Question

Almost three-fourths of US pecans are grown in the states of GA, NM, and TX. It is known that the average yield of pecans in Dona Ana County is μ = 1625 lbs/acre with a standard deviation of σ = 200 lbs/acre. If a RANDOM SAMPLE of 18 acres of land is taken, the probability of the average yield of these 18 acres being at most 1700 lbs/acre would be .9441.

T

F

Answer 1

Solution :

Given that,

mean = = 1625

standard deviation = = 200

= =1625

= / n = 200/ 18 = 47.14

P( <1700 )

= P[( - ) / < (1700-1625) /47.14 ]

= P(z <1.59 )

Using z table

=0.9441

this is true