A soft drink machine can be regulated so that it discharges an average of Mu ounces per cup. If the ounces of fill are Normally distributed with a standard deviation of 0.4 ounces, what value should Mu be set at so that 6-ounce cups will overflow only 2% of the time

5.0693 ounces

Step-by-step explanation:

For a 6 ounce cup to overflow, then the cup content is beyond 6 ounce, at 2%

P(x > 6) = 2% = 0.02; similarly

P(x<6) = 1 - P(x > 6)'.

P(x<6) = 1 - 0.02 = 0.98

The z-score for 0.98 = 2.326

The z - formular for a normally distributed goes thus :

Zscore =(observed score - mean) / standard deviation

Substituting the vamus :

2.326 = (6 - mean) / 0.4

2.326 × 0 4 = 6 - mean

0.9304 = 6 - mean

Mean = 6 - 0.9304

Mean = 5.0693

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