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A biologist found the wingspan of a group of monarch butterflies to be normally distributed. With a mean of 52. 2 mm and a standard deviation of 2. 3 mm. What percent of the butterflies had a wingspan less than 47. 6 mm?.

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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.

What is Z-score?

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:

  • The observed value (x) = 47.6 mm
  • The mean of the sample μ = 52.2 mm
  • The standard deviation of the sample = 2.3 mm

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

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