The percent of the butterflies that had a wingspan less than 47.6 mm is 2.3%
We need to understand the concept of z-score to be able to solve this question. By the word z-score, we need to know the mean and the standard deviation to estimate the percentage of the wingspan less than 47.6 mm.
The Z-score of a data set in a normal distribution is a statistical measurement used to illustrate the value of the data given with the mean of a group value.
From the parameters given:
As such; the Z-score of the distribution can be computed as:
[tex]\mathbf{Z = \dfrac{x - \mu}{\sigma}}[/tex]
[tex]\mathbf{Z = \dfrac{47.6 -52.2}{2.3}}[/tex]
[tex]\mathbf{Z =-2 }[/tex]
From the Z tables, the value of the z-score indicates that 0.477(47.7) of the data shows a normal distribution.
Thus;
the percent of data to the left of the distribution curve that had butterflies with a wingspan less than 47.6 mm is:
= 50% - 47.7%
= 2.3%
Learn more about Z-score here:
https://brainly.com/question/25638875
