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Hello!
First, you need to determine your study variable and summarize the information given. This way it'll be easier to answer all questions.
- Study variable X: "age of death of an American citizen" (measured in years)
- left-skewed distribution
- population mean (μ): 79.8 years
- population standard derivation (σ): 15.5 years
- sample taken (n) 100 death certificates
a. In this question you have to calculate the sample mean (x[bar]) and standard deviation (S), unfortunately, without the sample information, you cannot calculate these two values.
To calculate the sample mean you have to use the formula:
x[bar]: ∑xi/n
And the sample standard deviation
S²: ∑xi² - (∑xi)²/n
S:√S²
b. To answer this item you need to apply the Central Limit Theorem definition to the problem.
This theorem states that given a population with a probability function f (X; μ, σ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance σ²/ n when the sample size tends to infinity (or if the sample size is large enough; n≥30).
With this in mind, you can say that thanks to the Central Limit Theorem, in this example, the sample mean of the age of death of Americans has an approximately Normal distribution X[bar]≈N (79.8; 240.25/100)
c. The probability you need to calculate in this item is P(x(bar)<78)
For this, you need to standardize the variable using the statistical Z=(x[bar]-μ)/(σ²/ n) ≈ N(01;)
P(x[bar]<78) ⇒ P(Z<((78-79.8)/(15.5/√100) ⇒ P(Z<-1.8/1.55) ⇒ P(Z<-1.16) = 0.4364
The probability that the mean age at the time of death in a random sample of 100 American death certificates will be less than 78 years is 0.4364.
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Using the normal probability distribution and the central limit theorem, it is found that:
a) The mean is of 79.8 years and the standard deviation is of 1.55 years.
b) The Central Limit Theorem says that the sampling distribution of a mean has an approximately Normal distribution even if the population is not Normal, as long as the sample size is large enough.
c) There is a 0.1230 = 12.30% probability that the mean age at time of death in a random sample of 100 American death certificates will be less than 78 years.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n is approximately normal with standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex], if the underlying distribution is normal or if the sample size is of n > 30.
In this problem, we have that:
- The mean is of [tex]\mu = 79.8[/tex].
- The standard deviation is of [tex]\sigma = 15.5[/tex].
Item a:
Sample of 100, hence:
[tex]n = 100, s = \frac{15.5}{\sqrt{100}} = 1.55[/tex]
The mean is of 79.8 years and the standard deviation is of 1.55 years.
Item b:
Sample of 100 > 30, hence:
The Central Limit Theorem says that the sampling distribution of a mean has an approximately Normal distribution even if the population is not Normal, as long as the sample size is large enough.
Item c:
The probability is the p-value of Z when X = 78, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{78 - 79.8}{1.55}[/tex]
[tex]Z = -1.16[/tex]
[tex]Z = -1.16[/tex] has a p-value of 0.1230.
There is a 0.1230 = 12.30% probability that the mean age at time of death in a random sample of 100 American death certificates will be less than 78 years.
To learn more about the normal probability distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213