Two boys are trying to solve 17+36=?

16 is 10 +6, and 34 is 30+4
10*30=300
10*4=40
6*30=180
4*6=24
add them up: 300+40+180+24=544

Using an array to break
<span>https://www.lake.k12.fl.us/cms/lib05/FL01000799/Centricity/Domain/7908/Topic%204%20Centers%204th%20g...</span>

8 0

3 years ago

Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an

Sphinxa [80]

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

We have:

Mean $\mu$ , standard deviation $\sigma$ .

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

8 0

2 years ago