Respuesta :
Answer:
41.64% probability that a butterfly will live between 12.04 and 18.38 days.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 18.8, \sigma = 2[/tex]
Find the probability that a butterfly will live between 12.04 and 18.38 days.
This is the pvalue of Z when X = 18.38 subtracted by the pvalue of Z when X = 12.04. So
X = 18.38
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18.38 - 18.8}{2}[/tex]
[tex]Z = -0.21[/tex]
[tex]Z = -0.21[/tex] has a pvalue of 0.4168
X = 12.04
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12.04 - 18.8}{2}[/tex]
[tex]Z = -3.38[/tex]
[tex]Z = -3.38[/tex] has a pvalue of 0.0004
0.4168 - 0.0004 = 0.4164
41.64% probability that a butterfly will live between 12.04 and 18.38 days.
