The thorax lengths in a population of male fruit flies follow a normal distribution with mean 0.785 millimeters (mm) and standard deviation 0.07 mm. what are the median and the first and third quartiles of thorax length?

In this exercise we have to use the knowledge of distribution to calculate the value of the first and third quartiles, in this way we can say that:

We can estimate the expected value of the various percentiles assuming the normal distribution, so we have to make the conversion as:

 

The quartiles, 25th and 75th percentile, happen about 2/3 of a predictable difference external the mean. In other words ±.67 standard departure exist 50% of the likelihood of something happening:

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We can estimate the expected value of the various percentiles assuming the normal distribution.

The median, 50th percentile, is the mean, 0.785 mm.

Typically we remember ±1 standard deviation is 68% of the probability, so one standard deviation below the mean is 16th percentile (50-68/2) and one standard deviation above the mean is 84th percentile.

The quartiles, 25th and 75th percentile, are about 2/3 of a standard deviation away from the mean. In other words ±.67 standard deviations is 50% of the probability.  It's not a commonly remembered number like 68-95-99.7, but perhaps it should be.

The 25th percentile is 0.785 - (.67)(.07) = 0.7381 mm.

The 75th percentile is  0.785 + (.67)(.07) = 0.8319 mm.

I could only simplify it

3^2 + g^3 - 8^5

9+g^3-32768

-32759 + g^3