Probability Distributions Questions and Answers

Answer: B

About 68 per cent of scores lie within one standard deviation of the mean.

For a normal distribution the empirical rule gives the share of the area within one, two and three standard deviations of the mean as about 68.3%, 95.4% and 99.7%. So the interval \(\mu \pm \sigma\) holds roughly 68% of the scores.

34 is the share on one side only, from the mean to \(\mu + \sigma\). 95 is the share within \(\pm 2\sigma\), and 99 is close to the share within \(\pm 3\sigma\) (or exactly the share within about \(\pm 2.58\sigma\)).

Answer: D

The value is 0.398.

The normal density is \(f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^{2}/2\sigma^{2}}\). At the mean, \(x = \mu = 5\), the exponent is 0, so with \(\sigma = 1\):

\[ \begin{aligned} f(5) &= \frac{1}{\sqrt{2\pi}} \approx \frac{1}{2.5066} \\ &\approx 0.399 \end{aligned} \]

The density is greatest at the mean, so it cannot be 0.0. 0.159 is \(1/2\pi\), which omits the square root, and 0.282 is \(1/(2\sqrt{\pi})\), the value for \(\sigma = \sqrt{2}\), not 1.

Answer: A

The probability is \(\frac{2}{3}\).

The ball is put back after every draw, so the box always holds 1 red and 2 blue balls and each draw is independent of the earlier ones. The fact that blue came up twice changes nothing:

\[ P(\text{blue on draw 3}) = \frac{2}{3} \]

\(\frac{1}{3}\) is the chance of red. \(\frac{1}{2}\) treats the two colours as equally likely, and 1 assumes blue must repeat. A tempting slip is to compute \((2/3)^{3} = 8/27\), the chance of three blues in a row, but the question asks only about the third draw given the first two.

Answer: B

Y is normally distributed with mean 0 and variance 5.

Any linear combination of independent normal variables is itself normal. For Y = X₁ - X₂:

\[ \begin{aligned} E[Y] &= \mu_{1} - \mu_{2} = 1 - 1 = 0 \\ Var[Y] &= \sigma_{1}^{2} + \sigma_{2}^{2} = 1 + 4 = 5 \end{aligned} \]

Variances add even when the variables are subtracted, because Var(-X₂) = Var(X₂). Variance 1 comes from subtracting the variances (or the standard deviations), and the options saying Y is not normal deny the closure of the normal family under linear combination.

Answer: A

Y is exponentially distributed with mean 1/6.

An exponential variable with mean m has rate 1/m, so the rates are \(\lambda_{1} = 2\) and \(\lambda_{2} = 4\). For independent variables,

\[ \begin{aligned} P(Y \gt y) &= P(X_{1} \gt y)\,P(X_{2} \gt y) \\ &= e^{-2y}e^{-4y} = e^{-6y} \end{aligned} \]

That is the survival function of an exponential variable with rate 6, so Y is exponential with mean 1/6.

The minimum of exponentials is always exponential, never normal, which removes C and D. Mean 2 would be a rate of 1/2, which is lower than either component's rate, impossible for a minimum, which must occur sooner on average than either variable alone.

Answer: A

The mean and the variance are 3 and 3.

The Poisson parameter is the expected number of defectives in the box:

\[ \lambda = np = 200 \times 0.015 = 3 \]

A Poisson distribution has mean and variance both equal to \(\lambda\), so both are 3.

\(\sqrt{3}\) is the standard deviation, not the variance, and the mean is not \(\sqrt{3}\) either. 0.015 is the probability that a single component is defective, not the count in a box of 200. '3 and 9' squares the mean to get the variance, which confuses the Poisson variance \(\lambda\) with \(\lambda^{2}\).

Answer: D

The probability is 0.75.

The four equally likely outcomes are HH, HT, TH and TT. Only TT has no head, so

\[ P(\text{at least one head}) = 1 - \tfrac{1}{4} = 0.75. \]

0.25 is the probability of no head (or of two heads), 0.5 of exactly one head, and 0.675 does not correspond to any count of the four outcomes.

Answer: A

The probability is 79/120.

There are \(\binom{10}{3} = 120\) equally likely selections. Count the complements:

\[ \begin{aligned} \text{all different} &: 2 \times 3 \times 5 = 30 \\ \text{all same} &: \tbinom{3}{3} + \tbinom{5}{3} = 1 + 10 = 11 \\ \text{exactly two same} &: 120 - 30 - 11 = 79 \end{aligned} \]

Directly: two engineering and one other \(1 \times 8 = 8\); two science \(3 \times 7 = 21\); two arts \(10 \times 5 = 50\); total 79. 89/120 and 69/120 miscount one of the three cases, and 2/120 is far too small: the arts students alone give 50 favourable selections.

Answer: D

The probability is 5/8.

This is a conditional probability. The share who like at least one fruit is

\[ \begin{aligned} P(O \cup A) &= 0.5 + 0.7 - 0.4 = 0.8 \\ P(O \mid O \cup A) &= \frac{P(O)}{P(O \cup A)} = \frac{0.5}{0.8} = \frac{5}{8} \end{aligned} \]

since everyone who likes oranges is in the 'at least one' group. 1/2 ignores the condition and gives the unconditional probability. 1/8 is the share who like oranges only (0.1) divided by 0.8, and 5/12 divides 0.5 by 1.2, the sum of the two percentages without removing the overlap.

Answer: A

The correct statement is \(P(E_1 \cup E_2) = \frac{56}{81}\).

Each road is open with probability \(2/3\). Travelling via \(C_2\) needs both roads \(C_1C_2\) and \(C_2C_3\) open, so \(P(E_1) = (2/3)^{2} = 4/9\), and likewise \(P(E_2) = 4/9\). The two routes use different roads, so the events are independent:

\[ \begin{aligned} P(E_1 \cup E_2) &= \tfrac{4}{9} + \tfrac{4}{9} - \tfrac{4}{9} \cdot \tfrac{4}{9} \\ &= \tfrac{72}{81} - \tfrac{16}{81} = \tfrac{56}{81} \end{aligned} \]

\(\frac{8}{9}\) adds the two probabilities without removing the overlap. Independence means \(P(E_1|E_2) = P(E_1)\), so option C is false, and \(P(E_1 \cap E_2) = 16/81\), not 0.