Question

The average number of acres burned by forest and range fires in a large New Mexico county is 4,300 acres per year, with a standard deviation of 750 acres. The distribution of the number of acres burned is normal.

[4 decimal places]

What is the probability that between 2,500 and 4,200 acres will be burned in any given year? *

2 points

Your answer

What is the probability that less than 4000 acres were burned? *

2 points

Your answer

What is the probability that more than 4300 acres was burnt? *

2 points

Your answer

What number of burnt acres corresponds to the 28%? *

2 points

Your answer

Answer 1

Solution :

Given that ,

mean = = 4300

standard deviation = = 750

P( 2500 < x < 4200 )

= P[(2500 - 4300) / 750) < (x - ) /  < (4200 - 4300) / 750) ]

= P( -2.4 < z < -0.13 )

= P(z < -0.13) - P(z < -2.4)

Using z table,

= 0.4483 - 0.0082

= 0.4401

Probability = 0.4401

P(x < 4000)

= P[(x - ) / < (4000 - 4300) / 750]

= P(z < -0.4)

Using z table,

= 0.3446

Probability = 0.3446

P(x > 4300)

= 1 - P( x < 4300)

= 1 - P[(x - ) / < (4300 - 4300) / 750]

= 1 - P(z < 0)

Using z table,

= 1 - 0.5

= 0.5

Probability = 0.5

The z-distribution of the 28% is,

P(Z < z) = 28%

= P(Z < z ) = 0.28

= P(Z < -0.583 ) = 0.28

z = -0.583

Using z-score formula,

x = z * +

x = -0.583 * 750 + 4300

x = 3862.75

x = 3863