Hi, 

Using technology: Excel Function NORMSINV(0.9645)

z=NORMSINV(0.9645)= 1.8055  0r z = 1.806

Wish You the Best in your Studies. 

Answer by Edwin McCravy(20065)   (Show Source): You can put this solution on YOUR website!
 
Let's use a TI-84 graphing calculator.

press 2nd vars 3

Depending upon the age of your calculator, you may or may not get this helping
screen. If so make it read like this:

                            invNorm
               area:0.9645
               μ:0
               σ:1
               Tail: LEFT  CENTER  RIGHT
               Paste

Your calculator may or may not have the "Tail" line. If so highlight LEFT

Highlight Paste
      
Press enter

invNorm(0.9645,0,1,LEFT) or perhaps just invNorm(0.9645,0,1)

Or if your calculator is a very old version, you may have no helping screen. If so, just make it read:
 
invNorm(0.9645)

then press "enter"

Read 1.805477458

Then the area to the left of 1.805477458 is 0.9645.

This means that the left 96.45% of the normal curve below is shaded and the shading stops at 1.805477458 on the horizontal axis.




Edwin





 

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
 the z-score with .9645 of the area under the normal distribution curve to the left of it would be z = 1.805477458.





this can be seen visually as shown below:











i used the ti-84 plus calculator to get the more detailed answer.





if you were to use the z-score normal distribution tables, you would do the following:





z-score of 1.80 has area of .96407 to the left of it.

z-score of 1.81 has area of .96485 to the left of it.

.96485 - .96407 = .00078

.96450 - .96407 = .00043

.00043 / .00078 = .5512820513 * .01 = .005512820513

1.80 + .005512820513 = 1.805512820513.





that's not exactly equal to 1.805477458, but it's pretty close.

it is less than .002% higher.

that's less than 2/1000th of a percent.





the reason is that the interpolation is a straight line interpolation, whereas the calculator is looking at the actual curve itself between the two z-scores, which is not a straight line.





regardless, both methods will get you an answer that is perfectly acceptable.

in fact, they are the same when rounded to 4 decimal places.





since you are asked to round to 3 decimal places, then your answer would be 1.805 either way.





your answer should be a z-score of 1.805 that has .9645 of the area under the normal distribution curve to the left of it.













 

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