Statistics and Probability Questions and Answers

Answer: C

The second mean is 48.

Find \(x\) from the first mean:

\[ \begin{aligned} \frac{(12 + 17 + 25 + 34) + 4x}{4} &= 22 \\ 22 + x &= 22, \quad x = 0 \end{aligned} \]

Then the second mean is \(\frac{38 + 42 + 52 + 60}{4} = \frac{192}{4} = 48\).

A quicker way: adding the same \(x\) to every value adds \(x\) to the mean, so with \(x = 0\) the second mean is just the mean of 38, 42, 52 and 60. 22 repeats the first mean, 42 is just one of the four values, and 46 and 50 do not equal \(\frac{192}{4}\).

Answer: C

The new standard deviation is 27.1.

Each value in the second list is 10 times the corresponding value in the first. Multiplying every observation by a constant \(c\) multiplies the standard deviation by \(|c|\) (and the variance by \(c^{2}\)):

\[ \sigma_{\text{new}} = 10 \times 2.71 = 27.1 \]

271 multiplies by 100, the factor that applies to the variance, not the standard deviation. 0.271 divides instead of multiplying, and \(2.71\sqrt{10}\) would be right only if the values had been multiplied by \(\sqrt{10}\). 2.17 is the given value with digits swapped.

Answer: D

\(P(A) = \frac{6}{35}\).

Reading the stem as \(P(B) = \frac{1}{5}\): mutually exclusive events cannot happen together, so \(P(A \cap B) = 0\) and the addition rule becomes \(P(A \cup B) = P(A) + P(B)\):

\[ \begin{aligned} P(A) &= \frac{13}{35} - \frac{1}{5} \\ &= \frac{13}{35} - \frac{7}{35} \\ &= \frac{6}{35} \end{aligned} \]

\(\frac{1}{5}\) repeats \(P(B)\). \(\frac{1}{7} = \frac{5}{35}\), \(\frac{3}{35}\) and \(\frac{1}{35}\) do not satisfy \(P(A) + \frac{7}{35} = \frac{13}{35}\).

Answer: C

\(P(A \cup B) = 0.68\).

From \(P(A') = 0.8\), \(P(A) = 0.2\). Independence means \(P(A \cap B) = P(A)P(B)\):

\[ \begin{aligned} P(A \cup B) &= P(A) + P(B) - P(A)P(B) \\ &= 0.2 + 0.6 - 0.12 = 0.68 \end{aligned} \]

Check another way: \(P(A \cup B) = 1 - P(A')P(B') = 1 - 0.8 \times 0.4 = 0.68\).

0.8 adds the probabilities and forgets the overlap. 0.48 is \(P(A')P(B)\), and 0.86 and 0.52 do not follow from the addition rule with \(P(A) = 0.2\).

Answer: E

\(P(E) = \frac{2}{23}\).

Count favourable selections and divide by all selections of 3 from 25:

\[ \begin{aligned} \text{favourable} &= \binom{20}{1}\binom{5}{2} = 20 \times 10 = 200 \\ \text{total} &= \binom{25}{3} = 2300 \\ P(E) &= \frac{200}{2300} = \frac{2}{23} \end{aligned} \]

The option \(\frac{1}{23}\) halves the count of girl pairs (using 5 instead of \(\binom{5}{2} = 10\)), and the others do not come from a valid count.

Answer: A

\(P(A|B) = \frac{1}{6}\).

The two rolls are independent, so knowing the second roll was a 4 tells us nothing about the first.

\[ \begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac{1/36}{1/6} \\ &= \frac{1}{6} \end{aligned} \]

The option \(\frac{1}{36}\) is \(P(A \cap B)\), the probability of both events, not the conditional probability.

Answer: D

The variance is 8.

The mean is \(\frac{5 + 7 + 9 + 11 + 13}{5} = 9\). The deviations are \(-4, -2, 0, 2, 4\).

\[ \begin{aligned} \sigma^{2} &= \frac{16 + 4 + 0 + 4 + 16}{5} \\ &= \frac{40}{5} \\ &= 8 \end{aligned} \]

Dividing by \(n - 1 = 4\) (the sample variance) would give 10, which is an option; the population variance, dividing by \(n\), is the standard school definition and gives 8.

Answer: C

The mean deviation is 3.6.

The mean is 10, so the sum is 50:

\[ 4 + 7 + x + 13 + 16 = 50 \Rightarrow x = 10 \]

The absolute deviations from the mean are 6, 3, 0, 3 and 6.

\[ \begin{aligned} \text{M.D.} &= \frac{6 + 3 + 0 + 3 + 6}{5} \\ &= \frac{18}{5} \\ &= 3.6 \end{aligned} \]

Dividing by 4 instead of 5 gives 4.5, which is not offered. The mean deviation uses absolute values, so the deviations are not allowed to cancel.

Answer: A

\(P(A' \cup B')\) = 0.9.

By De Morgan's law, \(A' \cup B' = (A \cap B)'\). First find P(B) = 1 − 0.8 = 0.2 and the intersection.

\[ \begin{aligned} P(A \cap B) &= 0.7 + 0.2 - 0.8 = 0.1 \\ P(A' \cup B') &= 1 - 0.1 = 0.9 \end{aligned} \]

0.1 is \(P(A \cap B)\) itself, before taking the complement. Treating 0.8 as P(B) instead of P(B') would give an intersection of 0.7 and an answer of 0.3.

Answer: B

The mean deviation from the median is 6.

There are 10 values, so the median is the average of the 5th and 6th, \(\frac{82 + 82}{2} = 82\).

The absolute deviations from 82 are 10, 10, 8, 2, 0, 0, 4, 6, 10, 10, which add to 60.

\[ \text{M.D.} = \frac{60}{10} = 6 \]

The mean of the data is 82 as well, so taking the mean instead of the median gives the same 6 here; the other options come from miscounting the deviations.