Question

The World Health Organization reports that the distribution of ages at time of death in the United States has a mean u of 79.8 years and a standard deviation ơ of 15.5 years, based on death certificates. The distribution of ages at time of death is obviously left-skewed (thankfully, few people die young and most die old). Researchers plan to take a random sample of 100 American death certificates. (a) What are the mean and standard deviation of the sampling distribution of the mean age at death, x, for samples of 100 American death certificates? (Enter your answer rounded to two decimal places.) mean: years standard deviation years (b) Choose the correct explanation why the sampling distribution of is approximately Normal O The sampling distribution of is approximately Normal because the 100 death certificates represent a O The sampling distribution of is approximately Normal because the 100 death certificates do not represent a The Central Limit Theorem says that the sampling distribution of a mean does not have an approximately Normal Normal population. Normal population. distribution even if the population is not Norma The Central Limit Theorem says that the sampling distribution of a mean has an approximately Norm.l distribution even if the population is not Normal, (as long as the sample size is large enough) Scroll (c) What is the probability that the mean age at time of death in a random sample of 100 American death certificates will be less than 78 years? (Enter your answer rounded to four decimal places.)

Answer 1

Answer:

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Step-by-step explanation:

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.

I hope you have a SUPER day!