Use the distributive property to simplify -4(6-3y)
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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.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- 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:
.
The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:
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 brainly.com/question/24663213
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