5. The World Health Organization estimates (based on death certificates) that th

Question

5. The World Health Organization estimates (based on death certificates) that the distribution of ages at death in the United States has a mean of 79.8 years and standard deviation of 15.5 years. a Would you expect age at death to be a left-skewed,right-skewed, or symmetric distribution? Explain briefly. It may be useful to consider what ages of death are common, which are less common, and which are impossible. b) Researchers plan to collect random samples of 64 United States death certificates and calculate the sample mean age at death, , for each sample. Would you expect the sampling distribution of age at death, y, to have a left-skewed,right-skewed, or symmetric distribution? Explain briefly. c) In approximately what proportion of samples would be less than 78 years? d) Now suppose that the sample size is increased to 400 death certificates per sample. In what proportion of samples would be less than 78 years?

Explanation / Answer

(a)

The distribution will be skewed right, because most of the people live less than 80 years, and very few people live beyond 80 years. So more data will be towards the left of the mean value, while less will be on the right.

So the data will be skewed right.

(b)

The sampling distribution will have a symmterical normal distribution because of the Central Limit theorem, which states that whatever may be the shape of the parent distribution, the sampling distribution of sample means of the distribution will always be normally distributed.

For this sampling distribution:

mean = mean of parent distribution

standard deviation, also called as standard error = standard deviation of original distribution/(sample size^0.5)

(c)

For the sampling distribution we have:

Mean, m = 79.8

Standard error, SE = 15.5/(64^0.5) = 1.938

At y' = 78, we have:

z = (y'-m)/SE = (78-79.8)/1.938 = -0.928

Looking at the cumulative z-table we get:

P(y' < 78) = P(z < -0.928) = 0.177

So, proportion of samples with y' less than 78 is equal to 0.177

(d)

In this case:

Mean, m = 79.8

Standard error, SE = 15.5/(400^0.5) = 0.775

At y' = 78, we have:

z = (y'-m)/SE = (78-79.8)/0.775 = -2.32

Looking at the cumulative z-table we get:

P(y' < 78) = P(z < -2.32) = 0.177

So, proportion of samples with y' less than 78 is equal to 0.02

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