Question 1190332
mean = 55
standard deviation = 2
z-score = (x - m) / s
x = the raw score = 51
m = the mean = 55
s = standard deviation = 2


for part a ....


z = (x - m) / s becomes z = (51 - 55) / 2 = -4 / 2 = -2
area to the left of z-score of -2 = .02275
rounded to 4 decimal places to bet .0228 = 2.28%.


for part b ....


z = (x - m) / s becomes z = (61 - 55) / 2 = 6/2 = 3
area to the left of z-score of 3 = .99865.
area to the right of z-score of 3 = 1 - .99865 = .00135.
round to 4 decimal places to get .0014 = .14%.


i used the following z-score table to get these figures.


<a href = "https://www.rit.edu/academicsuccesscenter/sites/rit.edu.academicsuccesscenter/files/documents/math-handouts/Standard%20Normal%20Distribution%20Table.pdf" target = "_blank">https://www.rit.edu/academicsuccesscenter/sites/rit.edu.academicsuccesscenter/files/documents/math-handouts/Standard%20Normal%20Distribution%20Table.pdf</a>


using the ti-85 plus calculator, i got:


area to the left of z-score of -2 = .022750062.
rounded to 4 decimal places = .0228 = 2.28%.


area to the left of z-score of 3 = .9986500328.
area to the right of z-score of 3 = 1 minus .9986500328 = .0013499672.
rounded to 4 decimal places = .0013 = .13%.


the discrepancy between .13% and .14% has to do with the table figures being rounded to 5 decimal places while the calculator figures are being rounded to 7 or 8 decimal places.
the difference affected the rounding to 4 decimal places in this case.
it's a small difference, but it is a difference.
if you go by the table, then .14% is more accurate.