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Mathematics 10 Online
OpenStudy (pphalke):

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a. what percent of IQ scores are between 85 and 115?

OpenStudy (pphalke):

how do I go about doing that step @welshfella

OpenStudy (welshfella):

sorry i messed up there its been a long time.

OpenStudy (pphalke):

hey that's ok this standard deviation is new to me we never got taught this

OpenStudy (welshfella):

find the z scores first z1 = 85 - mean / STD DEV = 85-100 / 15 = -1 can you find the z score for IQ of 115?

OpenStudy (welshfella):

same method as for z1 z2 = (115 -100 )/ 15 = ?

OpenStudy (welshfella):

Oh - talking to myself again!

OpenStudy (pphalke):

sorry was writing it down and for z2 I got 1

OpenStudy (pphalke):

@welshfella

OpenStudy (welshfella):

thats correct now we find the area under the standard normal curve between z = -1 and 1 from the tables. Tgis is given as a fraction so we need to convert to a percentage.

OpenStudy (welshfella):

have you got normal distribution tables?

OpenStudy (welshfella):

in this case its the area between - 1 and 1 so look up the value for 1 and multiply it by 2.

OpenStudy (pphalke):

so I would look up that how?

OpenStudy (welshfella):

I guess you can find it online or in a math or statistics textbook

OpenStudy (welshfella):

OK heres a link But this table gives the area to the left of the z scores So you need to note the values in the table for -1 and 1 and subtract value for z = -1 from value for z = 1. http://math.arizona.edu/~rsims/ma464/standardnormaltable.pdf

OpenStudy (welshfella):

OK I'll read the value for z = 1 it is 0.84134 can you find the value for z =-1?

OpenStudy (welshfella):

(There's different variations of the standard normal curve ) Always read the instructions on the top of the graph before you use it.

OpenStudy (pphalke):

I am looking at that graph

OpenStudy (welshfella):

Its just a matter of finding -1.0 in the first (Z) column and reading the next number on its right.

OpenStudy (pphalke):

I see .15866 for the number to the right of 1.0

OpenStudy (welshfella):

correct so the required percentage is ( 0.84134 - 0.15866) * 100

OpenStudy (welshfella):

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OpenStudy (pphalke):

68%

OpenStudy (welshfella):

thats the area we require which is the area from left side of curve to z=1 - area from left side z = -1

OpenStudy (welshfella):

68% is correct 68 .27 to be more exact

OpenStudy (welshfella):

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