## A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample of 700 adul

Question

A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample of 700 adults is obtained. In a random sample of 700 adults, what is the probability that no more than 34% had received such an email?

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1 week 2021-11-17T14:18:26+00:00 1 Answer 0 views 0

5.05% probability that no more than 34% had received such an email.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this problem.

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:

In a random sample of 700 adults, what is the probability that no more than 34% had received such an email?

34% is 0.34*700 = 238

So this probability is the pvalue of Z when X = 238.

has a pvalue of 0.0505

5.05% probability that no more than 34% had received such an email.