I hope some one can help me!
The mean midday temperature recorded in June in a city in South California is 36ºC, and the standard deviation is 3ºC.
Assuming the data is normally distributed, the number of days in June when the midday temperature was between 39ºC and 42ºC is?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
standard deviation is 3 so it is 99.7%
OpenStudy (anonymous):
26-3=33
36+3=39
3/sqrt+(36)=1/2 = 0.5
OpenStudy (anonymous):
@jim_thompson5910 can u help please?
jimthompson5910 (jim_thompson5910):
x = 39
z = (x-mu)/sigma
z = (39-36)/3
z = 3/3
z = 1
So when the raw score is x = 39, the z score is z = 1
-------------------------------------------------------
x = 42
z = (x-mu)/sigma
z = (42-36)/3
z = 6/3
z = 2
So when the raw score is x = 42, the z score is z = 2
jimthompson5910 (jim_thompson5910):
What is the area under the standard normal distribution curve from z = 1 to z = 2 ?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
68
OpenStudy (anonymous):
Are you talking about the empirical rule?
OpenStudy (anonymous):
is that the z curve?
OpenStudy (anonymous):
I thought the difference was 1 under normal distribution but i still dont get this.
jimthompson5910 (jim_thompson5910):
do you have z table with you? or a TI calculator?
Still Need Help?
Join the QuestionCove community and study together with friends!