.6\).

Part (b)

Using the distribution table of \(Y\) alongside our mean \(E[Y] = 0.6\) yields:

Both computation paths lead to the same identical variance calculations.

Answer

(a) What is the mean of \(Y\)?

We can work this out two ways:

(1) \(E[Y] = E[(X - 2)^2] = \sum_{\text{all } x} (x - 2)^2 P[X = x] = (1-2)^2 \times 0.5 + (2-2)^2 \times 0.3 + (3-2)^2 \times 0.2 = 1 \times 0.5 + 0 \times 0.3 + 1 \times 0.2 = 0.5 + 0 + 0.2 = 0.7\).

(2) Alternatively, mapping out the state space for the distribution table of \(Y\):

• For \(x = 1 \implies y = (-1)^2 = 1\) and for \(x = 3 \implies y = (1)^2 = 1\). Combining these gives \(P(Y = 1) = 0.5 + 0.2 = 0.7\).
• For \(x = 2 \implies y = (0)^2 = 0\). This gives \(P(Y = 0) = 0.3\).

\(y\)	\(P(Y = y)\)
0	0.3
1	0.7

so \(E[Y] = \sum_{\text{all } y} y P[Y = y] = 0 \times 0.3 + 1 \times 0.7 = 0.7\).

(b) What is the variance of \(Y\)?

Finding the variance of \(Y\) using its direct distribution:

Answer

(a) This is $P(14700 < X < 15300)$. We can compute this by centering and standardizing and looking up the result in the tables. We have

$$P(14700 < X < 15300) = P(14700 - 15000 < X - 15000 < 15300 - 15000)$$ $$= P\left(\frac{14700 - 15000}{300} < \frac{X - 15000}{300} < \frac{15300 - 15000}{300}\right)$$ $$= P(-1 < Z < 1)$$ $$= 1 - 2 \times P(Z > 1)$$ $$= 1 - 2 \times 0.159$$ $$= 0.682$$

(b) This is asking us to work out $P(X > n) = 0.15$. Again by centering and standardizing this is

$$P(X > n) = P(X - 15000 > n - 15000)$$ $$= P\left(\frac{X - 15000}{300} > \frac{n - 15000}{300}\right)$$ $$= P\left(Z > \frac{n - 15000}{300}\right)$$ $$= 0.15$$

From the tables this means that

$$\frac{n - 15000}{300} = 1.04$$

so $n = 15000 + 300 \times 1.04 = 15312$ hours.

Answer

(a) This is $P(190 < X < 210)$. We can compute this by centering and standardizing and looking up the result in the tables. We have

$$P(190 < X < 210) = P(190 - 200 < X - 200 < 210 - 200)$$ $$= P\left(\frac{190 - 200}{5} < \frac{X - 200}{5} < \frac{210 - 200}{5}\right)$$ $$= P(-2 < Z < 2)$$ $$= 1 - 2 \times P(Z > 2)$$ $$= 1 - 2 \times 0.0228 $$ $$= 0.954$$

(b) This is asking us to work out $P(X > n) = 0.10$. Again by centering and standardizing this is

$$P(X > n) = P(X - 200 > n - 200)$$ $$= P\left(\frac{X - 200}{5} > \frac{n - 200}{5}\right)$$ $$= P\left(Z > \frac{n - 200}{5}\right)$$ $$= 0.10$$

From the tables this means that

$$\frac{n - 200}{5} = 1.28$$

so $n = 200 + 5 \times 1.28 = 206.4$ mL.