Item 8

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

Pls help me solve this

melamori03 [73]

Answer:

See below.

Step-by-step explanation:

1.) False - It contains 6 sig figs, not 2.

2.) False - It goes up to 2 decimal places, not 3.

3.) True

4.) False - It goes up to 1 sig fig, not 4.

8 0

3 years ago

An office manager has received a report from a consultant that includes a section on equipment replacement. the report indicates

wolverine [178]

Answer:

a) 22.663%

b) 44%

c) 38.3%

Explanation:

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates that scanners have a service life that is normally distributed with a mean of 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 41 months

σ is the population standard deviation = 4 months

a. Before 38 months of service

Before in z score score means less than 38 months

Hence,

z = 38 - 41/4

z = -0.75

Probability value from Z-Table:

P(x<38) = 0.22663

Converting to percentage = 0.22663 × 100

= 22.663%

b. Between 40 and 45 months of service

For x = 40 months

z = 40 - 41/4

z = -0.2

Probaility value from Z-Table:

P(x =40) = 0.40129

For x = 45

z = 45 - 41/4

z = 1

Probability value from Z-Table:

P(x = 45) = 0.84134

Between 40 and 45 months of service

= 0.84134 - 0.40129

= 0.44005

Converting to Percentage

= 0.44005 × 100

= 44.005%

= 44%

c. Wihin ± 2 months of the mean life

+ 2 months = 41 months + 2 months

= 43 month

- 2 months = 41 months - 2 months

= 39 months

For x = 43

z = 43 - 41 /4

z = 0.5

P-value from Z-Table:

P(x = 43) = 0.69146

For x = 39

z = 39 - 41/4

z = -2/4

z = -0.5

Probability value from Z-Table:

P(x = 39) = 0.30854

Within ± 2 months of the mean life

= 0.69146 - 0.30854

= 0.38292

= 38.3%

Learn more about z-score:

brainly.com/question/17436641

#SPJ4

4 0

2 years ago

What is the common ratio of the associated geometric sequence?

Sav [38]

Common Ratio. For a geometric sequence or geometric series, the common ratio is the ratio of a term to the previous term. This ratio is usually indicated by the variable r.

4 0

3 years ago

Cost To Store : $110 Markup : 140%

eimsori [14]

If you are talking about $110 being increased by 140% then the answer is $264

5 0

3 years ago