Question

The number of ants per acre in the forest is normally distributed with mean 45,000 and...

The number of ants per acre in the forest is normally distributed with mean 45,000 and standard deviation 12,444. 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 46,106 ants._____

c. Find the probability that a randomly selected acre has between 48,802 and 60,245 ants._____

d. Find the first quartile. _____ ants (round your answer to a whole number)

*** Enter an integral or decimal number, accurate to at least 4 decimal places.

Homework Answers

Answer #1

Solution :

(a)

X ~ N(45000 , 12444)

mean = = 45000

standard deviation = = 12444

(b)

P(x < 46106) = P[(x - ) / < (46106 - 45000) / 12144]

= P(z < 0.09)

= 0.5359

(c)

P(48802 < x < 60245) = P[(48802 - 45000)/ 12144) < (x - ) /  < (60245 - 45000) / 12144) ]

= P(0.31 < z < 1.26)

= P(z < 1.26) - P(z < 0.31)

= 0.8962 - 0.6217

= 0.2745

(d)

P(Z < -0.67) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 12144 + 45000 = 36864

first Quartile = 36864

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