In: Statistics and Probability
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 DA county is mean=1625 and standard dev.= 200 if a random
sample of 18 acres of a land is taken the probability of the
average yield of these 18 acres being at most 1700 lbs would
be
.9941?
Solution :
Given that ,
mean = 
 = 1625
standard deviation = 
 = 200
n = 18

=  
= 1625

= 
 / 
n = 200 / 
18 = 47.14
P(
1700 ) = P((
- 
) / 

(1700 - 1625) / 47.14)
= P(z 
 1.59)
Using z table
= 0.9441