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