Suppose the test scores of students in a class are normally distributed with a mean of 92 and a standard deviation of 3.

adell [148]

Answer:

The second one

Step-by-step explanation:

She started with x dollars and then used 8 dollars to buy a football game ticket, so x-8. Then, she is left with 56 dollars, so x-8=56. Therefore, the second story represents the equation.

5 0

3 years ago

Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly samp

aleksklad [387]

Answer:

95.44% probability the resulting sample proportion is within .04 of the true proportion.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sampling distribution of the sample proportion in sample of size n, the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$