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.
Answer:
<em>Probability ≈ 0.1071</em>
Step-by-step explanation:
Consider steps below;



<em>Solution; Probability ≈ 0.1071</em>
No they are not equivalent
Around 10.63 because 400/37.61=10.63
Estimated would be 700 x 80