Suppose that the probability of getting an interview when applying for a professional position is 0.15. You apply for 40 jobs. Use the normal distribution to estimate the probability that more than 5 of the applications result in interviews. Round to four decimal places.

0.6700 = 67% probability that more than 5 of the applications result in interviews.

Step-by-step explanation:

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

Probability that more than 5 of the applications result in interviews.

This is 1 subtracted by the pvalue of Z when X = 5. So

has a pvalue of 0.33

1 - 0.33 = 0.6700

0.6700 = 67% probability that more than 5 of the applications result in interviews.

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier