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The mean amount of time needed to microwave thaw a particular frozen dinner in preparation for convection cooking is 5.68 minutes. the standard deviation of thaw times is 0.95 minutes. what proportion of the meals will take between 4.36 and 7 minutes to thaw?

Respuesta :

BlueSky06
We need to determine the values of the z-scores of the given data.
             (4.36 minutes)
                         z-score = (4.36 - 5.68)/0.95 = -1.39
Which has an equivalent percentage of 8.2%.
            (7 minutes)
                         z-score = (7 - 5.68) / 0.95 = 1.39
Which has an equivalent percentile of 92%.

Subtracting the two values will give us an answer of 83.8%. 

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