A study of tang-distance phone cafe made from General Electric Corporate Headqua
ID: 3151347 • Letter: A
Question
A study of tang-distance phone cafe made from General Electric Corporate Headquarters m Fairfield. Connecticut, revealed the length of the cafe, in minutes, foiows the normal probability distnbufion The mean length of ton* per cal was 4.80 minutes and the standard deviation was 0 50 minutes a. What fraction of fte cals last between 4.80 and 5.60 m^utes9 (Round z-score computation to 2 decimal places and your final answer to A decimal places.) b. Wiat fraction of the cab last more than 5 60 minutes7 (Round/-score computation to 2 decimal places and your final answer to 4 decimal places) c. What fraction of the calb last between 5 80 and 8.50 minutes? (Round/-score computation to 2 decimal places and your final answer to 4 decimal places) d. Wiai fracaon of the calls last oetween 4.50 and 6 50 minutes? (Round z-score computation to 2 4. A charged hollow spherical shell has a charge density distribution^Pv= Pq0 - C/nr, where po is a constant. The inner shell radius is a. and the outer shell radius is b. Inside and outside the shell is free space.Explanation / Answer
we use here standard normal variate z=(x-mean)/sd
here mean=4.8 and sd=0.5
answer a) for x=4.8, z=(4.8-4.8)/0.5=0
for x=5.6,z=(5.6-4.8)/0.5=1.6
P(4.8<x<5.6)=P(0<z<1.6)=0.9452
answer b)
P(x>5.6)=P(z>1.6)=1-P(z<1.6)=1-0.9452=0.0548
answer c) for x=6.5, z=(6.5-4.8)/0.5=3.4
P(5.6<x<6.5)=P(1.6<z<3.4)=P(z<3.4)-P(z<1.6)=0.9998-0.9452=0.0546
answer d) for x=4.5, z=(4.5-4.8)/0.5=-0.6
P(4.5<x<6.5)=P(-0.6<z<3.4)=P(z<3.4)-P(z<-0.6)=0.9998-0.2743=0.7255
answer e) here P(z>z1)=0.04
therefore z1=1.75 or (x-mean)/sd=1.75 or x=1.75*0.5 + 4.8=5.68
length of longest 4% calls is 5.68
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