Question

Consider again the situation described in Example 8.2.3. How small wouldσ2 need to be in order for Pr(Y≤0.09)≥0.9?

Short answer

\({\sigma ^2} \le 0.0563\)

Step-by-step solution

Given information

The concentration of lactic acid in several chunks of cheese is independent of normal random variables with mean μand variance\({\sigma ^2}\).

Calculate the probability 

Generally, for \({\sigma ^2}\),

\(P\left( {Y \le 0.09} \right) = P\left( {W \le \frac{{10 \times 0.09}}{{{\sigma ^2}}}} \right)\)

Where \(W = \frac{{10Y}}{{{\sigma ^2}}}\)has \({\chi ^2}\)distribution with d.f 10.

\(P\left( {Y \le 0.09} \right)\) is at least 0.9 if \(\frac{{0.9}}{{{\sigma ^2}}}\) is at least the 0.9 quantile of the \({\chi ^2}\)distribution with d.f 10.

This quantile is 15.99, so\(\frac{{0.9}}{{{\sigma ^2}}} \ge 15.99\)is equivalent to\({\sigma ^2} \le 0.0563\).

Hence, \({\sigma ^2} \le 0.0563\).