Question

A report indicated that 5454​% of adults had received a bogus email intended to steal personal information. Suppose a random sample of 700700 adults is obtained. Complete parts ​(a) and ​(b) below. ​(a) In a random sample of 700700 ​adults, what is the probability that no more than 5050​% had received such an​ email? The probability is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Answer:

The probability that no more than 50​% in the sample of 700 had received such an​ email is 0.9830.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n ≥ 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 $\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

The information provided is:

p = 54% = 0.54

n = 700

Since the sample size selected is quite large, i.e. n = 700 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion.

$\hat p\sim N(p, \frac{p(1-p)}{n})$

Compute the probability that no more than 50​% in the sample of 700 had received such an​ email as follows:

$P(\hat p>0.50)=P(\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}>\frac{0.50-0.54}{\sqrt{\frac{0.54(1-0.54)}{700}}})$

                  $=P(Z>-2.12)\\=P(Z$

*use a z-table for the probability.

Thus, the probability that no more than 50​% in the sample of 700 had received such an​ email is 0.9830.