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:
$3291.60
Step-by-step explanation:
If the loan is amortized in the usual way, the monthly payment is ...
A = P(r/12)/(1 -(1 +r/12)^(-12t)) . . . . . loan of P at rate r for t years
A = $15,000(0.081/12)/(1 -(1 +0.081/12)^(-12·5)) ≈ $304.86
The total of payments is ...
(60 months) × ($304.86/month) = $18,291.60
Then the profit to the bank is ...
$18,291.60 -15,000 = $3,291.60 . . . bank profit
Answer:
25 students
Step-by-step explanation:
What we need to do here is to do some tracing. We simply need to go to the point on the concert band where we have the value 35.
After sighting this value, we then make a tracing to the line of best fit. Then from this line of best fit, we trace the point on the matching band that correlated with the value 35.
If properly traced, we would arrive at a value of 25 on the marching band
1/4 : 2 or 1/8 : 1. That is how I would have done it. I hope this helps you.