Normal Distribution Worded Problem

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  • 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.

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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.