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?

Answer 1

Answer:

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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

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

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 700, p = 0.37$

$\mu = E(X) = np = 700*0.37 = 259$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{700*0.37*0.63} = 12.77$

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.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{238 - 259}{12.77}$

$Z = -1.64$

$Z = -1.64$ has a pvalue of 0.0505

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