The distribution of the wing lengths of house flies, in millimeters (mm), is approximately normal with mean wing length 4.55 mm and standard deviation 0.39 mm. A scientist is conducting an experiment to determine the effect of wing length on flight range. The first phase of the experiment will test houseflies with wing lengths that are between 3.77 mm and 4.16 mm. Based on the information, if a housefly is to be selected at random, what is the probability that the housefly's wing length will meet the requirements of the first phase of the experiment?

Probability = 0.13591

Step-by-step explanation:

We are given;

Population mean; μ = 4.55

Population standard deviation: σ = 0.39

Now, we want to find the probability that the housefly's wing length will meet the requirements of the first phase of the experiment. This is;

P(3.77mm < X < 4.16mm)

Z-score formula is;

z = (x - μ)/σ

When x = 3.77mm

z = (3.77 - 4.55)/0.39

z = -2

When x = 4.16

z = (4.16 - 4.55)/0.39

z = -1

From online p-value calculator between 2 z-scores attached, we have;

P-value = 0.13591

answer: asdfghjkl,mnbv

step-by-step explanation:

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step-by-step explanation:

because that's the answer