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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Answer:
D
Step-by-step explanation:
5 ÷ 8 = 0.625 = 62.5%
Using addition of variables, it is found that the mean of S is of 73 and the standard deviation is of 8.5.
<h3>What happens to the mean and the standard deviation when two variables are added?</h3>
- The mean is the sum of the means.
- The standard deviation is the square root of the sum of variances.
In this problem, for variables A and B, we have that:


Variable S is the sum of A and B, hence:


The mean of S is of 73 and the standard deviation is of 8.5.
More can be learned about addition of variables at brainly.com/question/26156502
Answer: 20.5 units
Step-by-step explanation:
P=Perimeter
JI=3-(-3)
JI=3+3
JI=6
IK=7-1
IK=6
JK=(6^2+6^2)^1/2 ==> Distance Formula
JK=(36+36)^1/2
JK=(2*36)^1/2
JK=6(2)^1/2
P=JI+IK+JK
P=6+6+6(2)^1/2
P=12+6(2)^1/2
P=20.485 units
P=20.5 units