The activity of certain enzyme is measured by counting emissionsfrom a radioacti

Question

The activity of certain enzyme is measured by counting emissionsfrom a radioactively labeled molecule. For a given tissue specimen,the counts in consecutive 10sec time periods may be regarded(approximately) as repeated independent observations from a normaldistribution. Suppose the mean 10sec count for a certain tissuespecimen is 1200 and the standard deviation is 35. Let Y denote thecount in a randomly chosen 10sec time period.Find.

A)    Pr { Y >= 1250}

B)     Pr { Y <= 1175}

C)    Pr {1150 <= Y <= 1250}

D)    Pr {1150 <= Y<= 1175}

Explanation / Answer

average, = 1200 standard dev, = 35 A)    Pr{Y >= 1250} P( ( Y - ) / ) > P( Z> (1250- 1200 ) / 35))
= P( Z >1.43)
= 1- 0.9236= 0.0764 B)     Pr { Y <=1175} P( ( Y - ) / ) <P( Z< (1175 - 1200 ) / 35))
= P( Z<-0.71)
= 0.2389 C)    Pr {1150 <= Y<= 1250} P( ( Y - ) / ) <P( Z< (1150 - 1200 ) / 35))
= P( Z<-1.43)
= 0.0764 P( ( Y - ) / ) <P( Z< (1150 - 1200 ) / 35))
= P( Z<-1.43)
= 0.0764 P( ( Y - ) / ) <P( Z< (1250 - 1200 ) / 35))
= P( Z<1.43)
= 0.9236 Pr {1150 <= Y <=1250}=0.9236-0.0764 =0.8472 P( ( Y - ) / ) <P( Z< (1250 - 1200 ) / 35))
= P( Z<1.43)
= 0.9236 Pr {1150 <= Y <=1250}=0.9236-0.0764 =0.8472 D)    Pr {1150 <=Y<= 1175} P( ( Y - ) / ) <P( Z< (1150 - 1200 ) / 35))
= P( Z<-1.43)
= 0.0764 P( ( Y - ) / ) <P( Z< (1175 - 1200 ) / 35))
= P( Z<-0.71)
= 0.2389 Pr {1150 <= Y<= 1175}=0.2389-0.0764 =0.1625 P( ( Y - ) / ) <P( Z< (1150 - 1200 ) / 35))
= P( Z<-1.43)
= 0.0764 P( ( Y - ) / ) <P( Z< (1175 - 1200 ) / 35))
= P( Z<-0.71)
= 0.2389 Pr {1150 <= Y<= 1175}=0.2389-0.0764 =0.1625

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.