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Mathematics 7 Online
OpenStudy (anonymous):

Please Help Me! Multiple Choice!

OpenStudy (anonymous):

with what?

OpenStudy (anonymous):

For exercises 1 and 2, a set of data has a normal distribution with a mean of 120 and a standard deviation of 10. 1. Find the interval about the mean within which 90% of the data lie. a. 94.2-145.8 b. 103.5-136.5 c. 113.68-126.32 d. 100.4-139.6

OpenStudy (anonymous):

2. Find the probability that a value selected at random from this data is between 100 and 140. a. 99.9% b. 90% c. 99% 95.5%

OpenStudy (anonymous):

@Blank  @girlnotonfire @Destinymasha

OpenStudy (anonymous):

@iPwnBunnies @jim_thompson5910 @ganeshie8 @mathstudent55 @dan815

OpenStudy (anonymous):

@GeraldJRJunior

OpenStudy (anonymous):

@bloopman @shamim

jimthompson5910 (jim_thompson5910):

use a table like this one http://3.bp.blogspot.com/_5u1UHojRiJk/TEdJJc6of2I/AAAAAAAAAIE/Ai0MW5VgIhg/s1600/t-table.jpg to find the z score that corresponds to a 90% level

jimthompson5910 (jim_thompson5910):

tell me what z score you get

OpenStudy (anonymous):

Im sorry. I dont understand, can you explain?

jimthompson5910 (jim_thompson5910):

see the row that starts with infinity?

OpenStudy (anonymous):

yes

jimthompson5910 (jim_thompson5910):

which value is above 90%

OpenStudy (anonymous):

1.645?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

that means P( -1.645 < z < 1.645 ) = 0.90

jimthompson5910 (jim_thompson5910):

Basically, we have this normal distribution |dw:1401245391753:dw|

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