Statistics and Probability Questions and Answers

Answer: C

\(n\) is 11.

The variance of the first \(n\) natural numbers is \(\frac{n^{2} - 1}{12}\), since the mean of the squares is \(\frac{(n+1)(2n+1)}{6}\) and the square of the mean is \(\frac{(n+1)^{2}}{4}\).

\[ \begin{aligned} \frac{n^{2} - 1}{12} &= 10 \\ n^{2} &= 121, \quad n = 11 \end{aligned} \]

Check: for 1 to 11 the mean is 6 and the squared deviations add to 110, so the variance is \(110 / 11 = 10\). The other values give variances of 2, 6.67, 14 and 18.67.

Answer: A

\(P(A' / B') = \frac{27}{32}\).

\(A' \cap B'\) is the complement of \(A \cup B\).

\[ \begin{aligned} P(A \cup B) &= \frac{1}{4} + \frac{1}{5} - \frac{1}{8} = \frac{13}{40} \\ P(A' \cap B') &= 1 - \frac{13}{40} = \frac{27}{40} \\ P(A' / B') &= \frac{27/40}{4/5} = \frac{27}{32} \end{aligned} \]

The conditioning event is \(B'\), with probability \(1 - \frac{1}{5} = \frac{4}{5}\); dividing by \(\frac{3}{4}\) (that is, by \(P(A')\)) gives \(\frac{9}{10}\), which is the reverse conditional.

Answer: D

The combined variance is \(\frac{11}{2}\).

Work from the sums of squares. For each set, \(\sum x^{2} = n(\sigma^{2} + \bar{x}^{2})\):

\[ \begin{aligned} \text{set 1} &: 5(4 + 4) = 40 \\ \text{set 2} &: 5(5 + 16) = 105 \end{aligned} \]

The combined mean is \(\frac{5 \times 2 + 5 \times 4}{10} = 3\), so

\[ \begin{aligned} \sigma^{2} &= \frac{40 + 105}{10} - 3^{2} \\ &= 14.5 - 9 = 5.5 = \frac{11}{2} \end{aligned} \]

Simply averaging the two variances gives \(\frac{9}{2}\), which ignores the spread between the two means; the extra 1 comes from each mean being 1 away from the combined mean.

Answer: D

The percentage of boys is 60%.

Let there be b boys and g girls. The combined mean is the weighted mean.

\[ \begin{aligned} 14b + 9g &= 12(b + g) \\ 2b &= 3g \Rightarrow b : g = 3 : 2 \\ \frac{b}{b + g} &= \frac{3}{5} = 60\% \end{aligned} \]

40% is the percentage of girls. The combined mean of 12 lies nearer to the boys' 14 than to the girls' 9, which already shows boys are the majority. Check: with 3 boys and 2 girls, \(\frac{3(14) + 2(9)}{5} = \frac{60}{5} = 12\).

Answer: A

The sum of the squares is 5920.

The variance equals the mean of the squares minus the square of the mean.

\[ \begin{aligned} \sigma^{2} &= \frac{\sum x^{2}}{n} - \bar{x}^{2} \\ 16 &= \frac{\sum x^{2}}{10} - 576 \\ \sum x^{2} &= 10 \times 592 = 5920 \end{aligned} \]

Using the standard deviation 4 in place of the variance 16 gives 5800, and the other options come from similar slips. Check: \(\frac{5920}{10} - 24^{2} = 592 - 576 = 16 = 4^{2}\).

Answer: E

\(P(A \cup B)\) is \(\frac{2}{3}\).

Find P(A) and P(B) from the conditional probabilities.

\[ \begin{aligned} P(B) &= \frac{P(A \cap B)}{P(A|B)} = \frac{1/6}{1/2} = \frac{1}{3} \\ P(A) &= \frac{P(A \cap B)}{P(B|A)} = \frac{1/6}{1/3} = \frac{1}{2} \\ P(A \cup B) &= \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{6} = \tfrac{2}{3} \end{aligned} \]

Forgetting to subtract the intersection gives \(\frac{5}{6}\), which is not offered; \(\frac{1}{2}\) and \(\frac{1}{3}\) are the separate probabilities.

Answer: C

The probability is \(\frac{7}{10}\).

There are 10 balls, 9 of them not red. Choose all three balls from those nine.

\[ \begin{aligned} P &= \frac{\binom{9}{3}}{\binom{10}{3}} \\ &= \frac{84}{120} = \frac{7}{10} \end{aligned} \]

\(\frac{3}{10}\) is the probability of getting the red ball, the complement. A quick check: the red ball is among the three drawn with probability \(\frac{3}{10}\), so it is missed with probability \(\frac{7}{10}\).

Answer: B

The new 90th percentile is 3.

For 80 observations the 90th percentile sits at rank \(0.9 \times 80 = 72\), between the 72nd and 73rd smallest values. The 7 largest observations occupy ranks 74 to 80, so the percentile is determined by observations among the remaining 73.

Subtracting 3 from those 73 shifts them all down by 3 without changing their order, and adding 6 to the top 7 keeps those above them. So the ranks are unchanged and the percentile moves from 6 to

\[ 6 - 3 = 3 \]

12 would follow if the percentile point were among the 7 largest and the change were added, and 9 adds 3 instead of subtracting it.

Answer: B

The probability is 7/128.

Each guess is right with probability 1/2, so the number right out of 10 follows a binomial distribution. 'At least 8' means 8, 9 or 10 right:

\[ \begin{aligned} P &= \frac{\binom{10}{8} + \binom{10}{9} + \binom{10}{10}}{2^{10}} \\ &= \frac{45 + 10 + 1}{1024} = \frac{56}{1024} = \frac{7}{128} \end{aligned} \]

7/64 doubles the result and 7/256 halves it, so they come from a wrong power of 2.

Answer: C

The probability is 4/11.

PROBABILITY has 11 letters: P, R, O, B, A, B, I, L, I, T, Y. The vowels are O, A, I and I, four letters in all (each I counts, since a letter is picked from the 11 positions; Y is treated as a consonant). So P(vowel) = 4/11.

3/11 counts the distinct vowels O, A, I instead of the letters, 2/11 undercounts further, and 7/11 is the probability of a consonant.