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Sagot :
Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 2.8, \sigma = 0.7[/tex].
The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is one subtracted by the p-value of Z when X = 4, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 2.8}{0.7}[/tex]
Z = 1.71
Z = 1.71 has a p-value of 0.9564.
1 - 0.9564 = 0.0436.
0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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