Umder what conditions can you use a normal distribution to approximate the binomial distrubution

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Step-by-step explanation:

A binomial distribution is basically the probability of success of failure of an experiment that is repeated many times.

The parameters are "n" and "p".

Where

n is the number of times the experiment is performed, or the number of trials

p is the probability of success

Note: "q" is the probability of failure, or q = 1 - p

** Binomial Distribution is a discrete distribution

A normal distribution is a continuous distribution that is symmetric about the mean. That means, the data closer to mean occurs more frequently.

At times, we can use normal distribution to approximate a binomial distribution. The conditions can be said as:

" we can approximate a binomial distribution using normal distribution when n is large enough "

How large is large enough?

There isn't a precise answer, but we can take a rule of thumb as

If "n*p" and "n*q" is greater than 10, we can say the sample size, or n, is large enough.

We can approximate when this is less than 10 per say, but the approximation won't be that good. So, the more the value is greater than 10, the better the approximation.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

5104.38 inches

step-by-step explanation: