When you encounter questions asking for the mean (average) of a data set, you need to add all values and divide by the number of data points. This is one of the most fundamental statistical calculations you'll see on quantitative reasoning exams.

To find the mean number of patients per week, first add all six weekly totals: $$18 + 21 + 24 + 19 + 27 + 21 = 130$$. Then divide by the number of weeks: $$\frac{130}{6} = 21.666...$$ which rounds to 21.7.

Looking at the wrong answers: Choice B (22.0) might result from rounding errors or miscalculating the sum as 132 instead of 130. Choice C (21.0) could come from mistakenly using the mode (the most frequently occurring value, which is 21) instead of the mean, or from premature rounding. Choice D (22.5) likely results from a calculation error, perhaps getting a sum of 135 and dividing by 6.

The key trap here is confusing mean with other measures of central tendency. The mode is 21 (appears twice), but the question specifically asks for the mean. Also, be careful with your arithmetic—double-check your addition and don't round until the final step.

Remember: mean questions on the DAT are straightforward calculations, but accuracy in arithmetic is crucial. Always verify your sum before dividing, and round only your final answer to match the precision of the given choices.

The sum of the original 20 observations is $$20 \times 45 = 900$$. The sum of the 5 new observations is $$37 + 41 + 49 + 53 + 45 = 225$$. The total sum is $$900 + 225 = 1125$$. The mean of all 25 observations is $$\frac{1125}{25} = 45.0$$. Choice A incorrectly weights the original mean. Choice C assumes the new observations increase the mean proportionally. Choice D miscalculates the sum of new observations. Choice E uses an incorrect weighting formula.

The weighted mean is calculated as $$\frac{n_A \bar{x}_A + n_B \bar{x}_B}{n_A + n_B} = \frac{15(82) + 25(78)}{15 + 25} = \frac{1230 + 1950}{40} = \frac{3180}{40} = 79.5$$. Choice B incorrectly uses simple average of the means. Choice C weights incorrectly in favor of Class A. Choice D uses wrong calculation of weighted sums. Choice E assumes equal weighting despite different sample sizes.

First find the z-score: z = (350 - 500)/100 = -1.5. On the rescaled test with the same z-score: score = 75 + (-1.5)(15) = 75 - 22.5 = 52.5. The z-score remains constant during linear transformations.

Since each day has equal weight (20 transactions), the overall mean is the simple average: $$\frac{1200 + 1350 + 1100 + 1400 + 1250}{5} = \frac{6300}{5} = 1260$$. Choice B uses incorrect weighting. Choice C miscalculates the sum. Choice D omits one day's data. Choice E uses wrong division.

When converting units by multiplication, variance is multiplied by the square of the conversion factor. New variance = $$16 \times(2.54)^2 = 16 \times 6.4516 = 103.3$$ square centimeters. Choice A uses the conversion factor without squaring. Choice C adds instead of multiplies. Choice D uses an incorrect conversion. Choice E uses the wrong power of the conversion factor.

For the difference of two independent normal variables, the variance is the sum of individual variances: $$Var(X-Y) = Var(X) + Var(Y) = 10^2 + 15^2 = 100 + 225 = 325$$. Standard deviation = $$\sqrt{325} = \sqrt{25 \times 13} = 5\sqrt{13} \approx 18.0$$g. Choice A subtracts standard deviations. Choice B averages them. Choice D adds them directly. Choice E shows the calculation but rounds differently.

When data is transformed by $$y = ax + b$$, the variance is multiplied by $$a^2$$ (the additive constant $$b$$ doesn't affect variance). Here $$a = 2$$, so the new variance is $$2^2 \times 16 = 4 \times 16 = 64$$. Choice A incorrectly adds the constant. Choice B uses $$2 \times 16 + 3$$. Choice D uses $$(2+3) \times 16 - 16 + 1$$. Choice E uses $$4 \times 16 + 3$$.