Just need the answer to (4.35) 4.34 In the nerve-cell activity of a certain indi
ID: 2914267 • Letter: J
Question
Just need the answer to (4.35)
4.34
In the nerve-cell activity of a certain individual fly, the timeintervals between “spike” discharges followapproximately a normal distribution with mean15.6ms and standarddeviation 0.4ms. Let Y denote a randomly selected interspikeinterval. Find.
A) Pr{Y>15}
B) Pr { Y>16.5}
C) Pr { 15< Y< 16.5}
D) Pr {15<Y<15.5}
4.35
For the distribution of interspike-time intervals described inExercise 4.34, find the quartiles and the interquartile range.
Explanation / Answer
average, = 15.6ms standard dev, = 0.4ms A) Pr{Y>15} P( ( Y - ) / ) = P( Z> (15- 15.6 ) / 0.4 ))= P( Z >-1.5)
= 1- 0.0668 = 0.9332 B) Pr { Y>16.5} P( ( Y - ) / ) = P( Z> (16.5- 15.6 ) / 0.4 ))
= P( Z >2.25)
= 1- 0.9878= 0.0122 C) Pr{ 15< Y< 16.5} findP(Y<15) P( ( Y - ) /) = P( Z< (15 - 15.6 ) / 0.4 ))
= P( Z <-1.5)
= 0.0668 findP(Y<16.5) P( ( Y - ) /) = P( Z< (16.5 - 15.6 ) / 0.4 ))
= P( Z <2.25)
= 0.9878 hence, Pr { 15< Y<16.5}=0.9878-0.0668 =0.921 D) Pr{15<Y<15.5} findP(Y<15) P( ( Y - ) /) = P( Z< (15 - 15.6 ) / 0.4 ))
= P( Z <-1.5)
= 0.0668 findP(Y<15.5) P( ( Y - ) /) = P( Z< (15.5 - 15.6 ) / 0.4 ))
= P( Z <-0.25)
= 0.4013 Pr{15<Y<15.5}=0.4013-0.0668 =0.3345 C) Pr{ 15< Y< 16.5} findP(Y<15) P( ( Y - ) /) = P( Z< (15 - 15.6 ) / 0.4 ))
= P( Z <-1.5)
= 0.0668 findP(Y<16.5) P( ( Y - ) /) = P( Z< (16.5 - 15.6 ) / 0.4 ))
= P( Z <2.25)
= 0.9878 hence, Pr { 15< Y<16.5}=0.9878-0.0668 =0.921 D) Pr{15<Y<15.5} findP(Y<15) P( ( Y - ) /) = P( Z< (15 - 15.6 ) / 0.4 ))
= P( Z <-1.5)
= 0.0668 findP(Y<15.5) P( ( Y - ) /) = P( Z< (15.5 - 15.6 ) / 0.4 ))
= P( Z <-0.25)
= 0.4013 Pr{15<Y<15.5}=0.4013-0.0668 =0.3345 findP(Y<15) P( ( Y - ) /) = P( Z< (15 - 15.6 ) / 0.4 ))
= P( Z <-1.5)
= 0.0668 findP(Y<16.5) P( ( Y - ) /) = P( Z< (16.5 - 15.6 ) / 0.4 ))
= P( Z <2.25)
= 0.9878 hence, Pr { 15< Y<16.5}=0.9878-0.0668 =0.921 D) Pr{15<Y<15.5} findP(Y<15) P( ( Y - ) /) = P( Z< (15 - 15.6 ) / 0.4 ))
= P( Z <-1.5)
= 0.0668 findP(Y<15.5) P( ( Y - ) /) = P( Z< (15.5 - 15.6 ) / 0.4 ))
= P( Z <-0.25)
= 0.4013 Pr{15<Y<15.5}=0.4013-0.0668 =0.3345 findP(Y<15) P( ( Y - ) /) = P( Z< (15 - 15.6 ) / 0.4 ))
= P( Z <-1.5)
= 0.0668 findP(Y<15.5) P( ( Y - ) /) = P( Z< (15.5 - 15.6 ) / 0.4 ))
= P( Z <-0.25)
= 0.4013 Pr{15<Y<15.5}=0.4013-0.0668 =0.3345 findP(Y<15.5) P( ( Y - ) /) = P( Z< (15.5 - 15.6 ) / 0.4 ))
= P( Z <-0.25)
= 0.4013 Pr{15<Y<15.5}=0.4013-0.0668 =0.3345
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