Question

The number of ants per acre in the forest is normally distributed with mean 43,000 and standard deviation 12,405. 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 55,888 ants.

c. Find the probability that a randomly selected acre has between 32,267 and 37,554 ants.

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

Answer 1

Solution :

Given that ,

mean 43,000

standard deviation 12,405.

a) X ~ N(4300 ,12405 normal)

b) P(x < 55888 )

= P[(x - ) / < (55888 - 43000) / 12405 ]

= P(z < 1.04)

Using z table,

= 8508

c) P( 32267 < x < 37554 ) = P[(32267 - 43000) / 12405) < (x - ) /  < (37554 - 43000) / 12405) ]

= P(- 0.87 < z < - 0.44)

= P(z < - 0.44) - P(z < - 0.87)

Using z table,

= 0.3300 - 0.1922

= 0.1378

d) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = - 0.6745

Using z-score formula,

x = z * +

x = - 0.6745 * 12405 + 43000

x = 34632.83

First quartile =Q1 = 34632