According to the U. S. Bureau of the Census, about 75% of commuters in the United States drive to work alone. Suppose 150 U. S. commuters are randomly sampled. (a) What is the probability that fewer than 101 commuters drive to work alone

1.19% probability that fewer than 101 commuters drive to work alone

Step-by-step explanation:

I am going to use the binomial approximation to the normal 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:

(a) What is the probability that fewer than 101 commuters drive to work alone

Using continuity corretion, this is , which is the pvalue of Z when X = 100.5. So

has a pvalue of 0.0119

1.19% probability that fewer than 101 commuters drive to work alone

i think the correct option is the second option.