Nursing Elites

1. A teacher asked all the students in the class which days of the week they get up after 8 a.m. Which of the following is the best way to display the frequency for each day of the week?

• A. Histogram
• B. Pie chart
• C. Bar graph
• D. Scatter plot

Correct answer: A

Rationale: A histogram is the best way to display the frequency for each day of the week in this scenario. Histograms are ideal for showing the distribution of numerical data by dividing it into intervals and representing the frequency of each interval with bars. In this case, each day of the week can be represented as a category with the frequency of students getting up after 8 a.m. displayed on the vertical axis. Choice B, a pie chart, would not be suitable for this scenario as it is more appropriate for showing parts of a whole, not frequency distributions. Choice C, a bar graph, could potentially work but is more commonly used to compare different categories rather than displaying frequency distribution data. Choice D, a scatter plot, is used to show the relationship between two variables and is not the best choice for displaying frequency for each day of the week.

2. A patient was taking 310 mg of an antidepressant daily. The doctor reduced the dosage by 1/5, and then reduced it again by 20 mg. What is the patient’s final dosage?

• A. 20 mg
• B. 42 mg
• C. 228 mg
• D. 248 mg

Correct answer: C

Rationale: To calculate the final dosage, first find 1/5 of 310 mg, which is 62 mg, and subtract it from the original dosage. This gives 310 mg - 62 mg = 248 mg. Then, subtract an additional 20 mg from the result to get the final dosage: 248 mg - 20 mg = 228 mg. Therefore, the patient's final dosage is 228 mg. Choice A (20 mg) is incorrect because it only considers the second reduction of 20 mg and misses the initial reduction by 1/5. Choice B (42 mg) is incorrect as it miscalculates the reduction amounts. Choice D (248 mg) is incorrect as it does not account for the second reduction of 20 mg.

3. The force applied is directly proportional to the stretch of a coil. If a force of 132 Newtons stretches a coil 0.07 meters, what force would be needed to stretch a coil 0.1 meter? Round your answer to the nearest tenth.

• A. 92.4 Newtons
• B. 1885.7 Newtons
• C. 188.6 Newtons
• D. 136.0 Newtons

Correct answer: C

Rationale: To find the force needed to stretch the coil 0.1 meters, we can set up a proportion based on the given information. The initial force and stretch are in direct proportion, so we can use this relationship to determine the unknown force. (132 N / 0.07 m) = X / 0.1 m. Cross-multiplying, we get 132 N * 0.1 m / 0.07 m = 188.57 N, which rounds to 188.6 N. Therefore, the correct answer is 188.6 Newtons. Choice A is incorrect as it does not match the calculated answer. Choice B is significantly higher and does not align with the proportional relationship. Choice D is close but does not account for the correct rounding as specified in the question.

4. A lab technician took 500 milliliters of blood from a patient. The technician used 16.66% of the blood for further tests. How many milliliters of blood were used for further tests? Round your answer to the nearest hundredth.

• A. 83
• B. 83.3
• C. 83.33
• D. 83.34

Correct answer: C

Rationale: To find the amount of blood used for further tests, we multiply 500 mL by 0.1666 (equivalent to 16.66%). This calculation results in 83.3, which rounded to the nearest hundredth is 83.33. Therefore, 83.33 milliliters of blood were used for further tests. Choice A is incorrect as it does not consider rounding to the nearest hundredth. Choices B and D are slightly off due to incorrect rounding. Choice C is the correct answer after rounding to the nearest hundredth.

5. Given a double bar graph, which statement is true about the distributions of Group A and Group B?

• A. Group A is negatively skewed, Group B is normal.
• B. Group A is positively skewed, Group B is normal.
• C. Group A is positively skewed, Group B is neutral.
• D. Group A is normal, Group B is negatively skewed.

Correct answer: B

Rationale: The correct answer is B. In statistical terms, a positively skewed distribution means that the tail on the right side of the distribution is longer or fatter than the left side, indicating more high values. Therefore, Group A is positively skewed. Conversely, an approximately normal distribution, also known as a bell curve, is symmetrical with no skewness. Hence, Group B is normal. Choices A, C, and D are incorrect because they do not accurately describe the skewness of Group A and the normal distribution of Group B as depicted in a double bar graph.

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