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Chapter 5: Problem 4

Assume the given distributions are roughly normal. On average, every cigarette you smoke reduces your life expectancy by 11 minutes. Assuming an approximately normal distribution with a standard deviation of 1.5 minutes, \(95 \%\) of the reductions per cigarette are above what? (A) 0.95(11) (B) \(11-1.645(1.5)\) (C) \(11+1.645(1.5)\) (D) \(11-1.96(1.5)\) (E) \(11+1.96(1.5)\)

Short Answer

Expert verified
Option (D)

Step by step solution

01

- Identify the goal

Determine the reduction in life expectancy where 95% of the data lies above this value. This involves finding a value using the mean and the standard deviation.
02

- Understand the normal distribution property

In a normal distribution, 95% of the data lies within 1.96 standard deviations from the mean.
03

- Calculate the lower limit

Use the formula for the lower bound where the mean is subtracted by 1.96 times the standard deviation. \[11 - 1.96(1.5) = 11 - 2.94 = 8.06 \]
04

- Compare with the given options

From the calculated lower limit 8.06, we identify the correct option to be (D) \[11-1.96(1.5)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Life Expectancy Reduction
When we talk about life expectancy reduction due to smoking, it means how much time is lost from a person's life because of the habit. In this exercise, it is stated that every cigarette smoked shortens life expectancy by an average of 11 minutes.

It's important to understand that this is an average reduction, meaning it will vary slightly from person to person. This variation is where the concept of normal distribution and standard deviation comes in, which will be explained in the next sections.

Keep in mind, the reduction here is talked about in terms of minutes. It may not seem like a lot for one cigarette, but it adds up significantly over time.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it indicates how spread out the numbers are in a data set.

In this exercise, a standard deviation of 1.5 minutes is given. This tells us that the reductions due to smoking will vary by about 1.5 minutes from the average reduction of 11 minutes.

Using standard deviation helps us understand not just the average reduction, but how consistently this reduction happens. If the standard deviation were larger, it would imply more variation in the reduction times.
Normal Distribution Property
The normal distribution, often referred to as a bell curve, is a probability distribution that is symmetric about the mean. Most of the data points are concentrated around the mean, with fewer points as you move away from it.

In the context of our problem, the life expectancy reductions due to smoking follow a normal distribution. This allows us to use certain properties of the normal distribution to make calculations.

One key property is that in a normal distribution, 95% of the data lies within 1.96 standard deviations from the mean. Thus, we can calculate the reductions in life expectancy that fall within this range using the given standard deviation and mean.
Percentile Calculation
In this exercise, we aim to determine a value below which 95% of the reductions lie. We use the mean and standard deviation to find this cutoff point.

Since 95% of a normal distribution curve lies within 1.96 standard deviations from the mean, we can calculate the lower limit by subtracting 1.96 times the standard deviation from the mean.

The formula is:

Calculation

To determine this, you calculate: The mean reduction = 11 minutes The standard deviation = 1.5 minutes 1.96 times the standard deviation = 1.96 * 1.5 = 2.94 minutes Mean - 1.96 times the standard deviation = 11 - 2.94 = 8.06 minutes This means that 95% of the reductions per cigarette are above 8.06 minutes. By comparing this result with the given options, we can determine that the correct answer is (D)
This example makes the calculation of percentiles in a normal distribution clearer.

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Most popular questions from this chapter

Assume the given distributions are roughly normal. A factory dumps an average of 2.43 tons of pollutants into a river every week. If the standard deviation is 0.88 tons, what is the probability that in a week more than 3 tons are dumped? (A) \(P\left(z>\frac{3-2.43}{0.88}\right)\) (B) \(P\left(z>\frac{3-2.43}{0.88}\right)\) (C) \(P\left(z>\frac{3-2.43}{\frac{0.88}{\sqrt{7}}}\right)\) (D) \(P\left(z>\frac{3-2.43}{\frac{0.88}{\sqrt{7}}}\right)\) (E) \(P(z>(3-2.43))\)

Populations \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) are normally distributed and have identical means. However, the standard deviation of \(\mathrm{P}_{1}\) is twice the standard deviation of \(\mathrm{P}_{2}\). What can be said about the percentage of observations falling within two standard deviations of the mean for each population? (A) The percentage for \(\mathrm{P}_{1}\) is twice the percentage for \(\mathrm{P}_{2}\). (B) The percentage for \(\mathrm{P}_{1}\) is greater than but not twice as great as the percentage for \(\mathrm{P}_{2}\). (C) The percentage for \(\mathrm{P}_{2}\) is twice the percentage for \(\mathrm{P}_{1}\). (D) The percentage for \(\mathrm{P}_{2}\) is greater than but not twice as great as the percentage for \(\mathrm{P}_{1}\). (E) The percentages are identical.

Assume the given distributions are approximately normal. A trucking firm determines that its fleet of trucks averages a mean of 12.4 miles per gallon with a standard deviation of 1.2 miles per gallon on cross-country hauls. What is the probability that one of the trucks averages fewer than 10 miles per gallon? (A) \(P(z<-2.4)\) (B) \(P(z<-2)\) (C) \(P(z<10)\) (D) \(P\left(z<\frac{10}{1.2}\right)\) (E) \(P\left(z<\frac{12.4}{1.2}\right)\)

Suppose that, using accelerometers in helmets, researchers determine that boys playing high school football absorb an average of 355 hits to the head with a standard deviation of 8o hits during a season (including both practices and games). What is the probability on a randomly selected team of 48 players that the average number of head hits per player is between 340 and \(360 ?\) (A) \(P(340

Which of the following are unbiased estimators for the corresponding population parameters? I. Sample means II. Sample proportions III. Difference of sample means IV. Difference of sample proportions (A) None are unbiased. (B) I and II only (C) I and III only (D) III and IV only (E) All are unbiased.
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