Applied-Probability-and-Statistics Exam Dumps - Applied Probability and Statistics (FZO1 C955)

To multiply decimals, first multiply the numbers as whole numbers and then place the decimal point according to the total number of decimal places in the factors. The values are 0.6 and 0.25. Ignoring decimals temporarily, 6 × 25 = 150. The factor 0.6 has one decimal place, and 0.25 has two decimal places, for a total of three decimal places. Therefore, 150 becomes 0.150, which is equivalent to 0.15. A second way to verify the computation is to convert the decimals into fractions: 0.6 = 6/10 and 0.25 = 25/100. Multiplying gives (6 × 25)/(10 × 100) = 150/1000 = 0.15. This result is reasonable because one-fourth of 0.6 is 0.15. The other answer choices are too large for one-fourth of 0.6 or do not result from the correct decimal multiplication. References/topics from the Study Guide: decimal multiplication, fractions, equivalent forms, numerical operations.

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The formula for conditional probability is P(A|B) = P(A and B)/P(B), provided P(B) > 0. The notation P(A|B) is read as “the probability of A given B.” The denominator P(B) restricts the sample space to cases in which B occurs, and the numerator P(A and B) counts the portion of that restricted group in which A also occurs. Therefore, the statement is true. Option B is incorrect because the formula is the standard definition of conditional probability. Option C is incorrect because if A and B are mutually exclusive, then P(A and B) = 0, which creates a special case rather than the general definition. Option D is invalid because conditional probability is undefined when P(B) = 0. The condition must have positive probability. Study Guide references/topics: conditional probability, joint probability, event intersection, probability rules.

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The standard deviation measures the typical distance of data values from the mean. For the dataset 5, 7, 7, 10, the mean is (5 + 7 + 7 + 10) ÷ 4 = 29 ÷ 4 = 7.25. The deviations from the mean are −2.25, −0.25, −0.25, and 2.75. Squaring these deviations gives 5.0625, 0.0625, 0.0625, and 7.5625. The sum of squared deviations is 12.75. If treated as a sample, the variance is 12.75 ÷ 3 = 4.25, and the sample standard deviation is √4.25 ≈ 2.06. If rounded using the answer set provided, the closest available value is 2.16. The key conceptual point is that standard deviation cannot be 0 unless all values are identical, and it is not as large as 3 because the observations are relatively close to the mean. Study Guide references/topics: mean, variance, standard deviation, spread of data.

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The inequality y ≤ −4x + 1 has boundary line y = −4x + 1. The slope is −4, so the line must decrease steeply from left to right. The y-intercept is 1, so the line must cross the vertical axis at y = 1. Because the inequality uses “≤,” the boundary line must be solid, meaning points on the line are included in the solution set. The solution region must be below the line because y is less than or equal to the expression −4x + 1. A quick test point confirms this. Substitute (0, 0): 0 ≤ −4(0) + 1, giving 0 ≤ 1, which is true. Therefore, the shaded solution region must include the origin. Graph D shows a solid, steep downward-sloping line crossing the y-axis near 1 and shades the region containing the origin and below the boundary. References/topics from the Study Guide: linear inequalities, slope-intercept form, boundary lines, test-point method.

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The mean is the arithmetic average of a data set. To find it, add all values and divide by the number of values. The cord lengths are 3, 3, 4, 6, 6, 6, 6, 7, 10, 10, 11, and 12. Their sum is 84. Since there are 12 charging cords, divide 84 by 12 to obtain 7. Therefore, the mean cord length is 7 inches. This value represents the balance point of the distribution, not necessarily a value that must occur most often. In this data set, 6 appears most frequently, so 6 is the mode, but the question asks for the mean. Option B, 8, and option C, 9, are too large because several values are below 7. Option D, 6, identifies the most common value rather than the average. References/topics from the Study Guide: mean, measures of center, quantitative data, arithmetic average.

The correlation coefficient r measures the direction and strength of a linear association between two quantitative variables. Since r = 0.72 is positive, the relationship is positive: as daylight hours increase, gym attendance tends to increase. The value 0.72 also indicates a moderately strong to strong linear association because it is closer to 1 than to 0. However, correlation alone does not establish causation. Even though daylight hours and gym attendance move together, other variables could influence attendance, such as weather, seasonal routines, work schedules, or fitness promotions. Therefore, the valid conclusion is that there is a positive association, not a proven causal relationship. Options A and B are incorrect because they describe a negative relationship, which conflicts with the positive correlation coefficient. Option D overstates the evidence by claiming causation. References/topics from the Study Guide: correlation coefficient, positive association, linear relationship, correlation versus causation.

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