Normal Distribution Worded Problem 


*

*Standard deviation = 2.5 mL

*98% of bottles must be between 998 mL and 1000mL

*Pr( 998 < x < 1000) = 0.98


This is a technology exam question, therefore to find the mean I used the method:
Solve(normCDF(998,1000,m,2.5)=0.98,m) but this results in an error
Is my working out wrong? Ignore the calculator syntax, but please refer to my understanding of the question, and what it is asking for. 

Solve(normCDF(998,1000,m,2.5)=0.98,m)

I don´t have the 1000. 
I have $P(X > 998 )=1-P(X\leq 998)=1-\Phi \left( \frac{998-\mu}{2.5} \right) = 0.98$
$\Phi(.)$ is the cdf of the standard normal distribution.

•98% of bottles must be between 998 mL and 1000mL

No, 98% of the bottles must be above 998 ml.
