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.

answer: city needs 2.4 inches of rainfall more to match the record of 40 inches.

step-by-step explanation:

since we have given that

total annual rainfall a city record = 40 inches

this year, total rainfall recorded = 37.6 inches

number of inches of rainfall   is needed to match the record of 40 inches is given by

hence, city needs 2.4 inches of rainfall more to match the record of 40 inches.