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?
how do I go about doing that step @welshfella
sorry i messed up there its been a long time.
hey that's ok this standard deviation is new to me we never got taught this
find the z scores first z1 = 85 - mean / STD DEV = 85-100 / 15 = -1 can you find the z score for IQ of 115?
same method as for z1 z2 = (115 -100 )/ 15 = ?
Oh - talking to myself again!
sorry was writing it down and for z2 I got 1
@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.
have you got normal distribution tables?
in this case its the area between - 1 and 1 so look up the value for 1 and multiply it by 2.
so I would look up that how?
I guess you can find it online or in a math or statistics textbook
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
OK I'll read the value for z = 1 it is 0.84134 can you find the value for z =-1?
(There's different variations of the standard normal curve ) Always read the instructions on the top of the graph before you use it.
I am looking at that graph
Its just a matter of finding -1.0 in the first (Z) column and reading the next number on its right.
I see .15866 for the number to the right of 1.0
correct so the required percentage is ( 0.84134 - 0.15866) * 100
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68%
thats the area we require which is the area from left side of curve to z=1 - area from left side z = -1
68% is correct 68 .27 to be more exact
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