Question

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.

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