Let x be a ando variable that epresents hite blood cell count per cubic millilit
ID: 3060057 • Letter: L
Question
Let x be a ando variable that epresents hite blood cell count per cubic milliliter or hole blood. Assu me that leukopernia. This indicates bone marrow depression that may be the result of a viral infection has a distribution that is approximately n nal ith mean , 6950 and estimated standard de ation 285 A test esult of x 3500 i8 an indication o (a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.) (b) Suppcse a doctor uses the average x for two tests taken ebout a week apart. What can we say about the probability distribution of x? The probabity distribution of X is approximately normal with = 6950 and Og-2015.25. The probab Ity distribution cf x is nat normal. The probabity distribution of x is approximately normal with -6950 and 0 1425.0D. The probabilty distribution ofx is approximately normal with -6950 and -2850. what is the probability tfx 35007(Round your enswer to four deo mal places.) (c) Repeat pert (b; for 3 tests taken week spurt. [Round your answer to four cecin paces.) (d) Compare your enswers to parts () (), and () How did the probabilities cange es n increased? The probab lties decreased as n increased e probab!ties stayed the same as a increased. Tha prohabltas increased as a inceasad. If a person had x 3500 based on three tests, what conclusion would you draw as a doctor or a nurse? lt would be a common event fo person to have two or three tests below 3,sou purely by chunce. 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 has leukoperia It would be an extremely rare evert for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopeni obebly does not It would be a comman event for a person to have twa or three toets below 3,50purely by chancc. The person probably has lcukoperia.Explanation / Answer
Solution:- Given that mean = 6950 , sd = 2850
a) P(X < 3500) = P(Z < (X - )/)
= P(Z < (3500 - 6950)/2850)
= P(Z < -1.2105)
= 0.1131
b) option A. The probability distribution of X-bar is approximately normal with x = 6950 and X = 2015.25
P(X < 3500) = P(Z < (X - )/(/sqrt(n))
= P(Z < (3500 - 6950)/(2850/sqrt(2))
= P(Z < -1.7119)
= 0.0436
c) n = 3
P(X < 3500) = P(Z < (X - )/(/sqrt(n))
= P(Z < (3500 - 6950)/(2850/sqrt(3))
= P(Z < 2.0967)
= 0.0179
d) option A. The probabilites decreased as n increased.
=> option B. 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.
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