Nokea
In a study to investigate the rates of central line–acquired bacterial infections, it is discovered that patient length of stay (LOS) is not normally distributed but is highly right-skewed. What is the correct relationship between the mean, median, and mode of LOS?
- A. The mean is less than the median but greater than the mode.
- B. The mean is equal to the median and the mode.
- C. The mean is greater than the median and mode.
- D. The mean and median will both be less than the mode.
Correct Answer: C
Rationale: The correct relationship between mean, median, and mode in a highly right-skewed distribution is that the mean is greater than the median and mode. In a right-skewed distribution, the mean is pulled towards the longer tail, making it greater than the median, which is the middle value when the data is arranged in order. The mode is the most frequent value, but in a right-skewed distribution, it will be the smallest value, making the mean greater than both the median and mode. Therefore, choice C is correct. Choices A, B, and D are incorrect because they do not reflect the specific relationship between mean, median, and mode in a highly right-skewed distribution.
You may also like to solve these questions
Which of the following characteristics are similar with respect to Factor VIII and von Willebrand factor (vWF)?
- A. Both are made in endothelial cells and megakaryocytes.
- B. Both are activated by thrombin.
- C. They are present in normal to high relative amounts in newborns.
- D. They are stored in Weibel-Palade bodies in endothelial cells.
Correct Answer: C
Rationale: Rationale:
- Factor VIII and vWF are present in normal to high amounts in newborns due to the physiological immaturity of the hemostatic system.
- Choice A is incorrect as vWF is mainly produced in endothelial cells while Factor VIII is produced in both endothelial cells and liver.
- Choice B is incorrect as Factor VIII is activated by thrombin, but vWF is not.
- Choice D is incorrect as vWF is stored in Weibel-Palade bodies, but Factor VIII is not.
A 17-year-old female presents with cervical adenopathy and a history of daily fevers and drenching night sweats. A biopsy is performed and reveals classic Hodgkin lymphoma. Which of the following is least appropriate as part of the staging workup?
- A. Chest x-ray
- B. CT scan of chest, abdomen, and pelvis
- C. Functional imaging (PET scan)
- D. Lumbar puncture and cerebrospinal fluid (CSF) analysis
Correct Answer: D
Rationale: The correct answer is D, Lumbar puncture and cerebrospinal fluid (CSF) analysis. In Hodgkin lymphoma, central nervous system involvement is rare at presentation, making routine CSF analysis unnecessary. Staging workup typically includes imaging studies (A, B, C) to assess disease extent and involvement of distant organs. CSF analysis is reserved for cases with neurological symptoms or signs suggestive of CNS involvement. Therefore, in this case, the least appropriate option for staging workup is D.
The nurse is reviewing laboratory results and notes an aPTT level of 28 seconds. The nurse should notify the health care provider in anticipation of adjusting which medication?
- A. Aspirin
- B. Heparin
- C. Warfarin
- D. Erythropoietin
Correct Answer: B
Rationale: The correct answer is B: Heparin. An aPTT level of 28 seconds indicates a shorter clotting time than normal, suggesting that the patient may be at risk for bleeding due to excessive anticoagulation with heparin. The nurse should notify the healthcare provider to adjust the heparin dosage to prevent bleeding complications.
A: Aspirin is an antiplatelet medication and does not affect aPTT levels.
C: Warfarin is a vitamin K antagonist and primarily affects the PT/INR levels, not aPTT.
D: Erythropoietin is a hormone that stimulates red blood cell production and does not affect clotting parameters such as aPTT.
While taking a client history, which factor(s) that place the client at risk for a hematologic health problem will the nurse document? (Select all that apply.)
- A. Family history of military excellence
- B. Diet low in iron and protein
- C. Excessive alcohol consumption
- D. Family history of allergies
Correct Answer: C
Rationale: The correct answer is C: Excessive alcohol consumption. Alcohol can lead to various hematologic health problems such as anemia, bleeding disorders, and impaired immune function. Excessive alcohol intake can disrupt the production of red blood cells and platelets, leading to an increased risk of hematologic issues. Family history of military excellence (A) and family history of allergies (D) are not directly related to hematologic health problems. A diet low in iron and protein (B) can contribute to hematologic issues, but it is not as direct a risk factor as excessive alcohol consumption.
All produce microcytic anemia except:
- A. sideroblastic an.
- B. thalassemia
- C. pernicious anemia
- D. lead poisoning
Correct Answer: C
Rationale: The correct answer is C: pernicious anemia. Pernicious anemia is a macrocytic anemia caused by vitamin B12 deficiency, leading to large red blood cells. Sideroblastic anemia (A) is a type of microcytic anemia due to ineffective iron utilization. Thalassemia (B) causes microcytic anemia due to decreased production of hemoglobin. Lead poisoning (D) can also lead to microcytic anemia by disrupting heme synthesis. Therefore, pernicious anemia is the only choice that does not produce microcytic anemia.