The number of ants per acre in the forest is normallydistributed with mean 45,000 and standard deviation12,189. Let X = number of ants in a randomly selected acreof the forest. Round all answers to 4 decimal places wherepossible.a. What is the distribution of X? X - N(b. Find the probability that a randomly selected acre in theforest has fewer than 35,323 ants.c. Find the probability that a randomly selected acre hasbetween 44,652 and 55,584 ants.d. Find the first quartile.ants (round youranswer to a whole number)

Answered: The number of ants per acre in the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

The number of ants per acre in the forest is normally
distributed with mean 45,000 and standard deviation
12,189. Let X = number of ants in a randomly selected acre
of the forest. Round all answers to 4 decimal places where
possible.
a. What is the distribution of X? X - N(
b. Find the probability that a randomly selected acre in the
forest has fewer than 35,323 ants.
c. Find the probability that a randomly selected acre has
between 44,652 and 55,584 ants.
d. Find the first quartile.
ants (round your
answer to a whole number)

Expert Solution

Step 1

The random variable X is number of ants in a randomly selected acre of the forest which follows normal distribution with mean 45,000 and standard deviation 12,189.

That is, the distribution of X follows normal with mean and standard deviation.

Thus, the distribution of X is $X ~ N (45000, 12189)$ .

The probability that a randomly selected acre in the forest has fewer than 35,323 ants is,

$\begin{array}{rcl} P (X < 35323) & = & P (\frac{X - μ}{σ} < \frac{35323 - 45000}{12189}) \\ = & P (z < - 0.7939) \end{array}$

The probability of z less than –0.7939 can be obtained using the excel formula “=NORM.S.DIST(–0.7939,TRUE)”. The probability value is 0.2136.

The required probability value is,

$\begin{array}{rcl} P (X < 35323) & = & P (z < - 0.7939) \\ = & 0.2136 \end{array}$

Thus, the probability that a randomly selected acre in the forest has fewer than 35,323 ants is 0.2136.

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