Of the total population of the United States, 20% live in the northeast. If 200 residents of the United States are selected at random, approximate the probability that at least 50 live in the northeast. Use continuity correction. ROUND YOUR ANSWER TO 4 DECIMAL PLACES.

0.0465 = 4.65% probability that at least 50 live in the northeast.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

The standard deviation of the binomial distribution is:

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that , .

In this problem, we have that:

So

Approximate the probability that at least 50 live in the northeast.

Using continuity correction, this is , which is 1 subtracted by the pvalue of Z when X = 49.5. So

has a pvalue of 0.9535

1 - 0.9535 = 0.0465

0.0465 = 4.65% probability that at least 50 live in the northeast.

5.762159

step-by-step explanation:

it would be 9 because when it is the nine is over five so that would mean it rounds up to 90 or 9