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.)

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:

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

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.

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

*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.

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680