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