Let's say you want to compute the probability
where
converges in distribution to
, and
follows a normal distribution. The normal approximation (without the continuity correction) basically involves choosing
such that its mean and variance are the same as those for
.
Example: If
is binomially distributed with
and
, then
has mean
and variance
. So you can approximate a probability in terms of
with a probability in terms of
:
where
follows the standard normal distribution.
Step-by-step explanation:
A left Riemann sum approximates a definite integral as:
Given ∫₂⁸ cos(x²) dx:
a = 2, b = 8, and f(x) = cos(x²)
Therefore, Δx = 6/n and x = 2 + (6/n) (k − 1).
Plugging into the sum:
∑₁ⁿ cos((2 + (6/n) (k − 1))²) (6/n)
Therefore, the answer is C. Notice that answer D would be a right Riemann sum rather than a left (uses k instead of k−1).
You would look at number after the thousandths place to go up or down 72,000.
In any trapezoid, the area is given by
where B and b are the two bases, and h is the height. You're given all these elements already, so you only need to plug the values in:
