u + 1.645(.6) = 7.3
Solve for u.
1.645 is the z-score for the 95th percentile of the standard normal.
The height normally distributed with a standard deviation of .6 ounces. What should the average height be if the quality control specialist wants no more than 5% of the products to be more than 7.3 inches?
I am so confused and don't know where to start, any help would be appreciated. Here is what I have done so far, I know it probably isn't on the right path...
µ- x
Std dev- .6
X=7.3
Z= 7.3-x/.6=.05
.03=7.3-x
x=7.27 --> average height??
u + 1.645(.6) = 7.3
Solve for u.
1.645 is the z-score for the 95th percentile of the standard normal.
Thank you so much for the prompt response that was the only question on my homework I was having problems with for some reason. Your help is greatly appreciated as well as the steps for completion.
Thank you!
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