Based on Standard Normal Distribution

sensational

New member
Joined
Jun 4, 2015
Messages
21
Z is a random variable representing the standard normal distribution.
You are required to use the accompanying statistical table to answer these questions.
1. Find K such that P(0 < Z < 2k + 1)=0.4112.

By using the standard normal distribution I found out that 2k+1=1.348
finding k will gives 0.174.

I don't know what is wrong with my answer but the real answer is 1.734. Maybe theres something wrong with my z-value which is 1.348 but in my own understanding my z-value is correct since 1.34 has the probability of 0.4099 so we need 0.0013 or 13 in the diffrences column which falls on the 8th column of the differences so the final z-value is 1.348. PLEASE help me figure out whats wrong with my working out.:D
 
Z is a random variable representing the standard normal distribution.
You are required to use the accompanying statistical table to answer these questions.
1. Find K such that P(0 < Z < 2k + 1)=0.4112.

By using the standard normal distribution I found out that 2k+1=1.348
finding k will gives 0.174.

I don't know what is wrong with my answer but the real answer is 1.734. Maybe theres something wrong with my z-value which is 1.348 but in my own understanding my z-value is correct since 1.34 has the probability of 0.4099 so we need 0.0013 or 13 in the diffrences column which falls on the 8th column of the differences so the final z-value is 1.348. PLEASE help me figure out whats wrong with my working out.:D

A quick read for me gives not quite 1.35 [which would be 0.4115 instead of 0.4112] so I think you have the correct answer, certainly much closer than that 1.734. Are you sure it isn't 1.734 * 10-1:-D
 
A quick read for me gives not quite 1.35 [which would be 0.4115 instead of 0.4112] so I think you have the correct answer, certainly much closer than that 1.734. Are you sure it isn't 1.734 * 10-1:-D

Yes Ishuda, the answer is K=0.1734 by using the z-value=1.347 so why 1.347 and not 1.348?????
 
No idea. If I use the Excel NORMSINV I get 0.174063. If I do a linear interpolation I get 0.1740625

Thanks, Ishuda. So the correct answer is k=0.174. Maybe theres something wrong with the solution. This question is from a 2012 SPFSC (South Pacifc Form Seven Certificate) Examination and its solution shown below.
 

Attachments

  • kakaka.JPG
    kakaka.JPG
    43.4 KB · Views: 5
Top