Question
Attempts: Average: 16 Pica is a disorder in which individuals eat substances that are typically non-nutritive or inedible, such as dirt, paint chips, ice or frost, cday, paper, and so forth. To be diagnosed with pica, these cravings need to last longer than a month and be developmentally inappropriate. Pica is seen in all ages, but more commonly among pregnant women, small children, and those with developmental disabilities. Pica can lead to serious medical problems particularly when the substance ingested can lead to poisoning (such as lead poisoning from paint chips), parasites (for example, from soil), or ges. [ Source: American Psychiatric Association. Diagnostic and Statistical Manual of Mental Disorders. 4th ed. Text rev. (Washington, DC: American Psychiatric Association, 2000).] A medical researcher is studying the relationship between pica and iron deficiency. She measures the hemoglobin levels mean hemoglobin level in women over the age of 60 is 12.6 gm/dl with a standard deviation of 1.2 gm/di. The researcher compiles the following in five of her patients, all of She wants to compare these levels to the population norms. Suppose the where the first column of values is the patient's hemoglobin level (Hb level), the second column describes the location of that hemoglobin ird column contains the corresponding z-score of that patient's in the distribution of hemoglobin levels among women over the age of 60, and the th hemoglobin level. Fill in the missing values. Z-score Location Description Hb Level (gm/di) 10.2 13.2 two standard deviations below the mean Patient 1 Patient 2 Patient 3 Patient 4 Patient 5 +0.50 one half oras tandard deviation abovethe mean .] one-quarter of a standard deviation above the mean one standar one standard deviation below the mean [ -3.00 9.0 10:5 9/5
Explanation / Answer
Ans:
z score=(x-mean)/stdandard deviation
patient 1 10.2 two standard deviations below tghe mean -2
Patien 2 13.2 one half of standard deviation above the mean +0.5
Patient 3 12.6+0.25*1.2=12.9 one quarter of a standard devation above the mean +0.25
Patient 4 12.6-3*1.2=9 three standard deviations below the mean -3
Patient 5 12.6-1.2=11.4 one standard deviation below the mean -1
