Question

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?

Answer 1

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.

Answer 2

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:

$25^{th}= 0.7381 mm\\75^{th}= 0.8319 mm$

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

 

$50th =0.785 mm$

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:

$25th =0.785 - (.67)(.07) \\75th = 0.785 + (.67)(.07) = 0.8319 mm$

See more about distrribuition at brainly.com/question/1620226