Question

Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean u = 6900 and estimated standard deviation o = 2800. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection. (a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.) (b) Suppose a doctor uses the average for two tests taken about a week apart. What can we say about the probability distribution of ? The probability distribution of x is approximately normal with us = 6900 and o- = 2800. The probability distribution of is approximately normal with us = 6900 and 0 = 1979.90. The probability distribution of is approximately normal with us = 6900 and 0 = 1400.00. O The probability distribution of is not normal. What is the probability of < 3500? (Round your answer to four decimal places.) (c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.) (d) Compare your answers to parts (a), (b), and (c). How did the probabilities change as n increased? The probabilities decreased as n increased. O The probabilities stayed the same as n increased. O The probabilities increased as n increased. If a person had < 3500 based on three tests, what conclusion would you draw as a doctor or a nurse? It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia. It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia. O It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia. It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.

Answer 1

count Now, Z < 3500 - 6900 2800 Z We are given, X : Blood ceil per cubic milliliter of whole blood. XNN (l=6900 5 - 2800) a) p(x< 3500) p(z x-l Now, p( x< 3500) plz < -1.21429 p( x < 3500 = 0.112319 By Rounding it to four decimal point. |p(*< 3500) 10 = 0.11 23 15 taken about week Suppose X be the average for two tests apart. so the probability distribution of is approximately = li = 6900 ¢ 5x = 5 = 2800—1979 20 Noce J2 < normal with lly .899 3500 6900 2800 p( < 3500) - PC* =plz Z < -14 -1.71926 x < 3500 25 LpQ 0.042966 px # < 3500) = 0.0430 If n = 3 probability distribution of t is approximately normal with l = = 6900 ¢ 30 = 0 = 2800 J3 Therefore INN (6900 1616-581 Save the Children

MTWTFS Date: 7 / Page No. Now 20723500 (z Z < 3500 6900 1616.58) =plz Z< -2.1032 p(*<3500) 0,017724 0.0177 0.01 10 e Comparing part axt a, b, ofc The soba decreased can say that as n increased, a person had x < 3500 based on three test, It would be an Ex+8 emely rare event for ce person to have three tests below 35oo pusely by Chance. The person probably has leuko - penia two 20