Suppose that X is a binomial random variable with n=200 and p=0.4. Approximate the following pro

Suppose that X is a binomial random variable with n=200 and...

Key Concepts

Continuity Correction

When applying a continuous normal distribution to approximate a discrete distribution like the binomial, a continuity correction is used to improve accuracy. This involves adjusting the binomial event by 0.5 units in the desired direction to account for the fact that probabilities for discrete outcomes are estimated using the area under the curve of the continuous normal distribution.

Normal Approximation

The normal approximation is used when the number of trials is large and the product np and n(1-p) are sufficiently large, which ensures the binomial distribution is approximately symmetric. This approximation leverages the central limit theorem to use a normal distribution with mean np and variance np(1-p) to estimate binomial probabilities.

Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. Its key parameters are the number of trials (n) and the probability of success (p), and it gives the probability of getting exactly k successes in n trials.

Transcript

00:01 In this problem, we have the approximation to the binominal distribution.

00:07 So n is 20, 200, the probability is 0 .4, and 1 minus p is 0 .6, and n times p is 80.

00:19 So these are a very important statistic we need to remember so that we can calculate the probability easier.

00:27 And then we calculate, so part a, we want to calculate that the probability x is smaller than or equal than 70.

00:35 So this would be equal to the probability that x is squared than 70 .5.

00:41 This is true because x is always an integer and it's discrete.

00:47 So this would approximately equal to the probability that zay is small equal than 70 .5 minus 80 over square of, so this is mp times 6 point, 0 .6.

01:04 And if we calculate this part, we get the probability that the normal distribution variable is a more than negative, 1 .37.

01:14 And this probability, if we check the table, we can get this probability is 0 .085.

01:20 This is the problem a, finish it.

01:24 Problem b, we have something different.

01:28 So this is 70s, it's more than x, more than 90.

01:31 So note that here we don't have the equality.

01:37 We just have inequality.

01:39 And it's a little bit different.

01:40 So first we transform it to something like this because it's greater than 70 is equal to equivalent to greater than 70 .5 and it's more than that, it's more than 89 .5.

01:55 So again, we use the approximation of the random variable out of the normal distribution function...

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