A blueberry farmer finds that the diameters of blueberries harvested on his farm follow a normal distribution with a mean diameter of 5.85 mm and a standard deviation of 0.24 mm.
The middle 20% of blueberries from this farm have diameters between 5.79 and mm.

The middle 20% of blueberries from this farm have diameters between 5.79 and 5.91 mm

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Middle 20%

50 - 20/2 = 40th percentile to the 50 + 20/2 = 60th percentile.

We want the 60th percentile, which is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.253.

So the answer is:

The middle 20% of blueberries from this farm have diameters between 5.79 and 5.91 mm

the two may have that, even a breakdown is probably offered

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step-by-step explanation: