SOLUTION: The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 405 grams and a standard deviation of 21 grams. Find the weight that corresponds to

Question 1196953: The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 405 grams and a standard deviation of 21 grams. Find the weight that corresponds to each event. (Use Excel or Appendix C to calculate the z-value. Round your final answers to 2 decimal places.)
Middle 60 percent_____to_____
Highest 80 percent
Lowest 15 percent

Found 2 solutions by Theo, ewatrrr:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i used the ti-84 plus.
middle 60 percent has area between z-score = -.8416212335 and z-score = .8416212335
round to 3 decimal places to get from z-score -.842 to .842
highest 80 percent has area to the right of z-score = -.8416212335
round to 3 decimal places to get area to the right of z-score = -.842
lowest 15 percent has area to the left of z-score = - 1.03643330
round to 3 decimal places to get area to the left of z-score = -1.036
i confirmed these results were accurate using the online normal distribution calculator at https://davidmlane.com/hyperstat/z_table.html.
here are the results from using that calculator.

the commands you would use with excel and the result of using those commands is shown below.

normsinv() gets you the z-score that has the specified area to the left of it.
normsdist() gets you the area to the left of the specified z-score.
to get the z-scores that has have an area of .6 between them, you would do the following.
find the area to the left and right of .6.
since the whole area is 1, then .2 is the area to the left of .6 and .2 is the area to the right of .6 because .2 + .6 + .2 = 1
the low z-score will have an area of .2 to the left of it.
the high z-score will have an area of .8 to the left of it.
the .8 to the left of the high z-score contains .2 + .6 = .8

Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi Recommend the use of a calculator and understanding of the standard normal curve Normal Distribution: µ = 0 and σ = 1 z score represents the area under the curve to the LEFT of its value Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus Calculator function Invnorm(X) gives value to the LEFT of z Middle 60% |z= Invnorm(20%) = -.84 (to 2 decimal places) Middle 60%: -.84 to .84 Highest 80 percent |z= Invnorm(80%) = .84 Highest 80 percent: to the RIGHT of z = .84 Lowest 15 percent: |z= Invnorm(15%) = -1.04 Wish You the Best in your Studies.

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