Consider the following matrix A, and let B = A−1. Find b31, b32, and b33. (i.e., find the entries in the third row of A−1.
1 answer:
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Answer:
a) 0.0082
b) 0.9987
c) 0.9192
d) 0.5000
e) 1
Step-by-step explanation:
The question is concerned with the mean of a sample.
From the central limit theorem we have the formula:
a)
The area to the left of z=2.40 is 0.9918
The area to the right of z=2.40 is 1-0.9918=0.0082
b)
The area to the left of z=3.00 is 0.9987
c) The z-value of 1200 is 0
The area to the left of 0 is 0.5
The area to the left of z=1.40 is 0.9192
The probability that the sample mean is between 1200 and 1214 is
d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000
e)
The area to the left of z=-112.65 is 0.
The area to the right of z=-112.65 is 1-0=1