Question 430161: Question:study of long distance phone calls made from the corporate office revealed the length of calls in minutes and follows a normal distribution. Mean is 4.2 minutes and standard deviation is 0.60 minutes.
a) what fraction of the calls last between 4.2 and 5 minutes?
b) what fraction of the calls last more than 5 minutes?
c) what fraction of the calls last between 5 and 6 minutes?
d) what fraction of the calls last between 4 and 6 minutes?
For A I took
Z=4.2-4.2/.6 and got 0 with a p score of 0.0000
and then
Z=5-4.2/.6 and got 1.333 with a p score of 0.4082
So is the final answer the 0.4082 because both Zscore land on one side of the curve or do I need to take the .5000 for the other side and subtract the 0.4082?
For B I took
Z= 5-4.2/0.6 and got 1.333 with a p score of 0.4082
and then I .5000 minus 0.4082 and got 0.0918 and then took 1-0.0918 and got a final answer of 0.9082
For C I took
the z score for 5 and got 1.333 with probability of 0.4082 and z score for 6 and got 3.0 and probability of 0.4987. My question on C is what do I do next with the 2 probabilities, they are both on the same side of the curve so do I add them, or do I subtract them, and do I subtract anything from 1?
Any help is greatly appreciated, I am taking this class online in a different time zone, so getting help from the instructor is difficult.
Thank you:)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Question:study of long distance phone calls made from the corporate office revealed the length of calls in minutes and follows a normal distribution. Mean is 4.2 minutes and standard deviation is 0.60 minutes.
a) what fraction of the calls last between 4.2 and 5 minutes?
For A I took
Z=(4.2-4.2)/.6 and got 0 with a p score of 0.0000
and then Z=5-4.2/.6 and got 1.333 with a p score of 0.4082
So is the final answer the 0.4082 because both Z score land on one side of the curve or do I need to take the .5000 for the other side and subtract the 0.4082?
----
Response: The p-value for z = 0 is 0.5000
Once you have the z-values you get
P(4.2< x < 5) = P(0< z < 1.333) = 0.4082
So you are correct.
===================================
b) what fraction of the calls last more than 5 minutes?
For B I took Z= (5-4.2)/0.6 and got 1.333 with a p score of 0.4082
and then I .5000 minus 0.4082 and got 0.0918 and then took 1-0.0918 and got a final answer of 0.9082
----
Response:
P(x> 5) = P(z > 1.333) = the p-value of 1.333 = 0.0913
============================================================
c) what fraction of the calls last between 5 and 6 minutes?
For C I took the z score for 5 and got 1.333 with probability of 0.4082 and z core for 6 and got 3.0 and probability of 0.4987. My question on C is what do I do next with the 2 probabilities, they are both on the same side of the curve so do I add them, or do I subtract them, and do I subtract anything from 1?
---
Response:
P(5< x < 6) = P(1.333 < z < 3) = 0.0899
============================================================
d) what fraction of the calls last between 4 and 6 minutes?
Response:
z(4) = (4-4.2)/0.6 = -0.3333
Z(6) = (6-4.2)/0.6 = 3
---
P(4< x < 6) = P(-0.3333 < z < 3) = 0.6292
================
If you do not have a TI-83 or better calculator,
you should try to get one. Otherwise you will
get bogged down with the arithmetic of these
stat problems.
====================
Cheers,
Stan H.
======
|
|
|
