Question

The Bradenton Post polled 1,356 adults in the city to determine whether they wet their toothbrush before they put toothpaste on it. Of the respondents, 54% said they wet it first. Suppose 68% of all adults actually wet their toothbrush first. What are the mean and standard deviation of the sampling distribution?

The mean is 0.54 and the standard deviation is 0.0135.
The mean is 0.54 and the standard deviation is 0.0127.
The mean is 0.68 and the standard deviation is 0.0224.
The mean is 0.68 and the standard deviation is 0.0135.
The mean is 0.68 and the standard deviation is 0.0127.

Answer 1

Answer:

Number 2

Step-by-step explanation:

Answer 2

Answer:

The mean is 0.68 and the standard deviation is 0.0127.

Step-by-step explanation:

First see if it satisfies the 10 percent condition and normal condition.

10(1,356) = 13,560. Presumably there are more than 13,560 adults so this condition is met.

See if it satisfies the Normal Condition

1,356(1-0.68) = 433.92 >= 10 So this condition is met.

Now we can use the formula:

sample mean proportion = population proportion (Greek letters aren't working)

The sample mean is equal to the population proportion. In other words, the sample mean is equal to 68% or 0.68.

σp=√ p(1−p) / n

.68(1-.68) / 1,356 = 0.00016047197

$\sqrt{0.00016047197\\}$ = 0.126677534 the standard deviation,

That leaves us with (0.68, 0.0127) or answer 'E'