Statistics and Probability Questions and Answers

Answer: A

The standard deviation is \(\frac{1}{2}\sqrt{3333}\).

For the first n natural numbers the variance is \(\frac{n^{2} - 1}{12}\).

\[ \begin{aligned} \sigma^{2} &= \frac{100^{2} - 1}{12} = \frac{9999}{12} = \frac{3333}{4} \\ \sigma &= \frac{\sqrt{3333}}{2} \end{aligned} \]

The options with 4, 6 or 8 in front take the wrong root of the denominator 4, and \(\frac{1}{4}\sqrt{1111}\) divides 9999 by 9 instead of 3. The formula follows from \(\sigma^{2} = \frac{\sum x^{2}}{n} - \bar{x}^{2}\) with \(\sum x^{2} = \frac{n(n + 1)(2n + 1)}{6}\) and \(\bar{x} = \frac{n + 1}{2}\).

Answer: C

The probability is \(\frac{9}{100}\).

The multiples of 11 up to 100 are 11, 22, ..., 99. Since \(11 \times 9 = 99\) and \(11 \times 10 = 110\) is too large, there are 9 of them out of 100 equally likely numbers.

\(\frac{1}{11}\) is the long-run proportion of multiples of 11 among all integers, not the exact count up to 100; \(\frac{1}{10}\) and \(\frac{11}{100}\) miscount by one or two.

Answer: B

The mean deviation about the mean is 2.5.

\[ \begin{aligned} N &= 7 + 4 + 5 + 4 = 20 \\ \bar{x} &= \frac{14 + 16 + 30 + 40}{20} = 5 \\ \sum f|x - \bar{x}| &= 7(3) + 4(1) + 5(1) + 4(5) = 50 \end{aligned} \]

So MD = \(\frac{50}{20} = 2.5\). Dividing by the 4 distinct values instead of N, or measuring deviations from a wrong mean, gives the other options.

Answer: D

\(P(A / B)\) is 0.6.

\[ \begin{aligned} P(A \cap B) &= P(A) + P(B) - P(A \cup B) \\ &= 0.7 + 0.5 - 0.9 = 0.3 \\ P(A / B) &= \frac{0.3}{0.5} = 0.6 \end{aligned} \]

0.3 is \(P(A \cap B)\) without dividing by \(P(B)\). Dividing by \(P(A)\) instead would give \(P(B / A) \approx 0.43\), and 0.5 and 0.7 are just the given probabilities.

Answer: E

The variance is \(\frac{875}{4}\).

The mean is \(\frac{240 + 260 + 270 + 280}{4} = 262.5\). The deviations are −22.5, −2.5, 7.5 and 17.5, so

\[ \begin{aligned} \sigma^{2} &= \frac{506.25 + 6.25 + 56.25 + 306.25}{4} \\ &= \frac{875}{4} \end{aligned} \]

(Using deviations from 260 in units of 10 gives the same: \(100\left(\frac{4 + 0 + 1 + 4}{4} - \frac{1}{16}\right) = \frac{875}{4}\).) The options with 475 come from an error in the sum of squares.

A shortcut is to subtract 260 and divide by 10 first (giving −2, 0, 1, 2), find that variance, and multiply it by 100, since variance is unchanged by a shift and scales with the square of a scale factor.

Answer: C

The probability is \(\frac{11}{16}\).

With four fair coins there are 16 equally likely outcomes. 'At most two heads' means 0, 1 or 2 heads:

\[ \begin{aligned} \frac{\binom{4}{0} + \binom{4}{1} + \binom{4}{2}}{16} &= \frac{1 + 4 + 6}{16} \\ &= \frac{11}{16} \end{aligned} \]

\(\frac{5}{8} = \frac{10}{16}\) is one outcome short, \(\frac{15}{16}\) is 'at most three heads', and \(\frac{13}{16}\) and \(\frac{9}{16}\) miscount the binomial terms.

Equivalently, use the complement: 'at most two heads' fails only for 3 or 4 heads, which have probability \(\frac{4 + 1}{16} = \frac{5}{16}\), and \(1 - \frac{5}{16} = \frac{11}{16}\).

Answer: A

The mean deviation about the mean is 2.4.

The mean is \(\frac{2 + 4 + 6 + 8 + 10}{5} = 6\). The absolute deviations from 6 are 4, 2, 0, 2, 4:

\[ \begin{aligned} \text{M.D.} &= \frac{4 + 2 + 0 + 2 + 4}{5} \\ &= \frac{12}{5} = 2.4 \end{aligned} \]

0 is what you get by adding signed deviations, which always cancel about the mean. The standard deviation here is \(\sqrt{8} \approx 2.83\), a different measure; 4.8 doubles the answer.

Answer: A

The variance is 64.

Adding a constant shifts every value equally and does not change the spread; multiplying by 2 multiplies the standard deviation by 2 and the variance by \(2^{2} = 4\):

\[ \begin{aligned} \sigma_{\text{new}} &= 2 \times 4 = 8 \\ \sigma_{\text{new}}^{2} &= 64 \end{aligned} \]

8 is the new standard deviation, not the variance. 16 is the original variance, and 67 and 19 wrongly add the constant 3 to the variance.

Answer: D

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

Given that the die shows an odd number, the outcomes are reduced to {1, 3, 5}, all equally likely. 'Other than 5' leaves {1, 3}:

\[ P = \frac{2}{3} \]

As a conditional probability, \(P(\text{not 5} \mid \text{odd})\) is \(\frac{P(\{1, 3\})}{P(\text{odd})} = \frac{2/6}{3/6}\). \(\frac{5}{6}\) ignores the condition, and \(\frac{1}{3}\) is the probability of getting 5.

Answer: C

\(P(\overline{A \cap B})\) equals 0.64.

By the multiplication rule, \(P(A \cap B) = P(A)\,P(B/A)\), then take the complement:

\[ \begin{aligned} P(A \cap B) &= 0.4 \times 0.9 = 0.36 \\ P(\overline{A \cap B}) &= 1 - 0.36 = 0.64 \end{aligned} \]

0.36 is \(P(A \cap B)\) itself, without the complement, and 0.6 is \(P(\overline{A})\). 0.44 and 0.24 do not come from the given data.