Suppose that a poll taken 10 years ago found that 55% of parents spank their children. Suppose a recent poll of 500 adults with children finds that 242 indicated that they spank their children. Complete parts (a) and (b) below. (a) Assuming parents' attitude toward spanking has not changed since the original poll, how many of the 500 parents surveyed would be expected to spank their children?

From the 500 parents surveyed, the number of parents that will be expected to spank their children is 275 parents.

Number of parents surveyed = 500

Percentage of parents that spank their children = 55%

Therefore, from the 500 parents surveyed, the number of parents that will be expected to spank their children  will be:

= 55% × 500

= 0.55 × 500

= 275

Therefore, the number of parents will be 275 parents.

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Of the 500 parents surveyed 275 would be expected to spank their children.

Step-by-step explanation:

Let the random variable X represent the number of parents who spank their children.

It is provided that, from previous studies, p = 55% of parents spank their children.

A random sample of n = 500 adults were selected.

The event whether a parent spank their children is independent of other parents.

The random variable X thus follows a Binomial distribution with parameters n = 500 and p = 0.55.

The expected value of a Binomial random variable is:

Thus, of the 500 parents surveyed 275 would be expected to spank their children.

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9

78.59 is the mean

step-by-step explanation: