Few months ago, Frank purchased stock for

Rule for x intercepts at (4,0) and (-1,0) and opening upward

djverab [1.8K]

This is a quadratic equation:
y = a x² + b x + c, a > 0 ( opening upward )
The zeroes are: ( - 1 , 0 ) and ( 4, 0 ).
y = a ( x - x 1 ) ( x - x 2 )
y = a ( x + 1 ) ( x - 4 )
y = a ( x² - 4 x + x - 4 )
y = a ( x² - 3 x - 4 )

6 0

3 years ago

Suppose a large shipment of stereos contained 18% defectives. If a sample of size 306 is selected, what is the probability that

xxMikexx [17]

Answer:

99.36% probability that the sample proportion will differ from the population proportion by less than 6%

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 a sample proportion p in a sample of size n, we have that the sampling distribution of the sample proportions has $\mu = p, s = \sqrt{\frac{p(1-p)}{n}}$ .