Given a random sample size of n=1,600 from a binomial...

Given a random sample size of $n=1,600$ from a binomial probability distribution with $P=0.40$, do the following: a. Find the probability that the number of successes is greater than 1,650 . b. Find the probability that the number of successes is fewer than 1,530 . c. Find the probability that the number of successes is between 1,550 and 1,650 . d. With probability 0.09 , the number of successes is fewer than how many? e. With probability 0.20 , the number of successes is greater than how many?

Given a random sample size of $n=1,600$ from a binomial probability distribution with $P=0.40$, do the following:
a. Find the probability that the number of successes is greater than 1,650 .
b. Find the probability that the number of successes is fewer than 1,530 .
c. Find the probability that the number of successes is between 1,550 and 1,650 .
d. With probability 0.09 , the number of successes is fewer than how many?
e. With probability 0.20 , the number of successes is greater than how many?

Key Concepts

Binomial Distribution

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials, each with the same probability of success. It is defined by two parameters: the number of trials (n) and the probability of success (p) in each trial.

Mean and Variance of the Binomial Distribution

The mean of a binomial distribution is given by np, and the variance is given by np(1-p). These parameters are essential for understanding the central tendency and spread of the distribution, especially when approximating it using other distributions.

Normal Approximation to the Binomial Distribution

For large sample sizes, the binomial distribution can be approximated by a normal distribution, which simplifies the calculation of probabilities. This approximation is valid when both np and n(1-p) are sufficiently large.

Continuity Correction

When approximating a discrete distribution like the binomial distribution with a continuous distribution such as the normal distribution, a continuity correction is applied. This involves adjusting the boundaries of the interval by 0.5 to account for the discrete nature of the original distribution.

Standardization and Z-Scores

Standardization is the process of converting a normal random variable to a standard normal variable (z-score) by subtracting the mean and dividing by the standard deviation. This allows the use of standard normal tables or software to find probabilities.

Percentiles and Inverse Probability

This concept involves determining the value (or quantile) of the random variable corresponding to a given probability from the cumulative distribution function of the normal distribution. It is especially useful in solving problems where one needs to find thresholds corresponding to specific probability levels.

Transcript

00:01 Suspect they meant 650.

00:03 And if i quick look at the table in the book, that, uh, nope, that is what it says, greater than 1650.

00:12 So i thought maybe there was a typo from numerade.

00:16 In any case, we have the probability of that.

00:21 We'll convert that to a z value and we can see in our distribution that it's going to be way, way up there high and extremely unlikely.

00:29 So 16.

00:30 50 minus the mean of 640 divided by 19 .59 .9 .9.

00:37 So this difference is 110 divided by that value, gives us a z value of 53.

00:46 So this question is kind of ridiculous.

00:50 51 .54.

00:52 And that probability is almost zero.

00:54 It could happen.

00:55 It's extremely unlikely if that probability is 0 .4.

01:00 And i'm going to double check in the book and make sure that's what that probability was.

01:03 And it is that.

01:04 So kind of a silly question.

01:06 Now, next question is we find that the probability we want to determine that it is less than probability of number of successes is less than 1530.

01:21 And once again, this is going to be basically 100%.

01:24 Because when we convert that to a z value, the 1530 minus the 640 divided the 1 .40, by that standard deviation of 19 .559, the 1530 minus the 640, and then i'm going to divide it by that value that i stored.

01:44 That z value is huge.

01:47 Probability that z is less than 45 .4, and that is approximately 100%.

01:53 Almost everything will be below that.

01:56 Now on part c, what's the likelihood that it's going to be between 15 .4? 50 and 1650.

02:08 And again, this is extremely unlikely because it's kind of a silly question.

02:12 We can see from the picture that 1550 to 1650 when we have the mean being 640 is going to be extremely unlikely.

02:23 And again, i was just checking to make sure that there wasn't a mistake.

02:27 And so this z value is equivalent to 46 .5, or 46, i should see, 46 .44.

02:40 And this z value we know from up here is that 51 .5, i think 4 was my writing.

02:49 And this is still approximately zero...

Please give Ace some feedback