I need to find the IQR of WAIS scores with mean 110 and sigma 25 (normal distribution). Will my calculations be Q1 = 25 , Q3 = 75 so..25-110/25=-3.4 (z) = z score of .0003 and 75-110/25 = -1.4 (z) and .0808 (z score). IQR = .0808 - .0003 = 0.0805?
IQR deals with 25% to 75% of that data it looks like you may have confused some of the parts tho
\[Z_{.2500}=\frac{x_1-mean}{\sigma}\] \[Z_{.7500}=\frac{x_2-mean}{\sigma}\] you have to find the z scores and then aglebra out the x1 and x2 parts
the zscores can either be found with an invNorm function, or from your author approved ztable
I used Table A with my text to find that -3.4 = .0003 and -1.4 = .0808. IQR = .0808-.0003 = 0.0805
thats not correct, you are trying to assume the x1 = .25 and that x2 = .75 and finding zscores ... that is completly backwards to the process you need to do
what zscore has a value closest to: .2500 ? what zscore has a value closest to: .7500 ? actually, when you find one the other is just the negation
and can you tell me the legend that your table is using? is it a left tailed shade, or a "to the middle" shade?
Ok...I found 2514 = -0.67 and 7517 = -0.68. The table entry for z is left tailed shade.
So, my computations should be: Q1: (-0.67) (25) + 110 = 93.25 Q 3 = (0.68) x 25 + 110 = 126.75. Q3 - Q1 = 126.75 - 93.25 = 43.5
youve done well, but lets refine this. we want "the closest value" to .2500 so we have to determine which option: z = 0.67 or 0.68 is ore apt .7486 < .7500 < .7517 subtract .74 from all the parts .0086 < .0100 < .0117 -14 +17 since 14 is closer than 17 ... we should use the z=0.67 scores, positive and negative
i meant to use the .7500 but my fingers hate me :) youll see the -0.67 is closest to the .2500 by the same comparison
\[z=\frac{x-mean}{\sigma}\]solve for x \[z(\sigma)+mean=x\] \[-0.67(25)+100=x_1\]\[0.67(25)+100=x_2\] then determine the range between x1 and x2
ugh ... and that 110, not 100
im getting 93.25 and 126.75, giving me an IQR of 33.5
I just finished and got that also. I did this problem last week using the 0.67 values but found something in my text that confused me... again........Thank you ...again...I consider you a friend and hope you don't mind
youre welcome, and i dont mind :) good luck
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