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Mathematics 7 Online
OpenStudy (anonymous):

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?

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 ?

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?

OpenStudy (anonymous):

no just online one

jimthompson5910 (jim_thompson5910):

ok let's use this table https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf

jimthompson5910 (jim_thompson5910):

locate the row that has 1.0 locate the column that has .00 what is in that row/column combo?

OpenStudy (anonymous):

0.8413

jimthompson5910 (jim_thompson5910):

So P(Z < 1) = 0.8413

jimthompson5910 (jim_thompson5910):

what is the value of P(Z < 2) ?

OpenStudy (anonymous):

im trying to figure that out

jimthompson5910 (jim_thompson5910):

use the table locate 2.00

OpenStudy (anonymous):

0.9773

OpenStudy (anonymous):

P(Z < 2)=0.9773

OpenStudy (anonymous):

this is hard

OpenStudy (anonymous):

now what do I do?

jimthompson5910 (jim_thompson5910):

you now subtract the two values

jimthompson5910 (jim_thompson5910):

0.9773 - 0.8413 = 0.136

jimthompson5910 (jim_thompson5910):

So P( 1 < Z < 2 ) = 0.136

jimthompson5910 (jim_thompson5910):

So roughly 13.6% of the values are between z = 1 and z = 2

jimthompson5910 (jim_thompson5910):

which means 13.6% of the days in June are going to have days with temps between 39 degrees C and 42 degrees C

OpenStudy (anonymous):

my answers to choose are 3, 4, 7, or 12

jimthompson5910 (jim_thompson5910):

how many days are in June?

OpenStudy (anonymous):

i would have to choose 12 then. I think 30

jimthompson5910 (jim_thompson5910):

30 days in june 13.6% of 30 = ???

OpenStudy (anonymous):

so it would be 4

jimthompson5910 (jim_thompson5910):

yes

OpenStudy (anonymous):

crazy how do you know and remember this. so hard. thank you!

jimthompson5910 (jim_thompson5910):

with more practice, it should come easier

OpenStudy (anonymous):

wow I hope so thanks

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